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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} - 6 y = x e^{- 5 x}

y^{^{''}} + 7 y^{^{'}} + 12 y = t^{2} e^{- 3 t}

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

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

(d^{4} + d^{2}) y = 18 x - 4 sin (x)

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