Certamen 1 - Matemáticas III (2006).pdf - UNIVERSIDAD TÉCNICA FEDERICO SANTA MARÍA DEPARTAMENTO DE MATEMÁTICAS SOLUCIÓN DEL CERTAMEN N◦ 1

Certamen 1 - Matemáticas III (2006).pdf - UNIVERSIDAD...

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UNIVERSIDAD TÉCNICA FEDERICO SANTA MARÍA DEPARTAMENTO DE MATEMÁTICAS SOLUCIÓN DEL CERTAMEN N 1 MATEMÁTICA-023 7 de Abril de 2006 1. Sea u ( x ) una solución de la ecuación diferencial y 00 ( x ) - 4 y 0 ( x )+4 y ( x ) = 0 , y sea v ( x ) una solución de la ecuación diferencial y 00 ( x ) - 4 y 0 ( x )+29 y ( x ) = 0 . Determine u ( x ) y v ( x ) , si se sabe que ambas curvas se cortan en el origen y tienen la misma pendiente en el origen y además v 0 ( π ) = - 1 . Solución: Como se sabe que la gráfica de ambas funciones, u ( x ) y v ( x ) pasan por el origen, entonces, u (0) = v (0) = 0 . Además tienen la misma derivada en el origen, entonces u 0 (0) = v 0 (0) = m . Por lo tanto se deben resolver los problemas de valores iniciales u 00 ( x ) - 4 u 0 ( x ) + 4 u ( x ) = 0 u (0) = 0 v 0 (0) = m v 00 ( x ) - 4 v 0 ( x ) + 29 v ( x ) = 0 v (0) = 0 v 0 (0) = m La ecuación característica de la primera ecuación es λ 2 - 4 λ + 4 = ( λ - 2) 2 = 0 , cuyas raíces son λ 1 = λ 2 = 2 . Por lo tanto, la solución general de esta ecuación es u ( x ) = c 1 e 2 x + c 2 x e 2 x . Ahora, sustituyendo las condiciones iniciales podemos determinar el valor de las constantes c 1 y c 2 . u (0) = c 1 = 0 u ( x ) = c 2 x e 2 x u 0 ( x ) = c 2 ( e 2 x + 2 x e 2 x ) u 0 (0) = c 2 = m u ( x ) = mx e 2 x La ecuación característica de la segunda ecuación es λ 2 - 4 λ + 29 = λ 2 - 4 λ + 4 + 25 = ( λ - 2) 2 + 5 2 = 0 , cuyas
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