What is the solution for x^{''}+4x^'+4x=0 ?

The solution for x^{''}+4x^'+4x=0 is x=c_{1}e^{-2t}+c_{2}te^{-2t}

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x^{''}+4x^'+4x=0

x^{^{''}} + 4 x^{^{'}} + 4 x = 0

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

Solution steps

x^{''} (t) + 4 x^{'} (t) + 4 x = 0

Solve linear ODE:

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

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

\frac{60 t + 180}{t + 6}

\int_{2}^{3} \frac{x}{x ^{2} + 1} d x

\frac{d}{d x} (x ln (y))

\int - \frac{4}{3}

\frac{\partial}{\partial x} (2 x y + y)

What is the solution for x^{''}+4x^'+4x=0 ?
The solution for x^{''}+4x^'+4x=0 is x=c_{1}e^{-2t}+c_{2}te^{-2t}