tarea 2 ecuaciones
DESCRIPTION
Ecuaciones HomogeneasTRANSCRIPT
![Page 1: TAREA 2 ECUACIONES](https://reader036.vdocumento.com/reader036/viewer/2022072112/5695d4f01a28ab9b02a36235/html5/thumbnails/1.jpg)
Universidad Autónoma de Baja California
Facultad de Ingeniería
Ecuaciones Diferenciales
TAREA 2
Alumna:
Jauregui Padilla Monica Giselle
Matricula:
01135400
Profesora:
Rivera Castellón Ruth Elba
Grupo:
324
Mexicali Baja California Lunes 22 de septiembre del 2015
![Page 2: TAREA 2 ECUACIONES](https://reader036.vdocumento.com/reader036/viewer/2022072112/5695d4f01a28ab9b02a36235/html5/thumbnails/2.jpg)
Ejercicios sección 2.2
a) En los ejercicios siguientes determine si la función dada es homogénea. Si lo es indique
su grado de homogeneidad.
1.x
yxyx
32 2
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡2𝑥2 + 2𝑡2𝑥𝑦 −(𝑡3)𝑦3
(𝑡)𝑥
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡2(𝑥2 + 2𝑥𝑦 −𝑦3
𝑥)
Homogénea grado 2
2. )64( yxyx
𝑓(𝑡𝑥, 𝑡𝑦) = √𝑡𝑥 + 𝑡𝑦 (4𝑡𝑥 + 6𝑡𝑦)
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡32(√𝑥 + 𝑦)(4𝑥 + 6𝑦))
Homogénea grado 𝟑
𝟐
3. 2)1( yx
𝑥2 + 𝑦2 + 2𝑥𝑦 + 2𝑥 + 2𝑦 + 1
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡2𝑥2 + 𝑡2𝑦2 + 𝑡22𝑥𝑦 + 2𝑡𝑥+ 2𝑡𝑦 + 1
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡(𝑡𝑥2 + 𝑡𝑦2 + 2𝑡𝑥𝑦 + 2𝑥+ 2𝑦) + 1
No es Homogénea
4.yx
x
2
cos
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑐𝑜𝑠𝑡2𝑥2
