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

The solution for linear 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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linear y^{''}+4y^'+4y=t-2e-2t

linear

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}

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 11 e^{- 3 x}

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{- 2 x}

y \cdot u^{2} - y^{^{''}} = 1

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} e - 2 x + 1

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