195_pdfsam_math 54 differential equation solutions odd

195_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.4 15. For this problem, the corresponding homogeneous equation is y 0 5 y 0 +6 y = 0, which has the associated auxiliary equation r 2 5 r + 6 = 0. The roots of this equation are r =3and r = 2. Therefore, neither y = e x nor y = xe x satis±es the homogeneous equation, and in the expression y p ( x )= x s ( Ax + B ) e x for a particular solution we can take s =0. So y p ( x )=( Ax + B ) e x y 0 p ( x Ax + B + A ) e x y 0 p ( x Ax + B +2 A ) e x ( Ax + B A ) e x 5( Ax + B + A ) e x +6( Ax + B ) e x = xe x (2 Ax 3 A B ) e x = xe x 2 A =1 , 3 A B =0 A / 2 , B =3 / 4 , and y p ( x x/ 2+3 / 4) e x . 16. The corresponding homogeneous equation has the auxiliary equation r 2 1 = 0, whose roots are r = ± 1. Thus, in the expression θ p ( t A 1 t + A 0 )cos t +( B 1 t + B 0 )sin t none of the terms is a solution to the homogeneous equation. We ±nd θ p ( t A 1 t + A 0 t B 1 t + B 0 t θ 0 p ( t A 1 cos t ( A 1 t + A 0 t + B 1 sin t B 1 t + B 0 t =( B 1 t + A 1 + B 0 t A 1 t A 0 + B 1 t θ 0 p ( t B 1 cos t ( B 1 t + B 0 + A 1 t A 1 sin t
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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