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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 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 = 1 and 2 A 2 B = 0 . By solving these two equations simultaneously for A and B , we see that 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 find a function y ( t ). To do this, we subtract the second of the two differential
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