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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x - 1, 0

x \to 0 lim (\frac{2}{3 + 4 ^{\frac{1}{0}}})

y = \frac{16 x}{x ^{2} + 16}, (- 2, - \frac{8}{5})

y^{^{''}} - 9 y = 0, y (0) = 7

\int \frac{3}{( x + 3 ) ^{2}} d x

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