267_pdfsam_math 54 differential equation solutions odd

# 267_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.2 Substituting these expressions into the nonhomogeneous equation given in (5.3) yields 2 x 0 p 2 x p =2 ( A sin t + B cos t ) 2( A cos t + B sin t ) =( 2 B 2 A )cos t +( 2 A 2 B )sin t = cos t. By equating coeﬃcients we obtain 2 B 2 A = 1a n d 2 A 2 B =0 . By solving these two equations simultaneously for A and B ,weseethat A = 1 4 and B = 1 4 . Thus, a particular solution to the nonhomogeneous equation given in (5.3) will be x p ( t )= 1 4 cos t 1 4 sin t and a general solution to the nonhomogeneous equation (5.3) will be x ( t x h ( t )+ x p ( t C 1 e t + 1 4 cos t 1 4 sin t. We now must Fnd a function y ( t ). To do this, we subtract the second of the two di±erential
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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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