208_pdfsam_math 54 differential equation solutions odd

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Chapter 4 31. For the nonhomogeneous term sin t + t cos t , a particular solution has the form y p, 1 ( t )=( A 1 t + A 0 ) t s cos t +( B 1 t + B 0 ) t s sin t. For 10 t = e t ln 10 , a particular solution should be of the form y p, 2 ( t )= Ct p e t ln 10 = Ct p 10 t . Since the roots of the auxiliary equation, r 2 +1=0,are r = ± i ,wechoose s =1and p =0 . Thus, by the superposition principle, y p ( t )= y p, 1 ( t )+ y p, 2 ( t )=( A 1 t + A 0 ) t cos t +( B 1 t + B 0 ) t sin t + C · 10 t . 33. The roots of the auxiliary equation, which is r 2 r 2 = 0, are r = 1, 2. The right-hand side of the given equation is a sum of two terms, e t cos t and t 2 + t + 1. Corresponding particular solutions have the forms y p, 1 ( t )=( A cos t + B sin t ) t s e t and y p, 2 ( t )=( C 2 t 2 + C 1 t + C 0 ) t p , and we can take s = p = 0 since neither r =1+ i nor r
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Unformatted text preview: By the superposition principle, y p ( t ) = ( A cos t + B sin t ) e t + C 2 t 2 + C 1 t + C . 35. Since the roots of the auxiliary equation are r = 4 ± √ 16 − 20 2 = 2 ± i, which are di±erent from 5 and 3 i , a particular solution has the form y p ( t ) = ( A 1 t + A ) cos 3 t + ( B 1 t + B ) sin 3 t + Ce 5 t . (The last term corresponds to e 5 t in the right-hand side of the original equation, and the ²rst two come from t sin 3 t − cos 3 t .) 204...
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