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UNIVERSIDAD NACIONAL

DE INGENIERIacuteAFACULTAD DE INGENIERIacuteA MECAacuteNICA

Departamento Acadeacutemico de Ciencias BaacutesicasHumanidades y Cursos Complementarios

Profesores

Garrido Juaacuterez Rosa

Castro Salguero Robert

Obregoacuten Ramos Maacuteximo

2009- 1

METODOS NUMERICOS

(MB ndash536)

SOLUCION DE ECUACIONES

DIFERENCIALES ORDINARIAS

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Universidad Nacional de Ingenieriacutea EDOFacultad de Ingenieriacutea Mecaacutenica PA 2009-1DACIBAHCC

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Soluciones Numeacutericas de Ecuaciones Diferenciales Ordinarias

Muchos problemas en la Ingenieriacutea y otras ciencias pueden ser formulados en teacuterminosde ecuaciones diferenciales Por ejemplo trayectorias baliacutesticas teoriacutea de sateacutelitesartificiales estudio de redes eleacutectricas curvaturas de vigas estabilidad de avionesteoriacutea de vibraciones reacciones quiacutemicas y otras aplicaciones de aquiacute la importanciade su solucioacuten

Objetivo Estudiar los meacutetodos numeacutericos que permitan obtener soluciones numeacutericasaproximadas de las ecuaciones diferenciales ordinarias (EDOs)

Trataremos los meacutetodos para resolver el problema del valor inicial y tambieacuten el valorfrontera

Meacutetodos de solucioacuten de EDOrsquos ndash Problema del Valor InicialLa ecuacioacuten [ ]T tot t ut f t u ))(()( isin=prime (1)es llamada Ecuacioacuten diferencial de primer orden Aquiacute f des una funcioacuten realconocida de dos variables reales t y u y u es llamada funcioacuten incoacutegnita de variableindependiente t Ademaacutes de esto u y f pueden ser vectores este caso lo veremos maacutesadelante y corresponderaacute a un sistema de e EDOrsquos de primer orden

Resolver (1) corresponde a determinar una funcioacuten u=u(t ) diferenciable con [ ]T tot isin tal que ))(()( t ut f t u =prime Cualquier funcioacuten que satisfaga esta propiedad es una solucioacuten

de la ecuacioacuten diferencial (1) Por ejemplo una funcioacuten

t

Cet u =

)( es para cualquiervalor constante de C una solucioacuten de la ecuacioacuten diferencial ut u =prime )( Asimismo cadaecuacioacuten diferencial de primer orden posee un nuacutemero infinito de soluciones De aquiacutepodemos seleccionar una solucioacuten particular si junto con la EDO fuera dado un valorde la solucioacuten u(t) en un punto por ejemplo u(to)=uo (Condicioacuten Inicial)Si para la ecuacioacuten diferencial ut u =prime )( con condicioacuten inicial 1)0( =u entonces

obtenemos C=1 y la solucioacuten seriacutea t et u =)( Problema del Valor Inicial

Nota Antes de empezar a resolver el problema interesa garantizar que esta tienesolucioacuten uacutenica

Dados una funcioacuten f RRxRRrarrRR un intervalo real [t o T ] y un valor uo isin RR el problema de

valor inicial consiste en determinar una funcioacuten u[t o T ]rarr RR que verifique una ecuacioacutendiferencial de primer orden

[ ]T tot t ut f t u ))(()( isin=prime con la condicioacuten inicial

u(t 0)= u0

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Existencia y unicidad de la solucioacutenTeorema Sea f RRxRRrarrrarr RR una funcioacuten con las siguientes propiedades

1 f es continua en [t o T ] con respecto al primer argumento2 f es continua seguacuten Lipschitz con respecto al segundo argumento esto es existe una

constante Lgt=0 (llamada constante de Lipschitz) tal querealisinforallisinforallminusleminus 212121 ][)()( uuT tot uu Lut f ut f

Entonces el problema de valor inicial posee una solucioacuten u que es uacutenica Ademaacutes la solucioacutenu es continuamente diferenciable en [t o T ]

Soluciones Aproximadas

Definicioacuten

Se dice que un conjunto de puntos N iit 0=

forma una malla del intervalo [toT ] si

t olt t 1lthellipltt N = T Los puntos t i son llamados nodos de la malla la distancias hi=t i -t i-1 i = 1hellip N Se llama paso de la malla Una malla se dice uniforme si todas estas distancias son igualesEl valor

i N i

hmaxhlele

=1

es designado el paso de la malla

Los meacutetodos numeacutericos que seraacuten estudiados a continuacioacuten se caracterizan por proveervalores de la solucioacuten aproximada uh en un conjunto finito de puntos t i en el intervalo

[t o T ]El valor de la solucioacuten aproximada uh en el nodo t i seraacute representado por ui o seaui=uh(t i)

El error de aproximacioacuten eh es una funcioacuten definida por eh = u - uh La solucioacuten aproximada solo esta definida en los nodos t i Para obtener valores enpuntos intermedios puede utilizarse la interpolacioacuten por ejemplo lineal entre cada dosnodos consecutivos como se muestra en la figura

Toda vez que las soluciones de las ecuaciones diferenciales son funciones y losmeacutetodos numeacutericos producen soluciones aproximadas es importante tener una forma demedir la distancia entre dos funciones

Dada una funcioacuten continua v definida en el intervalo [t o T ] se define ||v|| llamada normamaacutexima de v por

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)(max][

t vvT t t oisin

=

Dadas las funciones v y w definidas y continuas en el intervalo [t o T ] la distancia entreellas seraacuten entonces definidas por

[ ]

)()( t wt vmaxwvT t t o

minus=minus

isin

Es claro que estas funciones seraacuten iguales si y solo si 0=minus wv

Convergencia del Meacutetodo

Un meacutetodo numeacuterico de solucioacuten de ecuaciones diferenciales se dice convergentecuando

0limlim 00 =equivminusrarrrarr hhhh euu

para todas las soluciones de EDOs que verifiquen las condiciones de existencia yunicidad anteriormente mencionadas y para todas las condiciones iniciales tales que

0)(0

=rarr

oht

t elim

Se dice que un meacutetodo convergente posee orden de convergencia igual a pgt0 si p

h chuu leminus

para todo h suficientemente pequentildeo y ] [+infinisin 0c independiente de h mas dependientede la funcioacuten f que caracteriza una EDO

Meacutetodos numeacutericos descripcioacuten general

La solucioacuten numeacuterica de un problema de valor inicial se realiza de un modo general de

la siguiente forma1 Definir una malla N

iit 0= en el intervalo [ ]T t 0

2 Para i de 1 hasta N determinar ui que seraacute el valor de la solucioacuten aproximada uh enel nodo t i

Cabe notar que el valor uo es una condicioacuten inicial conocida Los meacutetodos numeacutericos sedistinguen por la forma como son calculados los valores sucesivos de ui Los meacutetodos en donde el caacutelculo de ui es realizado solo con la informacioacuten en elintervalo [t i-1 t i] se llaman meacutetodos de un solo paso Los que recurren a informacioacutenfuera de este intervalo para determinar ui se llaman meacutetodos de muacuteltiple paso o de pasomuacuteltiple

Meacutetodos de un solo paso

Consideremos la ecuacioacuten diferencial][t))(()( 0T t t ut f t u isin=

Integrando ambos miembros entre t y t +h se obtiene la relacioacuten

int+

=minus+

ht

t

d u f t uht u ξ ξ ξ ))(()()(

se concluye asimismo que el valor de la solucioacuten exacta u en el punto t + h podriacutea sercalculado sumando al valor de la solucioacuten exacta en t el valor de la integral de

))(( ζ ζ u f en [t t + h]

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Una vez que el valor u(t 0) = u0 es conocido todos los valores requeridos podriacutean serobtenidos bastando para tal considerar una malla adecuadaNota Consideacuterese aquiacute que el paso es uniforme e igual a h

Dificultad La integral anterior involucra una funcioacuten propia a determinar que impidesu caacutelculo de modo expliacutecitoSolucioacuten Prescindir del valor exacto de la integral calculaacutendolo de una formaaproximadaDe esta forma los valores modales de u seraacuten aproximadosSe define entonces F (t u) como

int+

=

ht

t

d u f h

ut F ξ ξ ξ ))((1

)(

Y sea F h(t u) un valor aproximado de F (t u) o sea

int

+

asymp

ht

t

h d u f h

ut F ξ ξ ξ ))((1)(

Representando por T h(t u) el error asociado a esta aproximacioacuten se tiene que)()()( ut T ut F ut F hh +=

Este error se llamaraacute error de truncacioacuten

La ecuacioacuten ( ) ζ ζ ζ d u f t uht uht

t int +

=minus+ )()()( puede ahora ser escrita como

)()()()()(

ut T ut F ut F h

t uht uhh +==

minus+

(2)

