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

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

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

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

y = c_{1} e^{8 x} + c_{2} - \frac{2}{5} e^{- 2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{8 x} + c_{2} - \frac{2}{5} e^{- 2 x}

y = c_{1} e^{8 x} + c_{2} - \frac{2}{5} e^{- 2 x}

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

y^{^{''}} + 36 y = 6 tan (6 t) - 2 e^{3 t}

\frac{1}{2} y^{^{''}} + 10 y^{^{'}} + 100 y = 150

\frac{1}{6} y^{^{''}} + 6 y = 4 tan (6 t) - \frac{1}{2} e^{4 t}

y^{^{''}} + 2 y^{^{'}} + 10 y = e^{2 x} cos (3 x)

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