What is the solution for y^{''}+2y^'-8y=2e^{-3x}-e^{-x} ?

The solution for y^{''}+2y^'-8y=2e^{-3x}-e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-4x}-2/5 e^{-3x}+1/9 e^{-x}

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y^{''}+2y^'-8y=2e^{-3x}-e^{-x}

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 e^{- 3 x} - e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{2}{5} e^{- 3 x} + \frac{1}{9} e^{- x}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 8 y = 2 e^{- 3 x} - e^{- x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{2}{5} e^{- 3 x} + \frac{1}{9} e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{2}{5} e^{- 3 x} + \frac{1}{9} e^{- x}

n = 0 \sum \infty (x - 2)^{n}

\frac{\partial}{\partial y} (e^{- x} sin (y) z)

\frac{d}{d a} (4 sin (3 a) cos (3 a))

\frac{d}{d x} (ln (x) d x)

x \to 0 + lim (\frac{8}{x ^{2} - x})

What is the solution for y^{''}+2y^'-8y=2e^{-3x}-e^{-x} ?
The solution for y^{''}+2y^'-8y=2e^{-3x}-e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-4x}-2/5 e^{-3x}+1/9 e^{-x}