𝑡𝑥 + 𝑡𝑦
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑐𝑜𝑠𝑡2(𝑥2)
𝑡(𝑥 + 𝑦)
No es Homogénea
5.yx
xsen
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑠𝑒𝑛𝑡𝑥
𝑡𝑥 + 𝑡𝑦
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑠𝑒𝑛𝑡𝑥
𝑡(𝑥 + 𝑦)
Homogénea grado 0
6.yx
yxyx
8
223
𝑓(𝑡𝑥, 𝑡𝑦)
=((𝑡3𝑥3)(𝑡𝑦)) − ((𝑡2𝑥2)(𝑡2𝑦2)
((𝑡)(𝑥)) + (8(𝑡)(𝑦))
𝑓(𝑡𝑥, 𝑡𝑦) =𝑡4((𝑥3𝑦) − (𝑥2𝑦2))
𝑡((𝑥)) + (8𝑦)
Homogénea grado 3
7. 442 yxy
x
𝑓(𝑡𝑥, 𝑡𝑦) =𝑡𝑥
𝑡2𝑦2 + √𝑡4(𝑥4 + 𝑦4)
𝑓(𝑡𝑥, 𝑡𝑦) =𝑡𝑥
𝑡2((𝑦2 + √(𝑥4 + 𝑦4))
Homogénea grado 1
8. yLnxLn 22
𝐼𝑛𝑥2
𝑦2
𝑓(𝑡𝑥, 𝑡𝑦) = 𝐼𝑛𝑡2𝑥2
𝑡2𝑦2
𝑓(𝑡𝑥, 𝑡𝑦) = 𝐼𝑛(𝑡2)𝑥2
(𝑡2)𝑦2
Homogénea grado 0
![Page 3: TAREA 2 ECUACIONES](https://reader036.vdocumento.com/reader036/viewer/2022072112/5695d4f01a28ab9b02a36235/html5/thumbnails/3.jpg)
b) Resuelva las ecuaciones diferenciales siguientes usando la sustitución más
conveniente:
9. 0)( xdydxyx
((𝑥 − 𝑢𝑥)𝑑𝑥 + 𝑥(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 0
𝑥𝑑𝑥 − 𝑢𝑥𝑑𝑥 + 𝑢𝑥𝑑𝑥 + 𝑥2𝑑𝑢 = 0
∫ 𝑥
𝑥2𝑑𝑥 = ∫ 𝑑𝑢
∫ 𝑑𝑥
𝑥= ∫ 𝑑𝑢
Inǀxǀ = 𝑢+c
Sust. u=𝑦
𝑥
Inǀxǀ =𝑦
𝑥 +c
10. 03)( 233 dyxydxyx
(𝑥3 + 𝑢3𝑥3)𝑑𝑥 + 3𝑢2𝑥3(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 0
𝑥3𝑑𝑥 + 𝑢3𝑥3𝑑𝑥 + 3𝑢3𝑥3𝑑𝑥 + 3𝑢2𝑥4𝑑𝑢=0
𝑥3𝑑𝑥 + 4𝑢3𝑥3𝑑𝑥 + 3𝑢2𝑥4𝑑𝑢=0
𝑥3(1 + 4𝑢3)𝑑𝑥 = −3𝑢2𝑥4𝑑𝑢
𝑥3
𝑥4𝑑𝑥 =
−3𝑢2
(4𝑢3 + 1)𝑑𝑢
∫𝑑𝑥
𝑥= ∫
−3𝑢2
(4𝑢3 + 1)𝑑𝑢
∫𝑑𝑥
𝑥= ∫
1
12𝑢2
−3𝑢2
(4𝑢3 + 1)𝑑𝑢
∫𝑑𝑥
𝑥= −
1
4∫
𝑑𝑢
(4𝑢3 + 1)
𝐼𝑛ǀ𝑥ǀ = −1
4𝐼𝑛ǀ4𝑢3 + 1ǀ + 𝑐
Sust. u=𝑦
𝑥
𝐼𝑛ǀ𝑥ǀ = −1
