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364_pdfsam_math 54 differential equation solutions odd

# 364_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 (c) As we have mentioned in (a), the first equation in (6.7) implies that y = ( x + 5 x ) / 2. Substituting the solution x ( t ) yields y ( t ) = 1 2 c 1 cos t + c 2 sin t + c 3 cos 6 t + c 4 sin 6 t +5 c 1 cos t + c 2 sin t + c 3 cos 6 t + c 4 sin 6 t = 1 2 c 1 cos t c 2 sin t 6 c 3 cos 6 t 6 c 4 sin 6 t +5 c 1 cos t + c 2 sin t + c 3 cos 6 t + c 4 sin 6 t = 2 c 1 cos t + 2 c 2 sin t c 3 2 cos 6 t c 4 2 sin 6 t . (d) Initial conditions x (0) = y (0) = 1 and x (0) = y (0) = 0 imply the system of linear equations for c 1 , c 2 , c 3 , and c 4 . Namely, x (0) = c 1 + c 3 = 1 , y (0) = 2 c 1 ( c 3 / 2) = 1 , x (0) = c 2 + c 4 6 = 0 , y (0) = 2 c 2 ( c 4 6 / 2) = 0 c 1 = 3 / 5 , c 3 = 2 / 5 , c 2 = 0 , c 4 = 0 . Thus, the solution to this initial value problem is x ( t ) = 3 5 cos t + 2 5 cos 6 t , y ( t ) = 6 5 cos t 1 5 cos 6 t . 35.
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