What is the solution for y^{''}+2y^'-8y=xe^{-x}+e^{-x} ?

The solution for y^{''}+2y^'-8y=xe^{-x}+e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-4x}+(-e^{-x}x-e^{-x})/9

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y^{''}+2y^'-8y=xe^{-x}+e^{-x}

y^{^{''}} + 2 y^{^{'}} - 8 y = x e^{- x} + e^{- x}

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

Solution steps

y^{''} (x) + 2 y^{'} (x) - 8 y = x e^{- x} + e^{- x}

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (- u (x, y) cos (k x) sin (k y))

\frac{d}{d x} (\frac{1}{2} sec^{2} (x))

\frac{d}{d x} (x^{\frac{2}{3}} - a^{\frac{2}{3}})

\int_{0}^{1} x^{3} 1 - x^{4} d x

y (X) = \frac{2 X + 5}{X}

What is the solution for y^{''}+2y^'-8y=xe^{-x}+e^{-x} ?
The solution for y^{''}+2y^'-8y=xe^{-x}+e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-4x}+(-e^{-x}x-e^{-x})/9