4𝐼𝑛ǀ4
𝑦
𝑥3
3
+ 1ǀ + 𝑐
11. 0)( 22 dyxdxyxy
(𝑢2𝑥2 + 𝑢𝑥2)𝑑𝑥 − 𝑥2(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 0
𝑢2𝑥2𝑑𝑥 + 𝑢𝑥2𝑑𝑥 − 𝑢𝑥2𝑑𝑥 + 𝑥3𝑑𝑢 = 0
𝑢2𝑥2𝑑𝑥 = −𝑥3𝑑𝑢
−∫𝑥2
𝑥3𝑑𝑥 = ∫
𝑑𝑢
𝑢2
−∫𝑑𝑥
𝑥= ∫
𝑑𝑢
𝑢2
−𝐼𝑛ǀ𝑥ǀ = −1
𝑢+ 𝑐
𝐼𝑛ǀ𝑥ǀ =1
𝑢+ 𝑐
Sust. u=𝑦
𝑥
𝐼𝑛ǀ𝑥ǀ =𝑥
𝑦+ 𝑐
12. 02
2
2
dyy
xdx
y
x
2𝑥
𝑦𝑑𝑥 =
𝑥2
𝑦2𝑑𝑦
2𝑢𝑦
𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢) =
𝑢2𝑦2
𝑦2𝑑𝑦
2𝑢2𝑑𝑦 + 2𝑢𝑦𝑑𝑢 =𝑢𝑦2
𝑦2𝑑𝑦
2𝑢𝑦𝑑𝑢 = ( 𝑢2𝑦2
𝑦2− 2𝑢2) 𝑑𝑦
2𝑢𝑦𝑑𝑢 = ( 𝑢2 − 2𝑢2)𝑑𝑦
2𝑢𝑦𝑑𝑢 = −𝑢2𝑑𝑦
−2∫𝑑𝑢
𝑢𝑑𝑢 = ∫
𝑑𝑦
𝑦
−2𝐼𝑛ǀ𝑢ǀ + 𝑐 = 𝐼𝑛ǀ𝑦ǀ
Sust. u=𝑥
𝑦
−2𝐼𝑛ǀ𝑥
𝑦ǀ + 𝑐 = 𝐼𝑛ǀ𝑦ǀ
![Page 4: TAREA 2 ECUACIONES](https://reader036.vdocumento.com/reader036/viewer/2022072112/5695d4f01a28ab9b02a36235/html5/thumbnails/4.jpg)
13. 22 xyydx
dyx
𝑥𝑑𝑦 = (√𝑦2 + 𝑥2 + 𝑦)𝑑𝑥
𝑥(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = (√𝑢2𝑥2 + 𝑥2 + 𝑢𝑥)𝑑𝑥
𝑢𝑥𝑑𝑥 + 𝑥2𝑑𝑢 = (√𝑥2(𝑢2 + 1) + 𝑢𝑥)𝑑𝑥
𝑢𝑥𝑑𝑥 + 𝑥2𝑑𝑢 = (𝑥√𝑢2 + 1)𝑑𝑥 + 𝑢𝑥𝑑𝑥
𝑢𝑥𝑑𝑥 + 𝑥2𝑑𝑢 = 𝑥√𝑢2 + 1𝑑𝑥 + 𝑢𝑥𝑑𝑥
∫𝑑𝑢
√𝑢2 + 1= ∫
𝑑𝑥
𝑥2
𝐼𝑛 ǀ𝑢 + √𝑢2 + 1ǀ + 𝑐 = −1
𝑥
Sust. u=𝑦
𝑥
𝐼𝑛 ǀ𝑦
𝑥+ √
𝑦
𝑥2
2
+ 1ǀ + 𝑐 = −1
𝑥
14. xydydxyex xy )( 2/2
(𝑥2𝑒−𝑢𝑥
𝑥 + 𝑢2𝑥2)𝑑𝑥 = 𝑢𝑥2(𝑢𝑑𝑥 + 𝑥𝑑𝑢)
(𝑥2𝑒−𝑢)𝑑𝑥 + (𝑢2𝑥2)𝑑𝑥 = (𝑢2𝑥2)𝑑𝑥 + (𝑢𝑥3)𝑑𝑢
(𝑥2𝑒−𝑢)𝑑𝑥 = (𝑢𝑥3)𝑑𝑢
𝑥2
𝑥3𝑑𝑥 =
𝑢
𝑒−𝑢𝑑𝑢
∫𝑑𝑥
𝑥= ∫ 𝑢𝑒𝑢𝑑𝑢
𝐼𝑛ǀ𝑥ǀ = 𝑢𝑒𝑢 − ∫ 𝑒𝑢𝑑𝑢
𝐼𝑛ǀ𝑥ǀ = 𝑢𝑒𝑢 − 𝑒𝑢 + 𝑐
𝐼𝑛ǀ𝑥ǀ = 𝑒𝑢(𝑢 − 1) + 𝑐
Sust. u=𝑦
𝑥
𝐼𝑛ǀ𝑥ǀ = 𝑒𝑦𝑥(
𝑦
𝑥− 1) + 𝑐
16. 02)25( ' yyxy
(5𝑦 − 2𝑥)𝑑𝑦
𝑑𝑥− 2𝑦 = 0
(5𝑦 − 2𝑥)𝑑𝑦 = 2𝑦𝑑𝑥 (5𝑦 − 2𝑢𝑦)𝑑𝑦 = 2𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢)
