solpep1 - SOLUCION PRIMERA PRUEBA ECUACIONES DIFERENCIALES...

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SOLUCION PRIMERA PRUEBA ECUACIONES DIFERENCIALES Ingenier´ ıa Civil C´ odigo 10008-11039-19003 Primer Semestre 2006 Problema 1 . Para y = 0 considere la ecuaci´ on diferencial 2 y dy dx = 5 + 2 x - 2 y 2 . a) Encuentre la soluci´ on general impl´ ıcita usando factores integrantes. b) Encuentre la soluci´ on particular que pasa por el punto (0 , 5) y el inter- valo m´ aximo donde est´ a definida. Soluci´ on . a) La ecuaci´ on se puede poner de la forma M ( x, y ) dx + N ( x, y ) dy = 0 , con M ( x, y ) = 5 + 2 x - 2 y 2 y N ( x, y ) = - 2 y . Entonces ∂M ∂y - ∂N ∂x ( x, y ) = - 4 y y la ecuaci´ on no es exacta, pero 1 N ∂M ∂y - ∂N ∂x ( x, y ) = 1 - 2 y ( - 4 y ) = 2 , depende solo de x . Luego μ ( x, y ) = e 2 dx = e 2 x , es un factor integrante y la ecuaci´ on (5 + 2 x - 2 y 2 ) e 2 x dx - 2 y e 2 x dy = 0 es exacta. Buscamos entonces una funci´ on potencial de la forma u ( x, y ) = - 2 y e 2 x dy = - e 2 x y 2 + h ( x ) . Calculamos h ( x ) de la igualdad ∂u ∂x ( x, y ) = (5 + 2 x - 2 y 2 ) e 2 x . Por lo tanto tenemos - 2 e 2 x y 2 + h ( x ) = (5 + 2 x - 2 y 2 ) e 2 x que implica h ( x ) = (5 + 2 x ) e 2 x . Luego h ( x ) = (2 + x ) e 2 x , y u ( x, y ) = - e 2 x y 2 + (2 + x ) e 2 x = (2 + x - y 2 ) e 2 x .
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