What is the solution for y^{''}-y^'-2y=8e^{-t} ?

The solution for y^{''}-y^'-2y=8e^{-t} is y=c_{1}e^{2t}+c_{2}e^{-t}-(8e^{-t}t)/3

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y^{''}-y^'-2y=8e^{-t}

y^{^{''}} - y^{^{'}} - 2 y = 8 e^{- t}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{8 e ^{- t} t}{3}

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = 8 e^{- t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{8 e ^{- t} t}{3}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{8 e ^{- t} t}{3}

\frac{d}{d x} ((- 2)^{x})

4 x^{- 1}

\frac{\partial}{\partial y} (x^{2} - 4 y^{2})

(0, - 2), (- \frac{1}{2}, 3)

n = 5 \sum \infty \frac{1}{n ( n + 1 )}

What is the solution for y^{''}-y^'-2y=8e^{-t} ?
The solution for y^{''}-y^'-2y=8e^{-t} is y=c_{1}e^{2t}+c_{2}e^{-t}-(8e^{-t}t)/3