hw8solns - Section 4.5 1 y y = e-x x2 homogeneous solution...

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Section 4.5 1. y + y = e - x + x 2 ; homogeneous solution r 2 + 1 = 0 r = ± i ; fundamental set is { cos x, sin x } . Particular UC sets: { e x } , { x 2 , x, 1 } , so no modification is necessary. y p = Ae - x + Bx 2 + Cx + E 3. y + y = 4 sin x + e x cos x ; homogeneous solution r 2 + 1 = 0; linearly independent solutions { cos x, sin x } , y c = C 1 cos x + C 2 sin x . Particular solution: 4 sin x { sin x, cos x } ⇒ modify as { x cos x, x sin x } e x cos x { e x cos x, e x sin x } ⇒ no modification necessary Hence y p = A 1 x cos x + A 2 x sin x + B 1 e x cos x + B 2 e x sin x 6. y - y = e x + xe - x ; homogeneous solution r 2 - 1 = 0 r = ± 1; fundamental set is { e x , e - x } . Particular solution: UC sets are { e x } , { xe - x , e - x } , both need modification, to { xe x } , { x 2 e - x , xe - x } . Thus y p = Axe x + Bx 2 e - x + Cxe - x 9. y + 4 y + 13 y = e - 2 x . We have y p = Ae - 2 x y p = - 2 Ae - 2 x , y p = 4 Ae - 2 x y p + 4 y p + 13 yp = (4 Ae - 2 x ) + 4( - 2 Ae - 2 x ) + 13( Ae - 2 x ) = e - 2 x 4 A - 8 A + 13 A = 1 9 A = 1 A = 1 9 11. y - 4 y = cos x + 3 sin x , y p = A cos x + B sin x . Substitution gives ( - A - 4 B ) cos x + ( - B + 4 A ) sin x = cos x + 3 sin x Equating coefficients of like terms, - A - 4 B = 1 - B + 4
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