379_pdfsam_math 54 differential equation solutions odd

379_pdfsam_math 54 - Exercises 6.4 The auxiliary equation r 2(r 2 2 = 0 has roots r = 2 and a double root r = 0 Hence yh(t = c1 c2 t c3 e 2t c4 e

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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 ² = ( D 4 2 D 2 Ae 3 x ² =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 Fnd x ( t ) from the second equation in the original system. x ( t e 3 t + y ( t ) y 0 ( t ) = e 3 t + ³ 8 e 3 x 63 + c 1 + c 2 t + c 3 e 2 t + c 4 e 2 t ´ ³ 72 e 3 x 63 +2 c 3 e 2 t 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, Frst we have to Fnd a fundamental solution set for the corresponding homogeneous equation, which is y 0 3 y 0 +4 y
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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