What is the solution for y^{''}+2y^'-8y=5e^{-2x}-e^{-x} ?

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

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y^{''}+2y^'-8y=5e^{-2x}-e^{-x}

y^{^{''}} + 2 y^{^{'}} - 8 y = 5 e^{- 2 x} - e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{5}{8} e^{- 2 x} + \frac{1}{9} e^{- x}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 8 y = 5 e^{- 2 x} - e^{- x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{5}{8} e^{- 2 x} + \frac{1}{9} e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{5}{8} e^{- 2 x} + \frac{1}{9} e^{- x}

y^{^{''}} - y^{^{'}} - 2 y = 5 x + 3

6 y^{^{''}} + 6 y^{^{'}} + 73 y = 8 cos (4 x)

\frac{d ^{2} y}{d t ^{2}} + 5 y = 5 sin (t)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 6 y = - e^{x} + 12 x

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 cos (x) - 4 sin (x)

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