Haciendo hrarr0 y asumiendo que los limites existen se tiene que)(lim)(lim)( 00 ut T ut F t u hhhh rarrrarr

+=

y entonces si el error de truncamiento tiende a cero en h es legiacutetimo suponer que laeliminacioacuten de este teacutermino en (2) conduzca a las ecuaciones proacuteximas a la ecuacioacutenoriginal

Representando por uh la solucioacuten de (2) cuando se desprecia el error de truncacioacuten ysiendo que ui = uh(t i) los valores nodales de uh se verifica que estos satisfacen lasiguiente relacioacuten de recurrencia

110)(1 minus=+=+

N iut hF uu iihii K

Diferentes selecciones de la funcioacuten F h conducen a diferentes meacutetodos para la solucioacutennumeacuterica del problema del valor inicial

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Meacutetodos de Euler

Este meacutetodo rara vez se usa en la praacutectica debido a que es el menos preciso de los

meacutetodos que veremos Sin embargo su derivacioacuten es tan simple que permite ilustrar las

teacutecnicas que normalmente se utilizan en la construccioacuten de meacutetodos maacutes avanzados

Pretendemos resolver

=

ltlt==

)()(

))(()(

0

0

InicialCondicioacutent u

bt at t ut f t u

α

Como dijimos lo que obtendremos es una aproximacioacuten a la funcioacuten u para algunos

puntos pertenecientes al intervalo [ab] Una vez que tenemos estos puntos podemos

encontrar el valor aproximado de la solucioacuten en otros puntos empleando alguna de las

teacutecnicas de interpolacioacuten que ya vimos

La idea es la aproximacioacuten de la ecuacioacuten diferencial por el cociente de diferenciashacia adelante

))(()()()(

t ut f t uh

t uht u=asymp

minus+

Tomamos el tamantildeo de paso hgt0 (h = (b - a)n) definiendo

n j jht t j 2100 K=+=

y obtenemos de tal manera

(3)

siendo los ju aproximaciones para )( jt u

A la ecuacioacuten (3) se le conoce como Meacutetodo de Euler Progresivo

Para Euler Regresivo usar la foacutermula impliacutecita

)( 111 +++ += j j j j

ut hf uu (4)

Ejemplo 1Aplicar el meacutetodo de Euler progresivo para aproximar la solucioacuten del problema devalor inicial

=

lele++minus=

1)0(

101

u

t t uu

)(1 j j j j ut hf uu +=

+

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Solucioacuten

El primer paso es encontrar el tamantildeo de paso h eligiendo n = 10

Del ejemplo escogemos

1010

01 =minus=minus=n

abh ademaacutes

)10(0 j jhat j +=+=

Luego se obtiene la expresioacuten de recurrencia de Euler

Usando el hecho de que 1)( ++minus= t uut f

9101001090

)110(10)1(10)(

1

1

0

=++=

++minus+=++minus+=+=

=

+

j ju

juut u yut f huu

u

i

j j j j j j j j j

Evaluamos para j = 0 y obtenemos

110019010001090 01 =++times=+times+= uu

ahora para j = 1

0111001019010101090 12 =++times=+times+= uu

La solucioacuten exacta es t et t u minus

+=)( La siguiente tabla muestra la comparacioacuten entre

los valores aproximados ju y los valores exactos )( jt u

jt ju )( jt u Error=| ju - )(

jt u |

00 1000000 1000000 0001 1000000 1004837 000483702 1010000 1018731 000873103 1029000 1040818 001181804 1056000 1070320 001422005 1090000 1106531 001604106 1131441 1148812 001737107 1178297 1196585 001828808 1230467 1249329 001886209 1287420 1306570 001915010 1348678 1367879 0019201

Noacutetese que el error crece ligeramente conforme el valor de jt aumenta

Error de truncacioacuten de los meacutetodos de EulerEl error de truncacioacuten en el meacutetodos de Euler puede ser acotado por

[ ] ))((sup2 0

t ut f h

T T t t h isin=

Siendo f de clase C 1 las condiciones del teorema sobre existencia y unicidad de la

solucioacuten permiten concluir que ())(u f prime es continua por lo que su supremo existe y esfinito

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Asimismo el error de truncacioacuten de los meacutetodos de Euler satisfacechT h le

Donde c no depende de h aunque depende de los datos que caracterizan el problema devalor inicial una funcioacuten f el intervalo [t oT ] y el valor de uo

Consistencia de un meacutetodoDefinicioacuten Se dice que un meacutetodo es consistente si0lim 0 =

rarr hh T

Se dice que su orden de consistencia es p gt 0 siPara todo h suficientemente pequentildeo y cgt0 independiente de h

p

h chT le

Se verifica asiacute que ambos meacutetodos de Euler tienen un orden de consistencia igual a 1Nota El error de truncacioacuten (y luego del orden de consistencia) apenas caracterizan unerror local en cada paso y no el error global de aproximacioacuten de la solucioacuten exacta u poruh

Meacutetodos de Taylor de orden q

Podemos desarrollar la funcioacuten u(t) solucioacuten de la EDO en una serie de Taylor alrededordel punto jt de manera de poder encontrar u(t j+1)

( )

11

)1(1)(21 )(

1

1)(

1)(

2

1)()()(

++

++

+

ltltminus=

+++++++=

j j j j

qq

j

qq

j j j j

uu yt t hcon

uhq

t uhq

t uht hut ut u

ξ

ξ K(5)

Donde el uacuteltimo teacutermino es el error de truncamiento local

Las derivadas en la expresioacuten (5) no son conocidas expliacutecitamente como la solucioacutenexacta no es conocida Si f es suficiente derivable estas pueden ser obtenidasconsideraacutendose una derivada total de )()( ut f t u =prime con respecto a t teniendo en cuentaque f se una funcioacuten impliacutecita en u

El meacutetodo de Euler consiste en truncar esta serie de Taylor al primer orden yrecordando que ))(( j j t ut f u = podemos generar una sucesioacuten de valores

aproximados )( j j

t uu asymp

α =+=+ 01 )( uconut hf uu j j j j Si truncamos la serie de Taylor al orden q entonces tenemos el meacutetodo de Taylor deorden q

001 10)( t uconniut hT uu j j j j ==+=+

K

Donde

)(

)(2

)()( )1(1

)( j j

qq

j j j j j j

qut f

q

hut f

hut f ut T

minus

minus

++= K

Los meacutetodos de Taylor de orden elevado no son en general de aplicacioacuten muy praacutectica

Meacutetodo de Taylor de orden 2

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α =++=+ 0

2

1 )(2

)( uconut f h

ut hf uu j j j j j j

Ejemplo 2

Aplicar el meacutetodo de Taylor de orden 2 para aproximar la solucioacuten del problema devalor inicial

=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=

Luego se obtiene la expresioacuten de recurrencia de Taylor de orden 2

Usando el hecho de que 1)( ++minus= t uut f y

t ut uut u

dt

d ut f minus=+minusminus=+minus=++minus= 111)1()(

Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u

j j j j j j j j j j j


0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=

minus+++minustimes+=minus+++minus+= t uh

t uhuu

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Resumimos los resultados en la siguiente tabla

jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541

Tarea 1Sea el siguiente problema de valor inicial

=lele=

++minus=prime

103002)0(

2

h x y

x y y

Resolver usando el meacutetodo de Taylor de orden 3

Meacutetodos de Runge-Kutta

Los meacutetodos de Taylor esbozados en la seccioacuten anterior tienen la desventaja de requerir

el caacutelculo y la evaluacioacuten de las derivadas de )( ut f Este puede ser un procedimiento

muy complicado y que consuma mucho tiempo para una gran cantidad de problemas Es

por ello que los meacutetodos de Taylor se usan muy poco en la praacutectica Los meacutetodos de

Runge Kutta eliminan el caacutelculo y la evaluacioacuten de las derivadas de )( ut f

Consideremos nuevamente la ecuacioacuten diferencial ordinaria

=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α

Para calcular u j+1 en t j+1 = t j + h dado un valor de yi podemos integrar la ecuacioacutenanterior en el intervalo [u i u i+1]

int+

+=+

1

)(1

j

j

t

t

j j ut f uu

Los diferentes meacutetodos de Runge-Kutta surgen de las diferentes aproximacionestomadas para calcular la integral de f(tu)

Meacutetodo de Runge-Kutta de orden 2

Si aplicamos la regla del trapecio para integrar )( yt f y = la sucesioacuten de valoresaproximados a la solucioacuten real es ahora

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)]()([2 111 j j j j j j ut f ut f h

uu ++=+++

Este meacutetodo es conocido con el nombre de Meacutetodo de Euler Modificado El problema

con este meacutetodo es que del lado derecho aparece 1+ ju que es lo que queremos encontrarSi f(tu) es lineal en y entonces podemos despejar 1+ ju

