What is the solution for y^{''}+y^'=cos(3t) ?

The solution for y^{''}+y^'=cos(3t) is y=c_{1}+c_{2}e^{-t}+1/30 sin(3t)-1/10 cos(3t)

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y^{''}+y^'=cos(3t)

y^{^{''}} + y^{^{'}} = cos (3 t)

y = c_{1} + c_{2} e^{- t} + \frac{1}{30} sin (3 t) - \frac{1}{10} cos (3 t)

Solution steps

y^{''} (t) + y^{'} (t) = cos (3 t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + \frac{1}{30} sin (3 t) - \frac{1}{10} cos (3 t)

y = c_{1} + c_{2} e^{- t} + \frac{1}{30} sin (3 t) - \frac{1}{10} cos (3 t)

y^{^{''}} + 2 y^{^{'}} - 35 y = x^{2}

y^{^{''}} + 16 y = cos (3 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 6 y = - 14 t + 12 t^{2}

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e \frac{t}{2}

(D^{2} - 2 D + 1) y = x e^{x} sin (x)

What is the solution for y^{''}+y^'=cos(3t) ?
The solution for y^{''}+y^'=cos(3t) is y=c_{1}+c_{2}e^{-t}+1/30 sin(3t)-1/10 cos(3t)