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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 2 lim (\frac{1}{x + 2})

\int sin (x) - cos (x) + 4 d x

\int (e^{x} sin (y) + tan (y)) d x

\frac{\partial}{\partial y} (y ln (y) + y e^{x})

\frac{d}{d l} (\frac{d l ^{2} - d - a l ^{3} - b l ^{2}}{l ^{2}})

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