Si aproximamos a )( 11 ++ j j ut f mediante el meacutetodo de Euler entonces tenemos el


)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+

Este meacutetodo suele aparecer bajo lo que se conoce como notacioacuten canoacutenica que es lasiguiente

][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3

Aplicar el meacutetodo de Runge-Kutta de orden 2 para aproximar la solucioacuten del problemade valor inicial

=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=

Luego se obtiene la expresioacuten de recurrencia del meacutetodo de Runge-Kutta de orden 2

Usando el hecho de que 1)( ++minus= t uut f y ademaacutes

]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

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910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001

=+++minustimes=+++minus=+=

=++minustimes=++minus==

t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685


Este es el meacutetodo maacutes empleado en la praacutectica Surge de aproximar la integral de f(t u) por la

regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++

+++= j j j j j j j j ut f ut f ut f uu

En notacioacuten canoacutenica se escribe como

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[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4

Aplicar el meacutetodo de Runge-Kutta de orden 4 para aproximar la solucioacuten del problema de valorinicial

=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


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( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001

=++++minustimes=++++minus=++=

=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6

1432101 =+times+times++=++++= k k k k y y


Comparacioacuten entre meacutetodos ejemplo

jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

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ConvergenciaEn muchas situaciones interesa analizar no el error de truncamiento (que solo ofreceinformacioacuten local) sino mas bien el error de aproximacioacuten global definido por

eh = u - uh

En particular interesa saber si este error converge a cero a medida que h tiende a cero y en caso afirmativo cual es el orden de convergencia del meacutetodo

Definicioacuten Un meacutetodo de paso simple se dice que satisface las condiciones de Lipschitzsi F h verifica

para todo h gt0 suficientemente pequentildeo donde Lh es independiente de h El resultado siguiente presenta condiciones de equivalencia entre los conceptos deconsistencia y convergencia y establece una estimacioacuten para el error de aproximacioacuten

Teorema Si un meacutetodo de paso simple satisface las condicioacuten de Lipschitz entonces

seraacute consistente si y solo si fuera convergenteMas auacuten para h suficientemente pequentildeo se verifica que

Donde eo = u(t o) - uh(t o)Si f isinC

p y las hipoacutetesis de este teorema se verifican entonces los meacutetodos de Taylor deorden (de consistencia) p tienen orden de convergencia p Lo mismo se verifica para losmeacutetodos de Runge-KuttaEn el resultado anterior no son considerados cualquier error de redondeo Considerandoahora que para cada valor ui calculado esta asociado un error de redondeo δ i y siendo δ

un maacuteximo en valor absoluto de todos estos errores es entonces posible obtener lasiguiente relacioacuten

Se concluye entonces que cuando hrarr0 el termino δ h impide que el error se aproxime acero En teacuterminos praacutecticos esto significa que a partir de un valor dado de pasos loserrores de redondeo son dominantes relativamente a los errores de truncamiento nohabiendo cualquier ventaja en disminuir el paso

Sistemas de ecuaciones diferenciales o EDO de orden superior

Dadas las funciones f 1 f 2 f n de RRn+1 en RR un sistema de ecuaciones diferenciales deorden 1 definido por

El problema de valor inicial consiste ahora en determinar las funciones u1 u2 un en unintervalo [t oT ] en RR que satisfacen estas ecuaciones diferenciales y las condiciones

u1(t0)= u10 u2(t0)= u20 un(t0)= un0

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para u10 u20 un0 isin RR dadosEn una notacioacuten mas compacta el sistema de ecuaciones diferenciales se representa por

))(()( t t t uFu =prime

donde F RR

n+1 rarr

RR

nn

esta definida por FF == [[ f 1 f 2 f n]

T

y u es una funcioacuten de RR

en RR

nn

ddeef f iinniiddaa ppoorr uu == [[u1 u2 un]T

El problema de valor inicial consiste en determinar la funcioacuten u que satisface estaecuacioacuten diferencial vectorial en un intervalo [t oT ] y la condicioacuten inicial

u(t o) = uo para alguacuten uo isin R RnEs posible establecer condiciones de existencia y unicidad de solucioacuten para esteproblema anaacutelogas a las formuladas en el caso de una ecuacioacuten diferencial escalar

Sistemas de ecuaciones diferenciales ndash Meacutetodos numeacutericos

Los meacutetodos numeacutericos de solucioacuten aproximada de problemas de valor inicialestudiados para el caso escalar (una ecuacioacuten) pueden ser aplicados de una formainmediata para el caso vectorial (sistema de ecuaciones)

)(1 iihii t h uFuu +=+

Considerando una malla de N

iit 0= de paso h en el intervalo [t oT ] siendo uh una

solucioacuten aproximada del problema de valor inicial de un modo general los valoresui=uh(t i) pueden ser obtenidos por la expresioacuten de recurrenciapara i=01 N -1 siendo tambieacuten habitual considerar uh(t o)= uo

Es de notar la semejanza entre esta expresioacuten de recurrencia y la expresioacuten generalutilizada en el caso escalarUna funcioacuten Fh se define en teacuterminos de f de forma anaacuteloga al caso escalarLa principal diferencia reside en el hecho de que los valores ui a determinar por viacuteanumeacuterica seraacuten elementos de RRn siendo en cada paso necesario calcular n nuacutemerosrealesbull El meacutetodo de Euler progresivo conduce a la expresioacuten de recurrencia

bull El meacutetodo de Taylor de orden 2 tiene por expresioacuten de recurrencia

Nota El calculo de f prime f primeprime puede ser bastante complejo pues cada componente de f depende de t bien directamente bien indirectamente a traveacutes de los componentes de u

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Ejemplo 5Considere el siguiente problema de valor inicial

[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu

(a) Determinar una solucioacuten aproximada por el meacutetodo de Euler progresivo con paso 01(b)Determinar una solucioacuten aproximada por el meacutetodo de Taylor de orden 2 con paso

01

SolucioacutenSe definen f 1y f 2 por

minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f

La expresioacuten del meacutetodo de Euler progresivo)(1 iiii t h uFuu +=

+

Toma en este caso la forma

)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+

Para i =019 con las condiciones iniciales0)0(1)0( 202101 ==== uuuu

b) La expresioacuten del meacutetodo de Taylor de orden 2 es

)(2

)(2

1 iiiiii t h

t h uf uf uu prime++=+

siendo entonces necesario determinar 1 f prime y 2 f prime Estas funciones se obtienen de acuerdocon

Las expresiones de recurrencia toman entonces la forma

debiendo ser determinadas para i = 0 1hellip9 con las condiciones iniciales u10=1 y u20=0La tabla abajo representa los valores obtenidos

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Ecuaciones diferenciales de orden n ndash Problema de valor inicial

Consiste en determinar una funcioacuten uRRrarrRR que es solucioacuten de una ecuacioacuten diferencialdada de orden n En un intervalo dado [t oT ] y satisface las condiciones iniciales

para 0u 0uprime )1(0

minusnu isin RR dados

La ecuacioacuten diferencial de orden n se transforma de una forma simple en un sistema den ecuaciones diferenciales de orden 1Considerando las variables dependientes (esto es las funciones) u1 u2un definidas

porse concluye faacutecilmente que )()( 1 t ut u ii +=prime para i = 12 n-1

Utilizando estas nuevas se tiene entonces que

El sistema de ecuaciones diferenciales de orden 1 toma entonces la forma

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Debiendo su solucioacuten satisfacer las condiciones iniciales

Ejemplo 6Determinar por el meacutetodo de Euler progresivo con paso 005 una solucioacuten aproximadade

SolucioacutenDefiniendo u1=θ y u2=θ prime se obtiene elsistema de ecuaciones diferenciales

21 uu =prime

)sin(10 12 uu minus= Las expresiones de recurrencia seraacuten

con u10 = 01 y u20 = 0

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Problema del Valor FronteraSea el problema de valor frontera en una EDO de segundo orden

Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00

Meacutetodo del Disparo

Consiste en transformar el problema de valor fronteraen un problema de valor inicial suponiendo unapendiente s luego desarrolla con un meacutetodonumeacuterico para encontrar )(su N se compara con B

si estos valores no son aproximados se siguesuponiendo pendientes hasta dar al blanco (B)

Problema de Valor inicial resultante

st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00

Algoritmo del Meacutetodo del Disparo

1 Elija un valor inicial de s = s0 =0

0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus

= y los puntos jht t j += 0

3 Use un meacutetodo numeacuterico para solucionar la EDO (por ejemplo RK de orden 4) alobtener )( o N N suu = compaacuterelo con )(bu

4 Elija un segundo valor para s = s1

001

t b

u Bss

N

minus

minus+=

Luego se aplica un meacutetodo numeacuterico para obtener

)( 1su N 5 Utilice interpolacioacuten lineal a fin de obtenerelecciones subsecuentes valores para s esto es

