What is the solution for y^{''}+4y^'+4y=3e^{-2t} ?

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

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y^{''}+4y^'+4y=3e^{-2t}

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- 2 t}

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 3 e^{- 2 t}

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 12 y = 3 e^{5 t}

y^{^{''}} + 2 y^{^{'}} + y = 7 e^{- t}

y^{^{''''}} + 4 y = 2 e^{3 x}

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} + 4 y = 2 t

- y^{^{''}} + 2 y^{^{'}} + 3 y = 1, y (0) = 0, y (1) = 0

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