What is the solution for y^{''}+2y^'-8y=6e^{-3x}-e^{-x} ?

The solution for y^{''}+2y^'-8y=6e^{-3x}-e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-4x}-6/5 e^{-3x}+1/9 e^{-x}

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y^{''}+2y^'-8y=6e^{-3x}-e^{-x}

y^{^{''}} + 2 y^{^{'}} - 8 y = 6 e^{- 3 x} - e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{6}{5} e^{- 3 x} + \frac{1}{9} e^{- x}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 8 y = 6 e^{- 3 x} - e^{- x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{6}{5} e^{- 3 x} + \frac{1}{9} e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{6}{5} e^{- 3 x} + \frac{1}{9} e^{- x}

y^{^{''}} + y = \frac{1}{cos ( t )}

x^{^{''}} + 7 x = t e^{3 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 3 x^{2} + 4 e^{4 x}

y^{^{''}} + 6 y = - 294 x^{2} e^{6} x

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

What is the solution for y^{''}+2y^'-8y=6e^{-3x}-e^{-x} ?
The solution for y^{''}+2y^'-8y=6e^{-3x}-e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-4x}-6/5 e^{-3x}+1/9 e^{-x}