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Lista de ejercicios resueltos de Ecuaciones Diferenciales de segundo orden $9.55
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Lista de ejercicios resueltos de Ecuaciones Diferenciales de segundo orden

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Lista de 28 ejercicios resueltos de ecuaciones diferenciales de segundo orden explicados paso a paso.

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  • April 1, 2021
  • 35
  • 2020/2021
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Tarea 3 de Ecuaciones Diferenciales
Capitaine Martı́nez Jesús Adán

November 4, 2020


Encuentra la solución general de

1. x2 y 00 − 2xy 0 + 2y = 0 y una particular para y(1) = 3, y 0 (1) = 5

Se toma y = xn y se sustituye en la ecuación

x2 (n(n − 1)xn−2 ) − 2x(nxn−1 ) + 2xn = 0

⇒ n(n − 1)xn − 2nxn + 2xn = 0
⇒ n2 − 3n + 2 = 0
Factorizando
⇒ (n − 2)(n − 1) = 0
Entonces, se tiene como solución

n1 = 2

n2 = 1
Luego, obtenemos como soluciones particulares

y1 = x2

y2 = x
Teniendo ası́ como solución general

yg = c1 x2 + c2 x




1

, Ahora usando las soluciones particulares dadas con y(1) = 1 y y 0 (1) = 5

y(1) = c1 (1)2 + c2 (1) = 3

⇒ c1 + c2 = 3
Teniendo otra ecuación

y 0 (1) = 2c1 x + c2 = 5

⇒ 2c1 + c2 = 5
Resolviendo el sistema de ecuaciones obtenido, tenemos como soluciones
c1 = 2 y c2 = 1

La solución particular obtenida es

y = 2x2 + x




2

, 2. y 00 − 3y 0 + 2y = 0 y una particular para y(0) = −1, y 0 (0) = 1

Se toma y = eλx , con derivadas y 0 = λeλx y y 00 = λ2 eλx .

Se sustituye en la ecuación

λ2 (eλx ) − 3λ(eλx ) + 2eλx = 0

Eliminando eλx
⇒ λ2 − 3λ + 2 = 0
Factorizando
(λ − 2)(λ − 1) = 0
Teniendo ası́
λ1 = 2
λ2 = 1
Ahora, se tiene dos soluciones particulares

y1 = e2x

y2 = e x
Teniendo como solución general

yg = c1 e2x + c2 ex

Ahora para calcular la solución particular y(0) = −1 y y 0 (0) = 1 Se obtiene
el siguiente sistema de ecuaciones

c1 e2(0) + c2 e0 = −1

⇒ c1 + c2 = −1
2c1 e2(0) + c2 e0 = 1
⇒ 2c1 + c2 = 1
Resolviendo, se tiene c1 = 2 y c2 = −3

Teniendo ası́
y = 2e2x − 3ex



3

, 3. y 00 − 4y 0 + 4y = 0

Tomamos y = eλx
Se calculan las derivadas y = λeλx y y 00 = λ2 eλx .

Se sustituyen en la ecuación

λ2 eλx − 4λeλx + 4eλx

⇒ λ2 − 4λ + 4 = 0
⇒ (λ − 2)(λ − 2) = 0
Teniendo como valor de λ
λ1 = 2
λ2 = 2
Obteniendo ası́ la solución general

y = c1 e2x + c2 e2x x




4

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