# hw6 - 3.7.7. Consider y 00 + 4 y + 4 y = t- 2 e- 2 t , t...

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Unformatted text preview: 3.7.7. Consider y 00 + 4 y + 4 y = t- 2 e- 2 t , t > Characteristic equation is 0 = r 2 + 4 r + 4 = ( r + 2) 2 , so y 1 = e- 2 t , y 2 = te- 2 t . Thus, W { y 1 ,y 2 } ( t ) = e- 2 t te- 2 t- 2 e- 2 t (1- 2 t ) e- 2 t = e- 4 t . Thus, y ( t ) =- e- 2 t Z te- 2 t t- 2 e- 2 t e- 4 t dt + te- 2 t Z e- 2 t t- 2 e- 2 t e- 4 t dt =- e- 2 t Z dt t + te- 2 t Z dt t 2 =- e- 2 t (ln | t | + c 1- 1 + tc 2 ) = e- 2 t (- ln t + c 3 + tc 2 ) . 3.7.13. Consider t 2 y 00- 2 y = 3 t 2- 1, t > 0 with y 1 ( t ) = t 2 , y 2 ( t ) = t- 1 . Plugging in y 1 , we see t 2 2- 2 t 2 = 0, so y 1 is a solution of the homogeneous problem. Likewise, if we plug in y 2 , we see t 2 2 t 3- 2 t = 0, so y 2 also solves the homogeneous problem. Now, W { y 1 ,y 2 } ( t ) = t 2 t- 1 2 t- 1 /t 2 =- 1- 2 =- 3 . Before we can use formula (3.7.28), we must rewrite the equation in the appropriate form, namely y 00- 2 t 2 y = 3- 1 t 2 ....
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## This note was uploaded on 07/17/2008 for the course MATH 415 taught by Professor Costin during the Fall '07 term at Ohio State.

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hw6 - 3.7.7. Consider y 00 + 4 y + 4 y = t- 2 e- 2 t , t...

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