Download - Benalcazar. Villacres.deber6.1
-
Universidad de las Fuerzas ArmadasEcuaciones Diferenciales Ordinarias
Deber 6: Ecuaciones no exactas - primer y segundo caso
C. Benalcazar y T. Villacres
Para las siguientes ecuaciones diferenciales, identifique el caso de factor integrante y
encuentre una solucion general.
1. (3x + 2y2)dx + 2xydy = 0
My
= 4y ; Nx
= 2yPrimer Caso:My N
x
N
f(x) =4y 2y
2xy=
1
xu(x) = elnx = x
Aplicamos u(x)(3x2 + 2xy2)dx + 2x2ydy = 0
My
= 4yx ; Nx
= 4yx
F (x, y)/F1 =
(3x2 + 2xy2)dx = x3 + x2y2 y F2 =
2x2ydy = x2y2
F (x, y) = x3 + x2y2 = C
3. ydx + (3 + 3x y)dy = 0My
= 1 ; Nx
= 3Segundo Caso:
Nx M
y
M
g(y) =3 1y
=2
yu(y) = eln(y
2) = y2
Aplicamos u(y)y3dx + (3y2 + 3xy2 y3)dy = 0
My
= 3y2 ; Nx
= 3y2
F (x, y)/F1 =
y3dx = xy3 y F2 =
(3y2 + 3xy2 y3)dy = y3 + xy3 y
4
4
F (x, y) = y3 + xy3 y4
4= C
1
-
4. (3x2 + y + 3x3y)dx + xdy = 0
My
= 1 + 3x3 ; Nx
= 1Primer Caso:My N
x
N
f(x) =1 + 3x3 1
x= 3x2
u(x) = ex3
Aplicamos u(x)ex
3(3x2 + y + 3x3y)dx + xex
3dy = 0
My
= ex3
+ 3x3ex3
; Nx
= ex3
+ 3x3ex3
F (x, y)/F1 =
(3x2ex
3+ yex
3+ 3x3yex
3)dx
= ex3
+ yex
3dx + 3y
x3ex
3
si u = x3; x = u
1
3 ; dx =du
3u23
= ex3
+y
3
euduu
23
+ y ueudu
u23
= ex3
+y
3
euduu
23
+ yu
1
3 eudu
= yu
1
3 eu y euduu
13
+ yu
1
3 eu
F1 = ex3 + xyex
3y
F2 =xex
3dy = xyex
3
F (x, y) = ex3(xy + 1) = C
5. ydx (x3 3x)dy = 0My
= 1 ; Nx
= 3Segundo Caso:
Nx M
y
M
g(y) =3 1y
=2
yu(y) = eln(y
2) = y2
Aplicamos u(y)y3dx (y5 3y2x)dy = 0
My
= 3y2 ; Nx
= 3y2
F (x, y)/F1 =
y3dx = y3x y F2 =
(y5 + 3y2x) = y3x y
6
6
F (x, y) = y3x y6
6= C
6. (x2y + y3)dy xdx = 0
2
-
My
= 0 ; Nx
= 0Segundo Caso:
Nx M
y
M
g(y) =2xy
x = 2yu(y) = ey
2= x
Aplicamos u(y)(xey2)dx + ey2(x2y + y3)dy = 0
My
= 2yxey2
; Nx
= 2yxey2
F (x, y)/F1 =
(xey
2)dx =
xey22
y
u = y2 ; du = 2ydy ; dv =yey
2; v =
ey22
F2 =ey
2(x2y + y3)dy
= x2ey
2
2 y
2ey2
+
ydyey2
=ey2
2(x2 + y2 + 1) F (x, y) = e
y2
2(x2 + y2 + 1) = C
7. (x + x3sen(2y))dy 2ydx = 0My
= 2 ; Nx
= 1 + 3x2sen(2y)Primer Caso:My N
x
N
f(x) =2 1 3x2sen(2y)
x + x3sen(2y)=3(1 + x2sen(2y))x(1 + x2sen(2y)
=3x
u(x) = e3lnx =1
x3Aplicamos u(x)
2yx3
dx + (1
x2+ sen(2y))dy = 0
My
=2x3
; Nx
=2x3 F (x, y)/
F1 = 2yx3dx =
y
x2y F2 =
(
1
x2+ sen(2y))dy =
y
x2 cos(2y)
2
F (x, y) = yx2 cos(2y)
2= C
8. (y2cos(x) y)dx + (x + y2)dy = 0My
= 2ycos(x) 1 ; Nx
= 1Segundo Caso:
Nx M
y
M
g(y) =1 + 1 2ycos(x)y(ycos(x) 1) =
2(ycos(x) 1)y(ycos(x) 1) =
2y
u(y) = e2lny =1
y2
3
-
Aplicamos u(y)
(cos(x) 1y
)dx + (x
y2+ 1)dy = 0
My
=1
y2; Nx
=1
y2
F (x, y)/F1 =
(cos(x) 1
y)dx = sen(x) x
yy F2 =
2x2ydy = sen(x) x
y+ y
F (x, y) = sen(x) xy
+ y = C
12. Una ecuacion diferencial puede tener mas de un factor integrante. Pruebe que u1 =1
xy;u2 =
1
y2;u3 =
1
x2 + y2son factores integrantes de la ecuacion diferencial ydx + xdy = 0.
Demuestre que las soluciones son formalmente equivalentes.
Si aplicamos u1
dx
x dy
y= 0
My
= 0 ; Nx
= 0
F (x, y)/Fa1 =
dxx
= ln|x| y Fb1 = dy
= ln|y|
F1(x, y) =x
y= C
Si aplicamos u2
dx
ydx xdy
y2dy = 0
My
=1y2
; Nx
=1y2
F (x, y)/Fa2 =
dxydx =
x
yy Fb2 =
xdyy2
dy =x
y
F2(x, y) =x
y= C
Si aplicamos u3
ydx
x2 + y2 xdyx2 + y2
= 0
My
=x2 y2x2 + y2
; Nx
=x2 y2x2 + y2
F (x, y)/
Fa3 = ydxx2 + y2
= ln(x2 + y2)
1
2 y Fb3 = xdyx2 + y2
= ln(x2 + y2)
12
F3(x, y) = ln(1) = 0 = CF1(x, y) = F2(x, y) = F3(x, y)
4