354_pdfsam_math 54 differential equation solutions odd

354_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 The roots of the auxiliary equation, r 2 +4 r + 5 = 0, for this second order equation are r = 2 ± i . Therefore, { w 1 ( x ) ,w 2 ( x ) } = ± e 2 x cos x, e 2 x sin x ² form a fundamental solution set. Integrating, we get v 1 ( x )= Z w 1 ( x Z e 2 x cos xdx = e 2 x (sin x 2cos x ) 5 , v 2 ( x Z w 2 ( x Z e 2 x sin = e 2 x (2 sin x +cos x ) 5 , where we have chosen integration constants to be zero. Thus, functions f ( x e 2 x , y 1 ( x v 1 ( x ) f ( x e 2 x (sin x x ) 5 e 2 x = sin x x 5 , y 2 ( x v 2 ( x ) f ( x e 2 x (2 sin x x ) 5 e 2 x = 2sin x x 5 are three linearly independent solutions to the given diFerential equation. 35. ±irst, let us evaluate the Wronskian of the system { x, sin x, cos x } to make sure that the result of Problem 34 can be applied. W [ x, sin x, cos x ]= ³ ³ ³ ³ ³ ³ ³ ³ x sin x cos x 1c o s x sin x 0 sin x
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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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