Certamen 2 - Matemáticas III (2005-3).pdf - Universidad T´ ecnica Federico Santa Mar´ıa Departamento de Matem´ atica Campus Santiago Certamen 2

Certamen 2 - Matemáticas III (2005-3).pdf - Universidad...

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Universidad T´ ecnica Federico Santa Mar´ ıa Departamento de Matem´ atica Campus Santiago Certamen 2 - Mat-023 05 de Enero del 2006 Nombre: 1. Si y = - 4 y 3 y y (2) = 4 ; y (2) = 0 . Hallar y (4) . Ayuda: Haga el cambio z = y .
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2. Resolver y (6) + 2 y (4) + y (2) = 0 Soluci´ on: Ecuaci´ on caracteristica: m 6 + 2 m 4 + m 2 = 0 m 2 ( m 2 + 1) 2 = 0 Raices de la ecuaci´ on caracteristica: m = 0 de multiplicidad 2 y m = ± i tambien de multiplicidad 2. Soluci´ on general: y = a + bx + c cos( x ) + d sen( x ) + ex cos( x ) + fx sen( x )
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3. (a) Muestre que la ecuaci´ on y F ( xy ) dx + x G ( xy ) dy = 0 puede resolverse usando el cambio u = xy . (b) Resuelva: ( x 2 y 3 + 2 xy 2 + y ) dx + ( x 3 y 2 - 2 x 2 y + x ) dy = 0 . Soluci´ on: (a) Observar que: y F ( xy ) dx + x G ( xy ) dy = 0 x dy dx = - y F ( xy ) G ( xy ) Por otra parte, haciendo u = xy se tiene: du dx = y + x dy dx x dy dx = - y + du dx . Reemplazando: - y + du dx = - y F ( u ) G ( u ) - u x + du dx = - u x · F ( u ) G ( u ) du dx = u x 1 - F ( u ) G ( u ) Y esta ´ultima ecuaci´ on es de variable separada. (b) La ecuaci´ on se puede reescribir como y ( x 2 y 2 + 2 xy + 1) dx + x ( x 2 y 2 - 2 xy + 1) dy Tomando F ( xy ) =
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