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hw5sol

# hw5sol - Homework 5 solutions page 186 In these 8 problems...

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Homework 5 solutions page 186 In these 8 problems, the roots associated to the forcing term are indicated in the answer in those cases where the answer was yes. 1) No , the t - 1 is not a polynomial 2) Yes , the roots ± 4 i are repeated 4 times (polynomial of degree 3) 3) Yes , 3 t = e ln 3 t , so the root is ln 3 and it is not repeated 4) Yes , sin( x ) /e 4 x = e - 4 x sin x , so the roots are - 4 ± i , and they are not repeated. 5) No , the secant cannot be written in the required form. 6) Yes , 4 x sin 2 x + 4 x cos 2 x = 4 x = 4 xe 0 x , so the root is 0 and it is repeated twice (the polynomial is degree 1) 7) No , the diﬀerential equation does not have constant coeﬃcients 8) Yes , The roots are 4 ± 25 i , and since the polynomial has degree 100, the roots are repeated 101 times. 12) 2 x 0 + x = 3 t 2 , The right hand side corresponds to the root zero repeated 3 times (the polynomial is of degree 2). The left hand side has one root, r = - 1 / 2. There is no overlap, so we have x p = C 1 + C 2 t + C 3 t 2 . Plug this in to the equation and we get 2 x 0 p + x p = 2( C 2 + 2 C 3 t ) + ( C 1 + C 2 t + C 3 t 2 ) = (2 C 2 + C 1 ) + (4 C 3 + C 2 ) t + C 3 t 2 = 3 t 2 , leading to 2 C 2 + C 1 = 0 , 4 C 3 + C 2 = 0 , C 3 = 3 . Solving we get C 3 = 3 , C 2 = - 12 , C 1 = 24 so x p = 24 - 12 t + 3 t 2 . Of course we have x h = C 4 e - t/ 2 , so x = x h + x p = C 4 e - t/ 2 + 8 - 4 t + t 2 . 16) θ 00 - θ = ( D 2 - 1) θ = t sin t . The roots associated to the forcing term t sin t are ± i and they are repeated twice since the polynomial t has degree 1. The left hand side has roots ± 1. Thus θ = C 1 e t + C 2 e - t + ( C 3 + C 4 t ) sin t + ( C 5 + C 6 t ) cos t , and since θ h = C 1 e t + C 2 e - t , we should take θ p = ( C 3 + C 4 t ) sin t + ( C 5 + C 6 t ) cos t . Plugging in we have θ 0 p = ( C 4 - C 5 - C 6 t ) sin t + ( C 3 + C 4 t + C 6 ) cos t and θ 00 p = ( - C 3 - C 4 t - 2 C 6 ) sin t + (2 C 4 - C 5 - C 6 t ) cos t so θ 00 p - θ p = ( - 2 C 3 - 2 C 4 t - 2 C 6 ) sin t + (2 C 4 - 2 C 5 - 2 C 6 t ) cos t = t sin t. Thus we must solve - 2 C 4 = 1 , - 2 C 3 - 2 C 6 = 0 , C 6 = 0 , and 2 C 4 - 2 C 5 = 0 . That leads to C 3 = 0 , C 4 = - 1 2 , C 5 = - 1 2 , and C 6 = 0

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so that θ p = - 1 2 ( t sin t + cos t ) . We also know that θ = θ h + θ p = C 1 e t + C 2 e - t - 1 2 ( t sin t + cos t ) . 27) The roots from the forcing are ± 3 i repeated 4 times, from the left hand side we also have ± 3 i . Thus the total multiplicity of ± 3 i is 5 – corresponding to a polynomial of de- gree 4. y h = C 1 cos 3 t + C 2 sin 3 t , and we do not include terms already in y h in y p , so y p = ( C 3 t + C 4 t 2 + C 5 t 3 + C 6 t 4 ) cos 3 t + ( C 7 t + C 8 t 2 + C 9 t 3 + C 10 t 4 ) sin 3 t . 28) The root corresponding to the forcing term 1 repeated 5 times. The roots from the left
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hw5sol - Homework 5 solutions page 186 In these 8 problems...

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