What is the solution for y^{''}-3y^'-4y=2e^{-t} ?

The solution for y^{''}-3y^'-4y=2e^{-t} is y=c_{1}e^{4t}+c_{2}e^{-t}-(2e^{-t}t)/5

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y^{''}-3y^'-4y=2e^{-t}

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 e^{- t}

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{2 e ^{- t} t}{5}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 2 e^{- t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{2 e ^{- t} t}{5}

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{2 e ^{- t} t}{5}

y^{^{''}} + 8 y^{^{'}} + 2 y = cos (x), y (0) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x + 6

y^{^{''}} + y^{^{'}} - 6 y = x e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + y = sin (x)

What is the solution for y^{''}-3y^'-4y=2e^{-t} ?
The solution for y^{''}-3y^'-4y=2e^{-t} is y=c_{1}e^{4t}+c_{2}e^{-t}-(2e^{-t}t)/5