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

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

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

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

y = c_{1} + c_{2} e^{- 2 x} - 2 e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} - 2 e^{- x}

y = c_{1} + c_{2} e^{- 2 x} - 2 e^{- x}

y^{^{''}} + y^{^{'}} - 2 y = 13 + e^{x}

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 4 t^{2} - 1

y^{^{''}} + 11 y^{^{'}} + 24 y = 4 sin (4 t) + 2 e^{- 8 t}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 2 y = 2 e^{x} cos (x)

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