What is the solution for y^{''}-2y^'-3y=e^{-t} ?

The solution for y^{''}-2y^'-3y=e^{-t} is y=c_{1}e^{3t}+c_{2}e^{-t}-(e^{-t}t)/4

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y^{''}-2y^'-3y=e^{-t}

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{- t}

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = e^{- t}

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{e ^{2 x} - e ^{x} + 1}{e ^{x}})

\int_{0}^{1} 11 x x^{2} + 36 d x

x y^{^{'}} + 2 y^{^{'}} - (x + 2) sin (x) + y = 0

\frac{\partial}{\partial x} ((x^{2} y - y^{3})^{5})

- \frac{e ^{(- 2 s)}}{( s ^{2} - 4 s + 4 )}

What is the solution for y^{''}-2y^'-3y=e^{-t} ?
The solution for y^{''}-2y^'-3y=e^{-t} is y=c_{1}e^{3t}+c_{2}e^{-t}-(e^{-t}t)/4