What is the solution for y^{''}-y^'-2y=5e^{-t} ?

The solution for y^{''}-y^'-2y=5e^{-t} is y=c_{1}e^{2t}+c_{2}e^{-t}-(5e^{-t}t)/3

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y^{''}-y^'-2y=5e^{-t}

y^{^{''}} - y^{^{'}} - 2 y = 5 e^{- t}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{5 e ^{- t} t}{3}

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = 5 e^{- t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{5 e ^{- t} t}{3}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{5 e ^{- t} t}{3}

\frac{d}{d x} (1 - x ln (x))

e^{2} t^{2}

sin (x) + x cos (x)

\frac{4}{x}, at x = \frac{1}{9}

What is the solution for y^{''}-y^'-2y=5e^{-t} ?
The solution for y^{''}-y^'-2y=5e^{-t} is y=c_{1}e^{2t}+c_{2}e^{-t}-(5e^{-t}t)/3