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

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

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{e ^{- x} x}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{e ^{- x} x}{3}

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{e ^{- x} x}{3}

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

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 25 y = 30 x + 3

y^{^{''}} + 3 y^{^{'}} + 2 y = sin (2 x)

y^{^{''''}} + 18 y^{^{''}} + 81 y = 162 + cos (6 t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 3 x

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