What is the solution for y^{''}+9y^'=cos(3t) ?

The solution for y^{''}+9y^'=cos(3t) is y=c_{1}+c_{2}e^{-9t}+1/30 sin(3t)-1/90 cos(3t)

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y^{''}+9y^'=cos(3t)

y^{^{''}} + 9 y^{^{'}} = cos (3 t)

y = c_{1} + c_{2} e^{- 9 t} + \frac{1}{30} sin (3 t) - \frac{1}{90} cos (3 t)

Solution steps

y^{''} (t) + 9 y^{'} (t) = cos (3 t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 9 t} + \frac{1}{30} sin (3 t) - \frac{1}{90} cos (3 t)

y = c_{1} + c_{2} e^{- 9 t} + \frac{1}{30} sin (3 t) - \frac{1}{90} cos (3 t)

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

\frac{d ^{2} y}{d x ^{2}} - y = 1

y^{^{''}} + y^{^{'}} + 1.25 y = 3 cos (x)

y^{^{''}} = 9 + (y^{^{'}}) 2

y^{^{''}} + 6 y^{^{'}} - 27 y = - 105 t + 54 t^{2}

What is the solution for y^{''}+9y^'=cos(3t) ?
The solution for y^{''}+9y^'=cos(3t) is y=c_{1}+c_{2}e^{-9t}+1/30 sin(3t)-1/90 cos(3t)