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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (cos (x) - 5 tan (x))

\int x^{2} - 6 y d y

\frac{\partial}{\partial x} (2^{x + 3 y})

\frac{12}{x ^{2}}, - 24 x + 36, 0

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

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