(5𝑦)𝑑𝑦 − (2𝑢𝑦)𝑑𝑦 = (2𝑢𝑦)𝑑𝑦 + (2𝑦2)𝑑𝑢
(5𝑦)𝑑𝑦 − (4𝑢𝑦)𝑑𝑦 = (2𝑦2)𝑑𝑢
𝑦(5 − 4𝑢)𝑑𝑦 = (2𝑦2)𝑑𝑢
∫𝑦
2𝑦2𝑑𝑦 = ∫
𝑑𝑢
(5 − 4𝑢)
∫ 𝑑𝑦
2𝑦= ∫
𝑑𝑢
(5 − 4𝑢)
1
2𝐼𝑛 ǀ𝑦 ǀ = 𝐼𝑛 ǀ 5 − 4𝑢 ǀ + 𝑐
Sust. u=𝑥
𝑦
1
2𝐼𝑛 ǀ𝑦 ǀ = 𝐼𝑛 ǀ 5 − 4
𝑥
𝑦 ǀ + 𝑐
18.dyyxydx )(2
𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢) = 2(𝑢𝑦 + 𝑢)𝑑𝑦
𝑢𝑦𝑑𝑦 + 𝑦2𝑑𝑢 = 2𝑢𝑦𝑑𝑦 + 2𝑢𝑑𝑦
𝑦2𝑑𝑢 = 𝑢𝑦𝑑𝑦 + 2𝑢𝑑𝑦
𝑦2𝑑𝑢 = 𝑢(𝑦 + 2)𝑑𝑦
∫𝑑𝑢
𝑢= ∫
𝑑𝑦
𝑦+ 2∫
𝑑𝑦
𝑦2
𝐼𝑛 𝑢 + 𝑐 = 𝐼𝑛 𝑦 −2
𝑦
Sust. u=𝑥
𝑦
𝐼𝑛 𝑥
𝑦+ 𝑐 = 𝐼𝑛 ǀ𝑦ǀ −
2
𝑦
![Page 5: TAREA 2 ECUACIONES](https://reader036.vdocumento.com/reader036/viewer/2022072112/5695d4f01a28ab9b02a36235/html5/thumbnails/5.jpg)
19. xy
xy
dx
dy
(𝑦 + 𝑥)𝑑𝑦 = (𝑦 − 𝑥)𝑑𝑥
(𝑢𝑥 + 𝑥)(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = (𝑢𝑥 − 𝑥)𝑑𝑥
𝑥2𝑑𝑢 = −(𝑢2𝑥 − 𝑢𝑥 + 𝑥)𝑑𝑥
𝑥2𝑑𝑢 = −𝑥(𝑢2 − 𝑢 + 1)𝑑𝑥 𝑑𝑢
(𝑢2 − 𝑢 + 1)𝑑𝑢 = −
𝑥
𝑥2𝑑𝑥
∫𝑑𝑢
(𝑢2 − 𝑢 + 1)𝑑𝑢 = −∫
𝑑𝑥
𝑥𝑑𝑥
𝐼𝑛ǀ𝑢2 − 𝑢 + 1ǀ + 𝑐 = −𝐼𝑛ǀ𝑥ǀ
Sust. u=𝑦
𝑥
𝐼𝑛ǀ𝑦2
𝑥2−
𝑦
𝑥+ 1ǀ + 𝑐 = −𝐼𝑛ǀ𝑥ǀ
20. yx
yx
dx
dy
3
3
(3𝑥 + 𝑦)𝑑𝑦 = (𝑥 + 3𝑦)𝑑𝑥
(3𝑥 + 𝑢𝑥)(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 𝑥𝑑𝑥 + 3𝑢𝑥𝑑𝑥
(3𝑢𝑥𝑑𝑥 + 3𝑥2𝑑𝑢 + 𝑢2𝑥𝑑𝑥 + 𝑢𝑥2𝑑𝑢)= 𝑥𝑑𝑥 + 3𝑢𝑥𝑑𝑥
(3𝑥2𝑑𝑢 + 𝑢𝑥2𝑑𝑢) = 𝑥𝑑𝑥 − 𝑢2𝑥𝑑𝑥
𝑥2(𝑢 + 3)𝑑𝑢 = −𝑥(𝑢2 − 1)𝑑𝑥 (𝑢 + 3)
(𝑢2 − 1)= −
𝑥
𝑥2𝑑𝑥
∫𝑢
(𝑢2 − 1)𝑑𝑢 + 3∫
𝑑𝑢
(𝑢2 − 1)= −∫
𝑑𝑥
𝑥
1
2𝐼𝑛ǀ𝑢2 − 1ǀ +
3
2𝐼𝑛ǀ
1 + 𝑢
1 − 𝑢ǀ + 𝑐 = − 𝐼𝑛ǀ𝑥ǀ
Sust. u=𝑦
𝑥
1
2𝐼𝑛ǀ
𝑦2
𝑥2− 1ǀ +
3
2𝐼𝑛ǀ
1 +𝑦𝑥
1 −𝑦𝑥
ǀ + 𝑐 = − 𝐼𝑛ǀ𝑥ǀ
c) Resuelva cada ecuación diferencial sujeta a la condición inicial indicada.
