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

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

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y^{''}+4y^'+4y=e^{-x}cos(x)

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- x} cos (x)

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = e^{- x} cos (x)

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 4 y = e^{x}

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 10

y^{^{''''}} + 49 y^{^{''}} = sin (7 t) + t e^{8 t} + 4

y^{^{''}} = y^{^{'}} + x

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{3 x}

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