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247_pdfsam_math 54 differential equation solutions odd

# 247_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.9 Substitution into the original equation yields m ( 2 cos γt 2 sin γt ) + k ( A cos γt + B sin γt ) = F 0 cos γt A ( 2 + k ) cos γt + B ( 2 + k ) sin γt = F 0 cos γt A = F 0 / ( k 2 ) , B = 0 y p ( t ) = F 0 k 2 cos γt. Therefore, a general solution to the original equation is y ( t ) = C 1 cos ωt + C 2 sin ωt + F 0 k 2 cos γt . With the initial conditions, y (0) = y (0) = 0, we get y (0) = C 1 + F 0 / ( k 2 ) = 0 , y (0) = ωC 2 = 0 C 1 = F 0 / ( k 2 ) , C 2 = 0 . Therefore, y ( t ) = F 0 k 2 cos ωt + F 0 k 2 cos γt , which can also be written in the form y ( t ) = F 0 k 2 (cos γt cos ωt ) = F 0 m ( ω 2 γ 2 ) (cos γt cos ωt ) . (b) Here one can apply the “difference-to-product” identity
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