Pauta Certamen 2.pdf - Universidad T´ ecnica Federico Santa Mar´ıa Departamento de Matem´atica Certamen 2 MAT 023 1 El objetivo de esta pregunta es

Pauta Certamen 2.pdf - Universidad T´ ecnica Federico...

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Universidad T´ ecnica Federico Santa Mar´ ıa Departamento de Matem´ atica Certamen 2 MAT 023 1. El objetivo de esta pregunta es resolver la ecuaci´ on diferencial a 2 y ′′ ( a 2 + 1) y + y = x, siendo a R una constante, a negationslash = 0. (a) Resuelva la ecuaci´ on diferencial homog´ enea asociada. Note que debe estudiar los casos a 2 = 1 y a 2 negationslash = 1. (b) Hallar una soluci´on particular para la ecuaci´ on no homog´ enea. (c) Hallar la soluci´on general de la ecuaci´ on. Soluci´ on: (a) La ecuaci´ on homog´ enea asociada: a 2 y ′′ ( a 2 + 1) y + y = 0 es una ecuaci´ on de coeficientes constantes, cuya ecuaci´ on caracter´ ıstica asociada es a 2 k 2 ( a 2 + 1) k + 1 = 0 . La soluci´on de esta ecuaci´ on de segundo grado es k = ( a 2 + 1) ± radicalbig ( a 2 + 1) 2 4 a 2 2 a 2 = ( a 2 + 1) ± radicalbig ( a 2 1) 2 2 a 2 . Se tiene entonces: (i) Si a 2 = 1, es decir si a = 1 o a = 1, entonces la ecuaci´ on cuadr´atica tiene una soluci´on real de multiplicidad dos, k = 1 (en ambos casos). Por lo tanto, la soluci´on general de la ecuaci´ on homog´ enea es: y ( x ) = C 1 e x + C 2 xe x . (ii) Si a 2 negationslash = 1, es decir a negationslash = 1 y a negationslash = 1, entonces las soluciones de la ecuaci´ on caracter´ ıstica son k 1 = ( a 2 + 1) + ( a 2 1) 2 a 2 = 2 a 2 2 a 2 = 1 , k 2 = ( a 2 + 1) ( a 2 1) 2 a 2 = 2 2 a 2 = 1 a 2 y la soluci´on general de la ecuaci´ on homog´ enea en este caso es: y ( x ) = C 1 e x + C 2 e x/a 2 .
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  • Spring '16
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