What is the solution for y^{''}+4y^'+4y=sin(x) ?

The solution for y^{''}+4y^'+4y=sin(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+3/25 sin(x)-4/25 cos(x)

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y^{''}+4y^'+4y=sin(x)

y^{^{''}} + 4 y^{^{'}} + 4 y = sin (x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = sin (x)

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

y^{^{''}} + y^{^{'}} = x^{2} e^{3 x}

y^{^{'''}} - 4 y^{^{'}} = e^{x}

y^{^{''}} + y^{^{'}} + 2 y - x - 3 = 0

y^{^{''''}} + 2 y^{^{''}} + y = 9 + cos (3 t)

y^{^{''}} - y = 8 e^{9 t} + t e^{9 t} + 5 t^{2} e^{9 t}

What is the solution for y^{''}+4y^'+4y=sin(x) ?
The solution for y^{''}+4y^'+4y=sin(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+3/25 sin(x)-4/25 cos(x)