318_pdfsam_math 54 differential equation solutions odd

# 318_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 x 0 ( t )=( c 1 + c 2 ) e t c 2 te t c 3 sin t + c 4 cos t x 0 ( t )=( c 1 2 c 2 ) e t + c 2 te t c 3 cos t c 4 sin t. To fnd y ( t ), note that by the frst equation oF the system given in (5.34), we have y 0 ( t )= x 0 ( t )+ x 0 ( t )+ x ( t ) . Substituting x ( t ), x 0 ( t ), and x 0 ( t ) into this equation yields y 0 ( t )=( c 1 2 c 2 ) e t + c 2 te t c 3 cos t c 4 sin t +( c 1 + c 2 ) e t c 2 te t c 3 sin t + c 4 cos t + c 1 e t + c 2 te t + c 3 cos t + c 4 sin t y 0 ( t )=( c 1 c 2 ) e t + c 2 te t c 3 sin t + c 4 cos t. By integrating both sides oF this equation with respect to t ,weobta in y ( t )= ( c 1 c 2 ) e t c 2 te t c 2 e t + c 3 cos t + c 4 sin t + c 5 , wherewehaveintegrated c 2 te t by parts. SimpliFying yields y ( t )= c 1 e t c 2 te t + c 3 cos t + c 4 sin t + c 5 . To determine the fve constants, we will use the Four initial conditions and the second equation in system (5.34). (We used the frst equation to determine y ). Substituting into the second
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Unformatted text preview: − ± ( − c 1 + c 2 ) e − t − c 2 te − t − c 3 sin t + c 4 cos t ² + ± ( − c 1 + 2 c 2 ) e − t − c 2 te − t − c 3 cos t − c 4 sin t ² + ± ( c 1 − c 2 ) e − t + c 2 te − t − c 3 sin t + c 4 cos t ² + ± − c 1 e − t − c 2 te − t + c 3 cos t + c 4 sin t + c 5 ² = 0 , which reduces to c 5 = 0. Using the initial conditions and the Fact that c 5 = 0, we see that x (0) = c 1 + c 3 = − 2 , y (0) = − c 1 + c 3 = 0 , x (0) = ( − c 1 + c 2 ) + c 4 = 0 , y (0) = ( c 1 − c 2 ) + c 4 = 0 . 314...
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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 Berkeley.

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