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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{x}{x ^{3}} d x

\int (x^{3} + e^{x}) d x

n = 1 \sum \infty n^{0.5}

\frac{2 x}{x + 1}

\int_{- 1}^{1} x^{2} (1 - x^{2}) d x

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