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}+(3x)/8-(20e-3)/(32)

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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 x}{8} - \frac{20 e - 3}{32}

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 x}{8} - \frac{20 e - 3}{32}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} + \frac{3 x}{8} - \frac{20 e - 3}{32}

\frac{d}{d x} (\frac{a x}{b + c})

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\frac{\partial}{\partial y} (tan (x^{3} y^{2}))

\frac{\partial}{\partial x} (cos (x^{2} - y))

\int \frac{2 x}{x - 1} 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}+(3x)/8-(20e-3)/(32)