196_pdfsam_math 54 differential equation solutions odd

# 196_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 Equating the coeﬃcients, we see that 2 A 1 =0 A 1 =0 , 2 A 0 +2 B 1 =0 B 1 = A 0 , 2 B 1 =1 B 1 = 1 2 and so A 0 = 1 2 , 2 A 1 2 B 0 =0 B 0 =0 . Therefore, a particular solution of the nonhomogeneous equation θ 0 θ = t sin t is given by θ p ( t )= t sin t +cos t 2 . 17. The right-hand side of the original equation suggests that a particular solution should be of the form y p ( t )= At s e t .S in c e r = 1 is a double root of the corresponding auxiliary equation, r 2 2 r +1=( r 1) 2 =0,wetake s = 2. Hence y p ( t )= At 2 e t y 0 p ( t )= A ( t 2 +2 t ) e t y 0 p ( t )= A ( t 2 +4 t +2 ) e t . Substituting these expressions into the original equation, we Fnd the constant A . A ( t 2 +4 t +2 ) e t 2 A ( t 2 +2 t ) e t + At 2 e t =8 e t 2 Ae t =8 e t A =4 . Thus, y p ( t )=4 t 2 e t
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