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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1}{3 + ( x - 2 ) ^{2}} d x

\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}}

\int_{0}^{1} ln (1 + x^{2}) d x

1 + x, 1

\int \frac{1}{4} cos^{2} (2 x) d x

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