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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 8 y = 1 - 2 x^{2}

y^{^{''}} + 64 y = csc (8 x)

y^{^{''}} + y = t, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = 8 sec (3 x)

y^{^{''}} + 16 y = 3 e^{8 x}

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