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

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

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{3}{8} e^{- 2 x} + \frac{1}{9} e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{3}{8} e^{- 2 x} + \frac{1}{9} e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} - \frac{3}{8} e^{- 2 x} + \frac{1}{9} e^{- x}

y^{^{''}} - a^{2} y = sec (a x)

y^{^{''}} - 4 y^{^{'}} - 5 y = x^{2}

y^{^{''}} + 2 y^{^{'}} + 2 y = t^{2}, y (0) = \frac{1}{2}, y^{^{'}} (0) = 0

y^{^{'''}} + 265 y^{^{'}} = tan (16 x)

y^{^{''}} - 10 y^{^{'}} + 9 y = e^{2 x}

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