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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{1 + sin ( x )}{1 - sin ( x )})

sin (x - 5)

n = 1 \sum \infty \frac{3}{n ln ( 9 n )}

f (x) = (x + (x + sin^{2} (x))^{3})^{4}

\int x^{2} cos (21 x) d x

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