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

The solution for y^{''}-y^'-2y=(x+1)e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-x}-(e^{-x}x^2)/6-(4e^{-x}x)/9

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y^{''}-y^'-2y=(x+1)e^{-x}

y^{^{''}} - y^{^{'}} - 2 y = (x + 1) e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{e ^{- x} x ^{2}}{6} - \frac{4 e ^{- x} x}{9}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = (x + 1) e^{- x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{e ^{- x} x ^{2}}{6} - \frac{4 e ^{- x} x}{9}

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{e ^{- x} x ^{2}}{6} - \frac{4 e ^{- x} x}{9}

y^{^{''}} - 2 y^{^{'}} - 3 y = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3 x}}{( x ^{2} + 1 )}

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

y^{^{''}} + y = sin (t) + cos (t)

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

What is the solution for y^{''}-y^'-2y=(x+1)e^{-x} ?
The solution for y^{''}-y^'-2y=(x+1)e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-x}-(e^{-x}x^2)/6-(4e^{-x}x)/9