)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++

Con cada sk resolvemos el problema de valor inicialy comparamos )( k N su con B

6 Deteacutengase cuando |)(| Bsu k N minus sea suficientemente pequentildeo (Criterio de

convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s

Fig 1 Soluciones numeacutericas con pendientess0 y s1

1s

B

0s

)( 0su N

)( 1su N

s

u

2s

Fig 2 Interpolacioacuten de la pendiente s2

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Meacutetodo de las diferencias finitas

Dado el problema de valor inicial de segundo orden con valor frontera

β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele

se requiere aproximar uu primeprimeprime usando diferencias centrales

Para lo cual seleccionamos un nuacutemero Ngt0 y dividimos el intervalo [ab] en (N+1) sub-intervalos iguales cuyos puntos son los puntos de la retiacutecula

ihat i += para i=01N+1 donde1++++

minusminusminusminus==== N

a b h

)(

Esto nos lleva a formar un sistema lineal de orden NxN que puede ser resuelto por elmeacutetodo para resolver SEL discutidos anteriormenteEn los puntos interiores de la retiacutecula t i= 12 N la ecuacioacuten diferencial a aproximares

)()()()()()( iiiiii t r t ut qt ut pt u ++=primeprime (1)

Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime

+minusasympprimeprime

Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=

+minus minus+minus+ (2)

en la frontera β α == )()( buau

Definiendo β α == +10 N ww

la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=


La ec (3) se puede escribir como

( ) )()(212)()(21 2121 iiiiiii t r hwt phwt qhwt ph =

minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h

=i B 2)(2minusminus it qh

=iC

minus )(

21 it p

h

=i D )(2it r h

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minus

minus

=

+

minusminusminusminusminus

1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K

La solucioacuten del sistema lineal se puede realizar por reduccioacuten de Crout

Ejemplo 7Dada la EDO de tercer orden

107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=

+minusprimeminusprimeprime=primeprimeprime minus

t

u

u

u

ut ueuu t

Encuentre la curva integral aproximada usando RK de orden 2Usando 4 pasosSolucioacuten-Haciendo el cambio de variable u1=u u2= uprime u3=u primeprime Convirtiendo la EDO en un sistema de EDOrsquoS de primer orden con sus respectivascondiciones iniciales

7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t

El nuacutemero de pasos es 4 por lo tanto el ancho del paso(h) seraacuteh=(1 - 0)4 = 025Vector de condiciones inicialesZ= [ 5 3 7 ]

u1= Z(1) u2 = Z(2) u3=Z(3)

F = [u2 u3 sin(u3)-exp(-t)u2-2tcos(u1)+25]t

Algoritmo de RK2 en forma vectorialPaso 1

=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

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172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630

Runge-Kutta solution to ursquorsquorsquo = f(t u ulsquo ulsquolsquo)

t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver

Con la condicioacuten de frontera x=0 y=0x=05 y=1

Solucioacuten

Haciendo el cambio de variable u1=y u2= yprime Convirtiendo la EDO en un sistema de EDOrsquoS de primer orden con sus respectivascondiciones iniciales

2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu

Considerando h=01 por lo tanto el nuacutemero de pasos (N) seraacuteN=(05 - 0)01 = 5Vector de condiciones inicialesZ= [ 0 2 ]t

Estimar y(x) utilizando el meacutetodo del disparo (con Euler modificado o RK2 con h=01)

022

2

=++ x ydx

dye

dx

yd xy

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[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05

07693-12 =+ (estimando otra pendiente)

x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

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6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9

Resolver aplicando diferencias finitas

yrdquo = (-2x)yrsquo+(2x2)y+sin(Lnx)x2 y(1)=1y(2)=2

h=02Solucioacutenyrdquo = p(x)yrsquo+q(x)y+r(x)

i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

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i A =

+ )(

21 i x p

h

====i B ( )2)(2+minus i xqh

====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x

The finite difference solution to x`` = f(txx`)

t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000

Tarea 2Utilizando meacutetodo del disparo con (N=5 sub-intervalos y usando RK2 o RK4) resolverx2yrdquo + xyrsquo + (x2-3)y = 0y(1)=0 y y(2)=2

SolucioacutenS0= 20000 Y5(S0) = 14715

S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

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APLICACIONESEjemplo1Se ha comprobado que en algunas circunstancias el nuacutemero de individuos dedeterminadas poblaciones de bacterias se rige por la ley siguiente

)(20)( t yt y = La variable t es el tiempo medido en horas e y(t) es el nuacutemero de miles de bacteriasque hay en el instante t Al comienzo de un determinado experimento hay 30 mil bacterias iquestCuaacutentas habraacute 10horas mas tarde iquestEn que instante habraacute 100000 bacterias

SolucioacutenSi se mide el tiempo en horas y se empieza a contar al comienzo del experimentopuesto que se desea saber cuaacutentas bacterias habraacute pasadas 10 horas se tiene que elproblema a resolver es

=

isin=

10)0(

]100[20

y

t y y

gtgtf=inline(lsquo02yrsquorsquotrsquorsquoyrsquo)f =Inline functionf(ty) = 02ygtgt ode23(f[010]30)A simple vista se puede ver que el valor en t=10 es aproximadamente 220 lo quesignifica que pasadas 10 horas habraacute (aprox) 220000 bacterias

Si se desea auacuten mayor exactitud se puede usar la orden siguientegtgt [ty]=ode23(f[010]30)y ver cuanto vale la uacuteltima componente del vector ygtgt y(end)ans =2215562

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Ejemplo2Un ejemplo es el modelo de Lotka-Volterra tambieacuten conocido como modelo de presa-depredador ya que modeliza la situacioacuten en la que hay dos especies que conviven y unade ellas es depredadora de la otraSi denotamos por y1(t) el nuacutemero de presas en el instante t y por y2(t) el nuacutemero dedepredadores en el instante t el modelo de Lotka-Volterra establece que el nuacutemero deindividuos de cada especie evoluciona en el tiempo de acuerdo con el sistemadiferencial

+minus=

minus=

2122

2111

ydycy y

ybyay y

en el que las constantes a b c y d variacutean de un caso a otro ya que dependen de lanatalidad y agresividad de cada especie Obseacutervese que ahora se tienen dos incoacutegnitas ydos ecuacionesA este sistema habraacute que antildeadir como en el caso de una sola ecuacioacuten unascondiciones iniciales que indiquen cual es la situacioacuten de partida es decir cuaacutentosindividuos de cada especie hay en el instante inicial

Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111

Para resolver con MATLAB este sistema se debe en primer lugar escribir con notacioacutenvectorial

=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1

gtgt f=inline([ay(1)-by(1)y(2)-cy(2)+dy(1)y(2)]ty)Despueacutes la resolucioacuten es anaacuteloga observando que la condicioacuten inicial tambieacuten es ahoraun vectorgtgt ode23(f[t0tf][AB])

Ejemplo Calcular la solucioacuten de

4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y

gtgt f=inline([05y(1)-02y(1)y(2)-05y(2)+01y(1)y(2)]ty)gtgt ode23(f[060][44])gtgt axis([060020])

Obseacutervese el comportamiento perioacutedico de la solucioacuten tiacutepico de las soluciones de estemodelo

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Problemas PropuestosP1- La ecuacioacuten diferencial dvdt=2000(200-t) Nkg representa el movimiento de uncuerpo con una masa inicial de 200 kg bajo la accioacuten permanente de una fuerza de 2000N y que va perdiendo un kg de su masa por cada segundo Considere ahora que elmismo cuerpo en movimiento soporta la resistencia del aire numeacutericamente igual aldoble del valor de su velocidad (v) Se pide encontrar la velocidad para t=50 seg sicuando t=0 el cuerpo esta en reposo usando el meacutetodo de Euler modificado (o RK-2)con h=5

P2- Dada la siguiente ecuacioacuten yrsquo = 1 + xy + y2 y(0)=0

Se Pide

Encontrar el algoritmo de Taylor de orden 4Aproximar y(02) usando el algoritmo de a) h=01Calcular el error cometido

P3- Un proyectil de masa m=010 kg se lanza verticalmente hacia arriba con unavelocidad inicial de Vo=80 mseg y se va frenando debido a la fuerza de gravedad Fg = -mg y a la resistencia del aire F r = -kV2 donde g=98 mseg2 y k=002 kgm La ecuacioacutendiferencial para la velocidad V esta dada por mVrsquo = -mg - kV 2

Calcule la velocidad para t=005 010 y 015 segundos aplicando RK-2 o Eulermejorado

