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}

x^{'} (t) = \frac{1}{3 x ^{2} + 1}, x (0) = 1

\frac{\partial}{\partial x} (e^{- 6 t} cos (π x))

\frac{\partial}{\partial y} (9 z e^{x yz})

\int_{0}^{1} (e^{- 0.1 t}) d t

\int_{0}^{π} 2 sin (n x) d x

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