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Examen resuelto paso a paso de ecuaciones diferenciales de segundo orden
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Examen de Ecuaciones de Segundo OrdenCapitaine Martı́nez Jesús Adán
November 17, 2020
Encontrar la solución general de las siguientes ecuaciones diferenciales:
1. x2 y 00 − 2xy 0 + 2y = xe−x
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 = x 2
y2 = x
Teniendo ası́ como solución general
yg = c1 x2 + c2 x
1
, Por otro lado, tomamos y1 = x2 y y2 = x
Se calculan sus derivadas
y10 = 2x
y20 = 1
Calculando el wronskiano
w(y1 , y2 ) = y1 y20 − y10 y2
⇒ w = −x2
Calculando v1
−y2 R(x)
Z
v1 = dx
w
−x2 e−x
Z
v1 = dx
−x2
v1 = −e−x
Calculando v2 Z
y1 R(x)
v2 = dx
w
x3 e−x
Z
v2 = dx
−x2
v2 = (x + 1)e−x
Teniendo la solución particular
yp = v1 y1 + v2 y2
yp = −x2 e−x + x(x + 1)e−x
Teniendo como solución general
y = c1 x2 + c2 x − x2 e−x + x(x + 1)e−x
⇒ y = c1 x2 + c2 x + xe−x
2
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