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

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

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y^{''}+4y^'+4=e^{-t}

y^{^{''}} + 4 y^{^{'}} + 4 = e^{- t}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 3

(D^{2} + 4) y = sinh (x) sin (2 x)

y^{^{''}} + 4 y^{^{'}} - 12 y = 3 x^{2} + x

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

y^{^{'''}} + 3 y^{^{''}} - 10 y^{^{'}} = x - 3

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