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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = 3 x^{3}

\int_{0}^{8} - x^{2} + 2 x + 4 d x

f (x) = \frac{4 x}{x + 1}, at x = 3

x \to - 4 lim (\frac{∣ x + 4 ∣}{x + 4})

y = (e^{4 x} - 3)^{8}

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