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379_pdfsam_math 54 differential equation solutions odd

379_pdfsam_math 54 differential equation solutions odd -...

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Exercises 6.4 The auxiliary equation, r 2 ( r 2 2) = 0, has roots r = ± 2 and a double root r = 0. Hence, y h ( t ) = c 1 + c 2 t + c 3 e 2 t + c 4 e 2 t is a general solution to the homogeneous equation coresponding to (6.22). A particular solution to (6.22) has the form y p ( t ) = Ae 3 t . Substitution yields D 2 ( D 2 2 ) Ae 3 x = ( D 4 2 D 2 ) Ae 3 x = 81 Ae 3 x (2)9 Ae 3 x = 63 Ae 3 x = 8 e 3 x y p ( t ) = Ae 3 x = 8 e 3 x 63 , and so y ( t ) = y p ( t ) + y h ( t ) = 8 e 3 x 63 + c 1 + c 2 t + c 3 e 2 t + c 4 e 2 t is a general solution to (6.22). We find x ( t ) from the second equation in the original system. x ( t ) = e 3 t + y ( t ) y ( t ) = e 3 t + 8 e 3 x 63 + c 1 + c 2 t + c 3 e 2 t + c 4 e 2 t 72 e 3 x 63 + 2 c 3 e 2 t + 2 c 4 e 2 t = e 3 x 63 + c 1 + c 2 t c 3 e 2 t c 4 e 2 t . EXERCISES 6.4: Method of Variation of Parameters, page 341 1. To apply the method of variation of parameters, first we have to find a fundamental solution set for the corresponding homogeneous equation, which is
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