218_pdfsam_math 54 differential equation solutions odd

218_pdfsam_math 54 differential equation solutions odd - e...

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Chapter 4 This equation has associated homogeneous equation x 0 x 0 2 x = 0. The roots of the associated auxiliary equation, r 2 r 2 = 0, are r =2and r = 1. Therefore, a general solution to this equation is x h ( t )= c 1 e 2 t + c 2 e t . For the variation of parameters method, we let x p ( t )= v 1 ( t ) x 1 ( t )+ v 2 ( t ) x 2 ( t ) , where x 1 ( t )= e 2 t and x 2 ( t )= e t . Thus, x 0 1 ( t )=2 e 2 t and x 0 2 ( t )= e t . This means that we have to solve the system e 2 t v 0 1 + e t v 0 2 =0 , 2 e 2 t v 0 1 e t v 0 2 = e 3 t . Adding these two equations yields 3 e 2 t v 0 1 = e 3 t v 0 1 = 1 3 e t v 1 ( t )= 1 3 e t . Substututing v 0 1 into the ±rst equation, we get 1 3 e 3 t + e t v 0 2 =0 v 0 2 = 1 3 e 4 t v 2 ( t )= 1 12 e 4 t . Therefore, x p ( t )= 1 3 e t e 2 t 1 12 e 4 t e t = 1 4 e 3 t , and a general solution is x ( t )= c 1
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Unformatted text preview: e 2 t + c 2 e t + 1 4 e 3 t . 5. This equation has associated homogeneous equation y 2 y + y = 0. Its auxiliary equation, r 2 2 r +1 = 0, has a double root r = 1. Thus a general solution to the homogeneous equation is y h ( t ) = c 1 e t + c 2 te t . For the variation of parameters method, we let y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) , where y 1 ( t ) = e t and y 2 ( t ) = te t . 214...
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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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