21. 332 xydx
dyxy ; 2)1( y
𝑥𝑦2𝑑𝑦 = (𝑦3 − 𝑥3)𝑑𝑥
𝑢2𝑥3(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = (𝑢3𝑥3 − 𝑥3)𝑑𝑥
𝑢3𝑥3𝑑𝑥 + 𝑢3𝑥4𝑑𝑢 = 𝑢3𝑥3𝑑𝑥 − 𝑥3𝑑𝑥
𝑢3𝑥4𝑑𝑢 = −𝑥3𝑑𝑥
𝑢3𝑑𝑢 = −𝑥3
𝑥4𝑑𝑥
∫ 𝑢3𝑑𝑢 = −∫𝑑𝑥
𝑥
𝑢4
4+ 𝑐 = −𝐼𝑛ǀ𝑥ǀ
𝑐 = −𝐼𝑛ǀ𝑥ǀ −𝑢4
4
Sust. u=𝑦
𝑥
𝑐 = −𝐼𝑛ǀ𝑥ǀ −
𝑦4
𝑥4
4
𝑐 = −𝐼𝑛ǀ𝑥ǀ −𝑦4
4𝑥4
𝒚(𝟏) = 𝟐
𝑐 = −(𝐼𝑛ǀ1ǀ +(2)4
4(1)4) 𝒄 = −4
22. dyyxydyxxydx 222 ; 1)0( y
𝑥𝑦𝑑𝑥 = (𝑦√𝑥2 + 𝑦2 + 𝑥2)𝑑𝑦
𝑢𝑦2(𝑢𝑑𝑦 + 𝑦𝑑𝑢) = (𝑦√𝑢2𝑦2 + 𝑦2 + 𝑢2𝑦2)𝑑𝑦
𝑢2𝑦2𝑑𝑦 + 𝑢𝑦3𝑑𝑢 = 𝑦√𝑢2𝑦2 + 𝑦2𝑑𝑦 + 𝑢2𝑦2𝑑𝑦
𝑢𝑦3𝑑𝑢 = 𝑦√𝑢2𝑦2 + 𝑦2𝑑𝑦
𝑢𝑦3𝑑𝑢 = 𝑦√𝑦2(𝑢2 + 1)𝑑𝑦
𝑢𝑦3𝑑𝑢 = 𝑦2√(𝑢2 + 1)𝑑𝑦
𝑢
√(𝑢2 + 1)𝑑𝑢 =
𝑦2
𝑦3𝑑𝑦
∫𝑢
√(𝑢2 + 1)𝑑𝑢 = ∫
𝑑𝑦
𝑦𝑑𝑦
1
2∫ (𝑢2 + 1)−
12𝑑𝑢 = ∫
𝑑𝑦
𝑦𝑑𝑦
√(𝑢2 + 1) + 𝑐 = 𝐼𝑛ǀ𝑦ǀ
𝑐 = 𝐼𝑛ǀ𝑦ǀ − √(𝑢2 + 1)
Sust. u=𝑥
𝑦
𝑐 = 𝐼𝑛ǀ𝑦ǀ − √𝑥2
𝑦2+ 1
𝒚(𝟎) = 𝟏
𝑐 = 𝐼𝑛ǀ1ǀ − √(0)2
(1)2 + 1 𝒄 = −𝟏
![Page 6: TAREA 2 ECUACIONES](https://reader036.vdocumento.com/reader036/viewer/2022072112/5695d4f01a28ab9b02a36235/html5/thumbnails/6.jpg)
23. 0cos
dyx
y
xyydx ; 2)0( y
𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢) + ( 𝑦𝑐𝑜𝑠𝑢𝑦
𝑦− 𝑢𝑦 ) 𝑑𝑦
𝑢𝑦𝑑𝑦 + 𝑦2𝑑𝑢 + (𝑦𝑐𝑜𝑠𝑢)𝑑𝑦 − 𝑢𝑦𝑑𝑦
𝑦2𝑑𝑢 = −𝑦𝑐𝑜𝑠𝑢𝑑𝑦 𝑑𝑢
𝑐𝑜𝑠𝑢= −
𝑦
𝑦2𝑑𝑦
∫ 𝑠𝑒𝑐𝑢𝑑𝑢 = −∫𝑑𝑦
𝑦
𝑠𝑒𝑐𝑢𝑡𝑎𝑛𝑢 + 𝑐 = −𝐼𝑛ǀ𝑦ǀ 𝑐 = −(𝐼𝑛ǀ𝑦ǀ + 𝑠𝑒𝑐𝑢𝑡𝑎𝑛𝑢)
Sust. u=𝑥
𝑦
𝑐 = −(𝐼𝑛ǀ𝑦ǀ + 𝑠𝑒𝑐𝑥
𝑦𝑡𝑎𝑛
𝑥
𝑦)
𝒚(𝟎) = 𝟐
𝑐 = − (𝐼𝑛ǀ2ǀ + 𝑠𝑒𝑐0
2𝑡𝑎𝑛
0
2)
𝒄 = −𝟎. 𝟔𝟗𝟑𝟏
24. ydx
duxyyx )( 2 ; 1)2/1( y
(𝑥 + √𝑦2 − 𝑥𝑦) 𝑑𝑦 = 𝑦𝑑𝑥
(𝑢𝑦 + √𝑦2 − 𝑢𝑦2) 𝑑𝑦 = 𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢)
𝑢𝑦𝑑𝑦 + √𝑦2 − 𝑢𝑦2𝑑𝑦 = 𝑢𝑦𝑑𝑦 + 𝑦2𝑑𝑢
√𝑦2(1 − 𝑢)𝑑𝑦 = 𝑦2𝑑𝑢
𝑦√(1 − 𝑢)𝑑𝑦 = 𝑦2𝑑𝑢
𝑦
𝑦2𝑑𝑦 =
𝑑𝑢
√(1 − 𝑢)
∫𝑑𝑦
𝑦= ∫
𝑑𝑢
√(1 − 𝑢)
𝐼𝑛ǀ𝑦ǀ =1
2√(1 − 𝑢) + 𝑐
𝐼𝑛ǀ𝑦ǀ −1
2√(1 − 𝑢) = 𝑐
Sust. u=𝑥
𝑦
𝐼𝑛ǀ𝑦ǀ −1
2√(1 −
𝑥
𝑦) = 𝑐
𝒚(𝟏𝟐⁄ ) = 𝟏
𝐼𝑛ǀ1ǀ −1
2√(1 −
12⁄
1) = 𝑐
𝒄 = −𝟎. 𝟑𝟓𝟑𝟓
25. 0)( dyxedxyex xy
xy
; 0)1( y
(𝑥 + 𝑦𝑒𝑦
𝑥⁄ ) 𝑑𝑥 = 𝑥𝑒𝑦
𝑥⁄ 𝑑𝑦
(𝑥 + 𝑢𝑥𝑒𝑢𝑥
𝑥⁄ )𝑑𝑥 = 𝑥𝑒𝑢𝑥
𝑥⁄ (𝑢𝑑𝑥 + 𝑥𝑑𝑢)
𝑥𝑑𝑥 + 𝑢𝑥𝑒𝑢𝑑𝑥 = 𝑥𝑢𝑒𝑢𝑑𝑥 + 𝑥2𝑒𝑢𝑑𝑢
𝑥𝑑𝑥 = 𝑥2𝑒𝑢𝑑𝑢 𝑥
𝑥2𝑑𝑥 = 𝑒𝑢𝑑𝑢
∫ 𝑑𝑥
𝑥= ∫ 𝑒𝑢𝑑𝑢
𝐼𝑛ǀ𝑥ǀ = 𝑒𝑢 + 𝑐
𝐼𝑛ǀ𝑥ǀ − 𝑒𝑢 = 𝑐
Sust. u=𝑦
𝑥
𝐼𝑛ǀ𝑥ǀ − 𝑒𝑦
𝑥⁄ = 𝑐
𝒚(𝟏) = 𝟎
𝐼𝑛ǀ1ǀ − 𝑒0
1⁄ = 𝑐 𝒄 = −𝟏