Estime el tiempo que tarda el proyectil en alcanzar su maacutexima altura

P4- Dado el problema de valor inicial

yrsquo = (1t2) - (yt) - y2 y(1) = -1Se PideiquestCuaacutel seria el valor de h adecuado si se conoce que y1 = -095000 (aproximacioacuten en elprimer paso usando el algoritmo de Euler)Con h encontrado en a) calcular 4 pasos por el meacutetodo de Euler (i=01234)Usando la ecuacioacuten del error global de Euler determinar el valor de ldquohrdquo necesario para

que|y(120)-yN|lt=0005

P5- Sea el problema de valor inicial

xrsquo = 1 ndash xtx(2)=2

Demuestre que

xrdquo = (1-2xrsquo)t x(n) derivada de orden nx(n) = -nx(n-1)t

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Calcular x(21) aplicando Taylor de orden k (h=01) hasta tener 5 cde

iquestSi desea calcular x(21) con h=005 explique como calculariacutea el numero de teacuterminosnecesarios para tener la misma precisioacuten de la pregunta b)

P6- Usando Runge-Kutta de orden 4 hallar la solucioacuten de la ecuacioacuten diferencialyrsquo=025y2+x2 En el intervalo [005] teniendo en cuanta la condicioacuten y(0) = -1 (tomar h=01)

P7- Dada la ecuacioacuten diferencial dydx=01(y+x)08

Con la condicioacuten inicial y(1)=3 estimar y(175) usando el siguiente meacutetodo de RK decuarto orden y h=025

P8- En un canal abierto se desea medir el caudal de un flujo de agua que fluye a traveacutesde un vertedero en forma de ldquoVrdquo Se conoce que la altura del periacutemetro mojado delvertedero es 1 metro (H) y el ancho es de 15 metros (L) La ecuacioacuten diferencial queexpresa la relacioacuten entre el caudal y la altura es la siguientedQdy = (2gY)12 (LH) (H-Y) Q(0)=0Se pide encontrar por el meacutetodo de Milne el caudal a 15 cm de alturaSugerencia- Usar h=25 cm y el meacutetodo de Runge Kutta de orden 4 para caacutelculosprevios

P9- Resolver

2xrsquo + yrsquo ndash y = txrsquo + yrsquo = t2 x(0)=1 y(0)=0Aproximar y(02) considerando h=01 aplicando el meacutetodo de Runge Kutta de orden 3Aproximar x(02) considerando h=01 aplicando el meacutetodo de Runge Kutta de orden 2

P10- Resolver la ecuacioacuten diferencial de segundo ordenyrdquo - 2yrsquo + 2y = e2x senxx en [01]y(0) = -04yrsquo(0) = -06

Aplique Runge Kutta de orden 4 con h=01

P11- Se tienexrdquo = x2 ndash y + et yrdquo = x ndash y2 + et x(0) = xrsquo(0) = 0y(0) = 1 yrsquo(0) = -2Aplique Runge-Kutta de orden 2 para obtener x(025)y(025) x(05) y(05)

P12- Dada la ecuacioacuten de segundo orden

200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf


Universidad Nacional de Ingenieriacutea EDOFacultad de Ingenieriacutea Mecaacutenica PA 2009-1DACIBAHCCResolver la EDO no lineal utilizando el meacutetodo de las diferencias finitas con h=05P13- Resolver


Estimar y(x) utilizando el meacutetodo del disparo (con Euler modificado y h=01)

P14- El Problema de valor de frontera normalmente se expresayrdquo = P(x) yrsquo + Q(x) y + R(x)x en [ab] y(a)=m y(b)=nEstudie las modificaciones en el enfoque de diferencias finitas para los siguientes casos

La condicioacuten de frontera y(a)=m se reemplaza por yrsquo(a)=u

La condicioacuten de frontera y(b)=n se reemplaza por yrsquo(b)=wSe realizan las modificaciones a) y b)

P15- Utilizando diferencias finitas con (N=4 sub-intervalos) resolver x2yrdquo + xyrsquo + (x2-3)y = 0y(1)=0 y y(2)=2

P16- Resolver aplicando diferencias finitasyrdquo = (-2x)yrsquo+(2x2)y+sen(Lnx)x2 y(1)=1y(2)=2h=02P17 - Un sistema mecaacutenico compuesto por dos resortes en serie y unido a dos masasobedece a la siguiente ecuacioacuten diferencial

m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1

Prime = minus + minus

Prime = minus minus

( )

( )

Si k1 = 6 k2 = 4 m1 = 1 m2 = 1 x1(0)=0 x1rsquo(0)=1 x2(0)=0 x2rsquo(0)= -1Reducir a un sistema de ecuaciones diferenciales de primer ordenCalcular x1(01) y x2(01) utilizando Euler con h = 005

Estime el error de la aproximacioacuten obtenida y comente su respuestaP18 - El sistema de ecuaciones que describe las corrientes i1(t) e i2(t) en la red eleacutectricaque contiene una resistencia una inductancia y un condensador es

E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1

Calcular el valor de i1(003) e i2(003) aplicandoRunge-Kutta de orden 2 con h = 001 Donde E= 60 V L=1 H R=50 Ω C = 10-4 F y donde i

1

e i2 son inicialmente son inicialmente iguales acero

E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

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Soluciones Numeacutericas de Ecuaciones Diferenciales Ordinarias

Muchos problemas en la Ingenieriacutea y otras ciencias pueden ser formulados en teacuterminosde ecuaciones diferenciales Por ejemplo trayectorias baliacutesticas teoriacutea de sateacutelitesartificiales estudio de redes eleacutectricas curvaturas de vigas estabilidad de avionesteoriacutea de vibraciones reacciones quiacutemicas y otras aplicaciones de aquiacute la importanciade su solucioacuten

Objetivo Estudiar los meacutetodos numeacutericos que permitan obtener soluciones numeacutericasaproximadas de las ecuaciones diferenciales ordinarias (EDOs)

Trataremos los meacutetodos para resolver el problema del valor inicial y tambieacuten el valorfrontera

Meacutetodos de solucioacuten de EDOrsquos ndash Problema del Valor InicialLa ecuacioacuten [ ]T tot t ut f t u ))(()( isin=prime (1)es llamada Ecuacioacuten diferencial de primer orden Aquiacute f des una funcioacuten realconocida de dos variables reales t y u y u es llamada funcioacuten incoacutegnita de variableindependiente t Ademaacutes de esto u y f pueden ser vectores este caso lo veremos maacutesadelante y corresponderaacute a un sistema de e EDOrsquos de primer orden

Resolver (1) corresponde a determinar una funcioacuten u=u(t ) diferenciable con [ ]T tot isin tal que ))(()( t ut f t u =prime Cualquier funcioacuten que satisfaga esta propiedad es una solucioacuten

de la ecuacioacuten diferencial (1) Por ejemplo una funcioacuten

t

Cet u =

)( es para cualquiervalor constante de C una solucioacuten de la ecuacioacuten diferencial ut u =prime )( Asimismo cadaecuacioacuten diferencial de primer orden posee un nuacutemero infinito de soluciones De aquiacutepodemos seleccionar una solucioacuten particular si junto con la EDO fuera dado un valorde la solucioacuten u(t) en un punto por ejemplo u(to)=uo (Condicioacuten Inicial)Si para la ecuacioacuten diferencial ut u =prime )( con condicioacuten inicial 1)0( =u entonces

obtenemos C=1 y la solucioacuten seriacutea t et u =)( Problema del Valor Inicial

Nota Antes de empezar a resolver el problema interesa garantizar que esta tienesolucioacuten uacutenica

Dados una funcioacuten f RRxRRrarrRR un intervalo real [t o T ] y un valor uo isin RR el problema de

valor inicial consiste en determinar una funcioacuten u[t o T ]rarr RR que verifique una ecuacioacutendiferencial de primer orden

[ ]T tot t ut f t u ))(()( isin=prime con la condicioacuten inicial

u(t 0)= u0

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Definicioacuten




i N i

hmaxhlele

=1







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)(max][

t vvT t t oisin

=


[ ]


minus=minus

isin






0)(0

=rarr

oht

t elim


h chuu leminus











int+

=minus+

ht

t

d u f t uht u ξ ξ ξ ))(()()(


))(( ζ ζ u f en [t t + h]

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int+

=

ht

t

d u f h

ut F ξ ξ ξ ))((1

)(


int

+

asymp

ht

t

h d u f h

ut F ξ ξ ξ ))((1)(




t int +


)()()()()(

ut T ut F ut F h

t uht uhh +==

minus+

(2)


+=



110)(1 minus=+=+

N iut hF uu iihii K


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=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α






))(()()()(

t ut f t uh

t uht u=asymp

minus+


n j jht t j 2100 K=+=


(3)




)( 111 +++ += j j j j

ut hf uu (4)


