What is the solution for y^{''}+4y^'+4y=3t ?

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

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y^{''}+4y^'+4y=3t

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 t

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 3 t

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = 24 e^{3 x}

y^{^{''}} - y^{^{'}} - 2 y = 8 sin (t) - 2 cos (t)

y^{^{''}} + 6 y^{^{'}} + 13 y = 13 x^{2}

2 y^{^{''}} - y^{^{'}} - y = t + 1, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 4 y = sin (x) - cos (x)

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