What is the solution for y^{''}+2y^'+2y=e^{-x} ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) + f^{^{'}} (x) + f^{^{''}} (x) = x^{5} + 64

y^{^{''''}} + 9 y^{^{''}} = sin (3 t) + t e^{4 t} + 4

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 4 x^{2}

y^{^{'''}} - 4 y^{^{'}} = x + 3 cos (x)

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

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