=

lele++minus=

1)0(

101

u

t t uu


+

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Solucioacuten



1010

01 =minus=minus=n

abh ademaacutes

)10(0 j jhat j +=+=



9101001090

)110(10)1(10)(

1

1

0

=++=


=

+

j ju

juut u yut f huu

u

i

j j j j j j j j j


110019010001090 01 =++times=+times+= uu

ahora para j = 1

0111001019010101090 12 =++times=+times+= uu





jt u |

00 1000000 1000000 0001 1000000 1004837 000483702 1010000 1018731 000873103 1029000 1040818 001181804 1056000 1070320 001422005 1090000 1106531 001604106 1131441 1148812 001737107 1178297 1196585 001828808 1230467 1249329 001886209 1287420 1306570 001915010 1348678 1367879 0019201



[ ] ))((sup2 0

t ut f h

T T t t h isin=



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rarr hh T


p

h chT le




( )

11

)1(1)(21 )(

1

1)(

1)(

2

1)()()(

++

++

+

ltltminus=

+++++++=

j j j j

qq

j

qq

j j j j

uu yt t hcon

uhq

t uhq

t uht hut ut u

ξ

ξ K(5)




aproximados )( j j

t uu asymp



K

Donde

)(

)(2

)()( )1(1

)( j j

qq

j j j j j j

qut f

q

hut f

hut f ut T

minus

minus

++= K



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α =++=+ 0

2

1 )(2

)( uconut f h


Ejemplo 2


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



t ut uut u

dt


Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u



0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=


t uhuu

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jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




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)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

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910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



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[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


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( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

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eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

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donde F RR

n+1 rarr

RR

nn


T


en RR

nn













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[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





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21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



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Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


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β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



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minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

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172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



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[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

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6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

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i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



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=

isin=

10)0(

]100[20

y

t y y



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+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



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Se Pide







que|y(120)-yN|lt=0005



Demuestre que


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P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

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Definicioacuten




i N i

hmaxhlele

=1







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)(max][

t vvT t t oisin

=


[ ]


minus=minus

isin






0)(0

=rarr

oht

t elim


h chuu leminus











int+

=minus+

ht

t

d u f t uht u ξ ξ ξ ))(()()(


))(( ζ ζ u f en [t t + h]

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int+

=

ht

t

d u f h

ut F ξ ξ ξ ))((1

)(


int

+

asymp

ht

t

h d u f h

ut F ξ ξ ξ ))((1)(




t int +


)()()()()(

ut T ut F ut F h

t uht uhh +==

minus+

(2)


+=



110)(1 minus=+=+

N iut hF uu iihii K


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=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α






))(()()()(

t ut f t uh

t uht u=asymp

minus+


n j jht t j 2100 K=+=


(3)




)( 111 +++ += j j j j

ut hf uu (4)


=

lele++minus=

1)0(

101

u

t t uu


+

7252019 10_Sep_EDO_2009_1pdf



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Solucioacuten



1010

01 =minus=minus=n

abh ademaacutes

)10(0 j jhat j +=+=



9101001090

)110(10)1(10)(

1

1

0

=++=


=

+

j ju

juut u yut f huu

u

i

j j j j j j j j j


110019010001090 01 =++times=+times+= uu

ahora para j = 1

0111001019010101090 12 =++times=+times+= uu





jt u |

00 1000000 1000000 0001 1000000 1004837 000483702 1010000 1018731 000873103 1029000 1040818 001181804 1056000 1070320 001422005 1090000 1106531 001604106 1131441 1148812 001737107 1178297 1196585 001828808 1230467 1249329 001886209 1287420 1306570 001915010 1348678 1367879 0019201



[ ] ))((sup2 0

t ut f h

T T t t h isin=



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rarr hh T


p

h chT le




( )

11

)1(1)(21 )(

1

1)(

1)(

2

1)()()(

++

++

+

ltltminus=

+++++++=

j j j j

qq

j

qq

j j j j

uu yt t hcon

uhq

t uhq

t uht hut ut u

ξ

ξ K(5)




aproximados )( j j

t uu asymp



K

Donde

)(

)(2

)()( )1(1

)( j j

qq

j j j j j j

qut f

q

hut f

hut f ut T

minus

minus

++= K



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α =++=+ 0

2

1 )(2

)( uconut f h


Ejemplo 2


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



t ut uut u

dt


Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u



0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=


t uhuu

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jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




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)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



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910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



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[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



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( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

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eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

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donde F RR

n+1 rarr

RR

nn


T


en RR

nn













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[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





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21 uu =prime


con u10 = 01 y u20 = 0

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Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


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983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



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minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

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172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

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[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

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6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

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i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



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=

isin=

10)0(

]100[20

y

t y y



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+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



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Se Pide







que|y(120)-yN|lt=0005



Demuestre que


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P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

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)(max][

t vvT t t oisin

=


[ ]


minus=minus

isin






0)(0

=rarr

oht

t elim


h chuu leminus











int+

=minus+

ht

t

d u f t uht u ξ ξ ξ ))(()()(


))(( ζ ζ u f en [t t + h]

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int+

=

ht

t

d u f h

ut F ξ ξ ξ ))((1

)(


int

+

asymp

ht

t

h d u f h

ut F ξ ξ ξ ))((1)(




t int +


)()()()()(

ut T ut F ut F h

t uht uhh +==

minus+

(2)


+=



110)(1 minus=+=+

N iut hF uu iihii K


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=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α






))(()()()(

t ut f t uh

t uht u=asymp

minus+


n j jht t j 2100 K=+=


(3)




)( 111 +++ += j j j j

ut hf uu (4)


=

lele++minus=

1)0(

101

u

t t uu


+

7252019 10_Sep_EDO_2009_1pdf



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Solucioacuten



1010

01 =minus=minus=n

abh ademaacutes

)10(0 j jhat j +=+=



9101001090

)110(10)1(10)(

1

1

0

=++=


=

+

j ju

juut u yut f huu

u

i

j j j j j j j j j


110019010001090 01 =++times=+times+= uu

ahora para j = 1

0111001019010101090 12 =++times=+times+= uu





jt u |

00 1000000 1000000 0001 1000000 1004837 000483702 1010000 1018731 000873103 1029000 1040818 001181804 1056000 1070320 001422005 1090000 1106531 001604106 1131441 1148812 001737107 1178297 1196585 001828808 1230467 1249329 001886209 1287420 1306570 001915010 1348678 1367879 0019201



[ ] ))((sup2 0

t ut f h

T T t t h isin=



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rarr hh T


p

h chT le




( )

11

)1(1)(21 )(

1

1)(

1)(

2

1)()()(

++

++

+

ltltminus=

+++++++=

j j j j

qq

j

qq

j j j j

uu yt t hcon

uhq

t uhq

t uht hut ut u

ξ

ξ K(5)




aproximados )( j j

t uu asymp



K

Donde

)(

)(2

)()( )1(1

)( j j

qq

j j j j j j

qut f

q

hut f

hut f ut T

minus

minus

++= K



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α =++=+ 0

2

1 )(2

)( uconut f h


Ejemplo 2


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



t ut uut u

dt


Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u



0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=


t uhuu

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jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




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)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



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910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



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[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


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( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

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eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

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donde F RR

n+1 rarr

RR

nn


T


en RR

nn













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[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





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21 uu =prime


con u10 = 01 y u20 = 0

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Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


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β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

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172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



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[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

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6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

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i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



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=

isin=

10)0(

]100[20

y

t y y



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+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



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Se Pide







que|y(120)-yN|lt=0005



Demuestre que


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P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

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int+

=

ht

t

d u f h

ut F ξ ξ ξ ))((1

)(


int

+

asymp

ht

t

h d u f h

ut F ξ ξ ξ ))((1)(




t int +


)()()()()(

ut T ut F ut F h

t uht uhh +==

minus+

(2)


+=



110)(1 minus=+=+

N iut hF uu iihii K


7252019 10_Sep_EDO_2009_1pdf



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=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α






))(()()()(

t ut f t uh

t uht u=asymp

minus+


n j jht t j 2100 K=+=


(3)




)( 111 +++ += j j j j

ut hf uu (4)


=

lele++minus=

1)0(

101

u

t t uu


+

7252019 10_Sep_EDO_2009_1pdf



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Solucioacuten



1010

01 =minus=minus=n

abh ademaacutes

)10(0 j jhat j +=+=



9101001090

)110(10)1(10)(

1

1

0

=++=


=

+

j ju

juut u yut f huu

u

i

j j j j j j j j j


110019010001090 01 =++times=+times+= uu

ahora para j = 1

0111001019010101090 12 =++times=+times+= uu





jt u |

00 1000000 1000000 0001 1000000 1004837 000483702 1010000 1018731 000873103 1029000 1040818 001181804 1056000 1070320 001422005 1090000 1106531 001604106 1131441 1148812 001737107 1178297 1196585 001828808 1230467 1249329 001886209 1287420 1306570 001915010 1348678 1367879 0019201



