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199_pdfsam_math 54 differential equation solutions odd

# 199_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.4 To choose s , we find the roots of the characteristic equation, which is r 2 + 9 = 0. Since r = ± 3 i are its simple roots, we take s = 1. Thus y p ( t ) = t ( A 3 t 3 + A 2 t 2 + A 1 t + A 0 ) cos 3 t + t ( B 3 t 3 + B 2 t 2 + B 1 t + B 0 ) sin 3 t. 29. The characteristic equation r 2 6 r + 9 = ( r 3) 2 = 0 has a double root r = 3. Therefore, a particular solution is of the form y p ( t ) = t 2 ( A 6 t 6 + A 5 t 5 + A 4 t 4 + A 3 t 3 + A 2 t 2 + A 1 t + A 0 ) e 3 t . 31. From the form of the right-hand side, we conclude that a particular solution should be of the form y p ( t ) = t s ( A 3 t 3 + A 2 t 2 + A 1 t + A 0 ) cos t + ( B 3 t 3 + B 2 t 2 + B 1 t + B 0 ) sin t e t . Since r = 1 ± i are simple roots of the characteristic equation, r 2 + 2 r + 2 = 0, we should take s = 1. Therefore, y p ( t ) = t ( A 3 t 3 + A 2 t 2 + A 1 t + A 0 ) cos t + ( B 3 t 3 + B 2 t 2 + B 1 t + B 0 ) sin t e t . 33. The right-hand side of the equation suggests that y p ( t ) = t s ( A cos t + B sin t ). By inspection, we see that r =
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