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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = x 3 - 2 x^{2}

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

g (x) = \frac{2 x + 2}{x ^{2} + 2}

n = 0 \sum \infty n

(x, y) \to (- 6, 5) lim (\frac{x ^{2} cos ( π y )}{e ^{x + y}})

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