[ ] ))((sup2 0

t ut f h

T T t t h isin=



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rarr hh T


p

h chT le




( )

11

)1(1)(21 )(

1

1)(

1)(

2

1)()()(

++

++

+

ltltminus=

+++++++=

j j j j

qq

j

qq

j j j j

uu yt t hcon

uhq

t uhq

t uht hut ut u

ξ

ξ K(5)




aproximados )( j j

t uu asymp



K

Donde

)(

)(2

)()( )1(1

)( j j

qq

j j j j j j

qut f

q

hut f

hut f ut T

minus

minus

++= K



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α =++=+ 0

2

1 )(2

)( uconut f h


Ejemplo 2


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



t ut uut u

dt


Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u



0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=


t uhuu

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jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




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)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



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910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



7252019 10_Sep_EDO_2009_1pdf



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[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

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eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

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donde F RR

n+1 rarr

RR

nn


T


en RR

nn













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[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





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21 uu =prime


con u10 = 01 y u20 = 0

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Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


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β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

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minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



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[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



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6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



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i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



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=

isin=

10)0(

]100[20

y

t y y



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+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



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P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983093






=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α






))(()()()(

t ut f t uh

t uht u=asymp

minus+


n j jht t j 2100 K=+=


(3)




)( 111 +++ += j j j j

ut hf uu (4)


=

lele++minus=

1)0(

101

u

t t uu


+

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983094

Solucioacuten



1010

01 =minus=minus=n

abh ademaacutes

)10(0 j jhat j +=+=



9101001090

)110(10)1(10)(

1

1

0

=++=


=

+

j ju

juut u yut f huu

u

i

j j j j j j j j j


110019010001090 01 =++times=+times+= uu

ahora para j = 1

0111001019010101090 12 =++times=+times+= uu





jt u |

00 1000000 1000000 0001 1000000 1004837 000483702 1010000 1018731 000873103 1029000 1040818 001181804 1056000 1070320 001422005 1090000 1106531 001604106 1131441 1148812 001737107 1178297 1196585 001828808 1230467 1249329 001886209 1287420 1306570 001915010 1348678 1367879 0019201



[ ] ))((sup2 0

t ut f h

T T t t h isin=



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rarr hh T


p

h chT le




( )

11

)1(1)(21 )(

1

1)(

1)(

2

1)()()(

++

++

+

ltltminus=

+++++++=

j j j j

qq

j

qq

j j j j

uu yt t hcon

uhq

t uhq

t uht hut ut u

ξ

ξ K(5)




aproximados )( j j

t uu asymp



K

Donde

)(

)(2

)()( )1(1

)( j j

qq

j j j j j j

qut f

q

hut f

hut f ut T

minus

minus

++= K



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α =++=+ 0

2

1 )(2

)( uconut f h


Ejemplo 2


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



t ut uut u

dt


Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u



0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=


t uhuu

7252019 10_Sep_EDO_2009_1pdf



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jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983088

)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983089

910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



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[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

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eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

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donde F RR

n+1 rarr

RR

nn


T


en RR

nn













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[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





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983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









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21 uu =prime


con u10 = 01 y u20 = 0

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Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


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β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



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minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



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172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



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[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

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i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



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=

isin=

10)0(

]100[20

y

t y y



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+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



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Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



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P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

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Solucioacuten



1010

01 =minus=minus=n

abh ademaacutes

)10(0 j jhat j +=+=



9101001090

)110(10)1(10)(

1

1

0

=++=


=

+

j ju

juut u yut f huu

u

i

j j j j j j j j j


110019010001090 01 =++times=+times+= uu

ahora para j = 1

0111001019010101090 12 =++times=+times+= uu





jt u |

00 1000000 1000000 0001 1000000 1004837 000483702 1010000 1018731 000873103 1029000 1040818 001181804 1056000 1070320 001422005 1090000 1106531 001604106 1131441 1148812 001737107 1178297 1196585 001828808 1230467 1249329 001886209 1287420 1306570 001915010 1348678 1367879 0019201



[ ] ))((sup2 0

t ut f h

T T t t h isin=



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rarr hh T


p

h chT le




( )

11

)1(1)(21 )(

1

1)(

1)(

2

1)()()(

++

++

+

ltltminus=

+++++++=

j j j j

qq

j

qq

j j j j

uu yt t hcon

uhq

t uhq

t uht hut ut u

ξ

ξ K(5)




aproximados )( j j

t uu asymp



K

Donde

)(

)(2

)()( )1(1

)( j j

qq

j j j j j j

qut f

q

hut f

hut f ut T

minus

minus

++= K



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α =++=+ 0

2

1 )(2

)( uconut f h


Ejemplo 2


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



t ut uut u

dt


Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u



0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=


t uhuu

7252019 10_Sep_EDO_2009_1pdf



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jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983088

)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983089

910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



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[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

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eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

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donde F RR

n+1 rarr

RR

nn


T


en RR

nn













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[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





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21 uu =prime


con u10 = 01 y u20 = 0

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Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


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β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

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minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

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172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



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[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

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6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

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i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983095




rarr hh T


p

h chT le




( )

11

)1(1)(21 )(

1

1)(

1)(

2

1)()()(

++

++

+

ltltminus=

+++++++=

j j j j

qq

j

qq

j j j j

uu yt t hcon

uhq

t uhq

t uht hut ut u

ξ

ξ K(5)




aproximados )( j j

t uu asymp



K

Donde

)(

)(2

)()( )1(1

)( j j

qq

j j j j j j

qut f

q

hut f

hut f ut T

minus

minus

++= K



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983096

α =++=+ 0

2

1 )(2

)( uconut f h


Ejemplo 2


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



t ut uut u

dt


Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u



0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=


t uhuu

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983097


jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983088

)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983089

910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983090

[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983092


eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983096

α =++=+ 0

2

1 )(2

)( uconut f h


Ejemplo 2


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



t ut uut u

dt


Luego

910)(2

)1()(2

)(

122

1

0

=minus+++minus+=++=

=

+ jt u

ht uhuut f

hut f huu

u



0051005001

)01(2

)10()101(101)(

2)1(

2

00

2

0001

=++=


t uhuu

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983097


jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983088

)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983089

910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983090

[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983092


eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983097


jt ju

00 1000000

01 100500002 101902503 104121804 107080205 110707606 114940407 119721108 124997609 130722810 1368541


=lele=

++minus=prime

103002)0(

2

h x y

x y y









=

ltlt==

)()(

))(()(

0

0


bt at t ut f t u

α


int+

+=+

1

)(1

j

j

t

t

j j ut f uu




7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983088

)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983089

910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983090

[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983092


eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983088

)]()([2 111 j j j j j j ut f ut f h

uu ++=+++





)]()([2

)(

111

1

j j j j j j

j j j j

ut f ut f h

uu

ut hf uu

++=

+=

+++

+


][2

1

)(

)(

211

112

1

k k uu

k ut hf k

ut hf k

j j

j j

j j

++=

+=

=

+

+

Ejemplo 3


=

lele++minus=

1)0(

101

u

t t uu

Solucioacuten



1010

01=

minus

=

minus

= n

ab

h ademaacutes

)10(0 j jhat j +=+=



]1)([)(

)1()(

11112

1

+++minus=+=

++minus==

++ j j j j

j j j j

t k uhk ut hf k

t uhut hf k

Luego

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983089

910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983090

[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983092


eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983089

910][2

1

1

211

0

=++=

=

+ jk k uu

u

j j


( ) 010110)01(10]1)([)(

0)101(10)1()(

1101012

00001



t k uhk ut hf k

t uhut hf k

0051]00100[2

11][

2

12101 =++=++= k k uu


( ) 00186110)009500051(10]1)([)(

00095)1100051(10)1()(

1111112

11111



t k uhk ut hf k

t uhut hf k

10190]0186000950[2

10051][

2

12112 =++=++= k k uu


jt ju

00 1000001 1005002 1019003 10412

04 1070805 1107106 1149407 1197208 1250009 1307210 13685



regla 13 de Simpson

[ ])()(4)(6

1112 12 11 +++++



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983090

[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983092


eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983090

[ ]43211

34

23

12

1

226

1

)(

)22(

)2

2

(

)(

k k k k uu

k uht hf k

k uht hf k

k u

ht hf k

ut hf k

j j

j j

j j

j j

j j

++++=

++=

++=

++=

=

+

Ejemplo 4


=

lele++minus=

1)0(101

u

t t uu

Solucioacuten



1010

01=

minus=

minus=

n

abh ademaacutes

)10(0 j jhat j +=+=



( )1)()()(

1)2

()2

()2

2

(

1)2

()2

()2

2

(

)1()(

334

223

112

1

++++minus=++=

++++minus=++=

++++minus=++=

++minus==

ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

j j j j

j j j j

j j j j

j j j j

Luego

[ ] 910226

1

1

43211

0

=++++=

=

+ jk k k k uu

u

j j


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983092


eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983091

( )

