338_pdfsam_math 54 differential equation solutions odd

338_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 Substituting x ( t ) into the frst equation oF the system (5.41) yields y 0 y = C 1 sin t + C 2 cos t. (5.42) The general solution to the corresponding homogeneous equation, y 0 y =0 ,is y h ( t )= C 3 e t . We look For a particular solution to (5.42) oF the Form y p ( t C 4 cos t + C 5 sin t . Di±erentiating, we obtain y 0 p ( t C 4 sin t + C 5 cos t . Thus the equation (5.42) becomes C 1 sin t + C 2 cos t = y 0 p y =( C 4 sin t + C 5 cos t ) ( C 4 cos t + C 5 sin t ) C 5 C 4 )cos t ( C 5 + C 4 )sin t. Equating the coefficients yields C 5 C 4 = C 2 , C 5 + C 4 = C 1 (5.43) (by adding the equations) 2 C 5 = C 1 + C 2 C 5 = C 1 + C 2 2 . ²rom the second equation in (5.43), we fnd C 4 = C 1 C 5 = C 1 C 2 2 . ThereFore, the general solution to the equation (5.42) is
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