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 5 y + 6 y = 0, which has the associated auxiliary equation r 2 5 r + 6 = 0. The roots of this equation are r = 3 and r = 2. Therefore, neither y = e x nor y = xe x satisfies 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 p ( x ) = ( Ax + B + A ) e x y p ( x ) = ( Ax + B + 2 A ) e x ( Ax + B + 2 A ) e x 5( Ax + B + A ) e x + 6( Ax + B ) e x = xe x (2 Ax 3 A + 2 B ) e x = xe x 2 A = 1 , 3 A + 2 B = 0 A = 1 / 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 find θ p ( t ) = ( A 1 t + A 0 ) cos t + ( B 1 t + B 0 ) sin t θ p ( t ) = A 1 cos t ( A 1 t + A 0 ) sin t + B 1 sin t + ( B 1 t + B 0 ) cos t = ( B 1
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