( )

( ) ( ) 0095250)1)100()0047501(101)()()(

0047501)0500()002501(101)2()2()22(

00501)0500()01(101)2

()2

()2

2

(

0)101(10)1()(

0303004

0202003

01

01

002

00001


=++++minustimes=

++++minus=++=

=++++minustimes=

++++minus=++=


ht k uhk uht hf k

ht

k uh

k u

ht hf k

ht

k uh

k u

ht hf k

t uhut hf k

[ ] [ ] 004837501009525000475020050206

1122

6




jt ju

00 1000000

01 10048375002 10187309003 10408184204 10703202805 11065309306 11488119307 11965856108 12493292809 13065699910 136787977

jt ju Euler

ju Taylor 2 ju RK2

ju RK4 )( jt u

00 1000000 1000000 10000000 1000000 100000001 1000000 1005000 100500000 100483750 100483702 1010000 1019025 101902500 101873090 101873103 1029000 1041218 104121762 104081842 1040818

04 1056000 1070802 107080195 107032028 107032005 1090000 1107076 110707577 110653093 110653106 1131441 1149404 114940357 114881193 114881207 1178297 1197211 119721023 119658561 119658508 1230467 1249976 124997526 124932928 124932909 1287420 1307228 130722761 130656999 130657010 1348678 1368541 136854098 136787977 1367879

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983092


eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983092


eh = u - uh













u1(t0)= u10 u2(t0)= u20 un(t0)= un0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983093



donde F RR

n+1 rarr

RR

nn


T


en RR

nn













7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983094


[ ]10212

211isin

minus+=prime

=primet

uut u

uuu

0)0(1)0( 21 == uu


01


minus+=

=

21212

21211

)(

)(

uut uut f

uuuut f


+


)(

)(

212212

211111

iiiii

iiiii

uut hf uu

uut hf uu

+=

+=

+

+

Ademaacutes

)(10

10

212212

21111

iiiii

iiii

uut uu

uuuu

minus+times+=

timestimes+=

+

+



)(2

)(2

1 iiiiii t h





7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983095









7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983096




21 uu =prime


con u10 = 01 y u20 = 0

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983089983097


Bbu

ut u

uut gu

=

=

=primeprime

)(

)(

)(

00





st u

ut u

uut gu

asympprime

=

=primeprime

)(

)(

)(

0

00



0

t b

u B

t

u

minus

minus=

∆

∆

2 Elija N

t bh 0minus




001

t b

u Bss

N

minus

minus+=



)()(

)()(

112

k N k N

k N k k k k

susu

su Bssss

minus

minusminus+=

+++



convergencia)

0t b

B

1s)( 0su N

)( 1su N

t

u

00 )( ut u =

0s


1s

B

0s

)( 0su N

)( 1su N

s

u

2s


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983088



β α ==

=++primeprime

)()(

)()()(

buau

t r ut qut pu

bt a lele





a b h

)(



Sabemos que

h

t ut ut u

h

t ut ut ut u

iii

iiii

2

)()()(

)()(2)()(

11

211

minus+

minus+

minusasympprime


Resultando

)()()(2

)()(

)(

)()(2)( 11

2

11iii

ii

i

iii

t r t ut qh

t ut u

t ph

t ut ut u++

minus=




la ec(2) queda

)()(2

)(2 11

211

iiiii

iiii t r wt q

h

wwt p

h

www++

minus=




minus+minusminus+

+ +minus

Haciendo

i A =

+ )(

21 it p

h


=iC

minus )(

21 it p

h

=i D )(2it r h

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983089

minus

minus

=

+


1

1

2

11

1

2

1

111

222

11

0

0

N N N

N

o

N

N

N N

N N N

wC D

D

D

w A D

w

w

w

w

B A

C B A

C B A

C B

M

K

OOOM

M

K



107)0(

3)0(

5)0(

25)cos(2)sin(

lele

=primeprime

=prime

=


t

u

u

u

ut ueuu t


7)0(3)0(

5)0(

25)cos(2)sin( 3

2

1

123

3

2

3

2

1

=

=

=

+minusminus=

=

=

prime

prime

prime

minusu

u

u

ut ueu

u

u

u

u

u

t


u1= Z(1) u2 = Z(2) u3=Z(3)



=

+

+

=

453112

45805

96885

24205

16613

18751

56642

751

750

2

1

7

3

5

1u

Paso 2

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983090

172640

92013

77224

581664

373324

142832

040195

113283

364511

2

1

453112

45805

96885

2

=

+

+

=u

Paso 1J 1 2 3K1J= [075000000000000 175000000000000 566424664967970]K2J = [118750000000000 316606166241992 524195497779435]

Paso2K1J = [136450770780249 311327520343426 504019199473519]K2J = [ 214282650866105 437332320211805 458166399404627]

Paso4

K1J = 352220957785520 551279173822577 472868904881154K2J = 490040751241164 669496400042865 467017188981630


t (k) u(k)0 5000000025 5968750050 7722417075 1056225110 14773559

Ejemplo 8Resolver


Solucioacuten


2050

01)0(

0)0(

02

1

212

2

2

1

1 =minus

minusasymp=

=

minusminus=

=

prime

prime

su

u

tuue

u

u

utu



022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983091

[TZ] = rks2(Fn1005[0s0]5)

K1 = 02000 -02000K2 = 01800 -01840

Z(1) = 01900 18080

K1 = 01808 -01846K2 = 01623 -01776

Z(2) = 03616 16269

K1 = 01627 -01775K2 = 01449 -01779

Z(3) = 05154 14492K1 = 01449 -01771K2 = 01272 -01831Z(4) = 06514 12691

K1 = 01269 -01817K2 = 01087 -01908

Z = 07693 10829

x y y00 0 20000 -- (S0)01 01900 1808002 03616 1626903 05154 1449204 06514 1269105 07693 10829

Y5(S0)

S1 = 4620-05


x000102030405

y y0 24600 S1

02337 2223204446 1997806333 1772707992 1539509413 12932

Y5(S1)

Interpolando

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983092

6172)()(

)()(

0515

050102 =

minus

minusminus+=

sY sY

sY Bssss

x

000102030405

y y

0 26170 S2 02486 2364904730 2124106735 1882308495 1629809996 13615

Y5(S2)

6182

)()(

)()(

1525

151213 =

minus

minusminus+=

sY sY

sY Bssss

x000102030405

y y0 26180 S3

02487 2365804731 2124906737 1883008498 1630410000 13619

Y5(S3)

Error| B - Y5(S3) | = 0

Ejemplo 9




i

i x

x p2

)( minus=

=)( i xq 2

2

i x

)( i

xr 2

))sin(ln(

i

i

x

x=

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983093

i A =

+ )(

21 i x p

h


====iC

minus )(21 i x p

h

====i D )(2i xr h

minus

minus

=

21542

00710

00670

82830

20247-0888900

1125020313-087500

01142920408-08571

001166720556-

4

3

2

1

x

x

x

x


t(k) y(k)=X1100000000000000 100000000000000120000000000000 118692106203224140000000000000 138127313638853160000000000000 158226407575024180000000000000 178883227403986200000000000000 200000000000000



S1 = 25285 Y5(S1) = 1860371822 =s Y5(S2) = 19998

718623 =

s Y5(S3) = 20002

Error | B - Y5(S3) | = 00002

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983094




=

isin=

10)0(

]100[20

y

t y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983095


+minus=

minus=

2122

2111

ydycy y

ybyay y


Bt y

At y

ydycy y

ybyay y

=

=

+minus=

minus=

)(

)(

02

01

2122

2111


=

+minus

minus

=

B

A

t y

t y

ydycy

ybyay

y

y

)(

)(

02

01

212

211

2

1



4)(

4)(

1050

2050

02

01

2122

2111

=

=

+minus=

minus=

t y

t y

y ycy y

y y y y



7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983096



Se Pide







que|y(120)-yN|lt=0005



Demuestre que


7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf



983120983154983151983142983141983155983151983154983141983155983098 983122983111983114983085983122983107983123983085983117983119983122 983090983097







P9- Resolver






200)2()0( )(10)(

20501

lele==

=+minus +

xe x x

φ φ φ φ

φ

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

7252019 10_Sep_EDO_2009_1pdf










m x k x k x x

m x k x x

1 1 1 1 2 2 1

2 2 2 2 1


Prime = minus minus

( )

( )



E t Ldi

dt i R

i RC

i i dt

( )

( )

= +

= minusint

12

2 1 2

1


1


E

i1

i2

RC

L

i1-i2

022

2

=++ x ydx

dye

dx

yd xy

10_sep_edo_2009_1.pdf

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