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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (tan (x^{2} - 1))

y^{^{''}} - y^{^{'}} - 20 y = 0

\frac{d ^{2}}{d x ^{2}} (3 x - 4 x^{\frac{9}{10}})

\frac{d y}{d x}, e^{y} = x + y

0.4 x - 1.295, - 1.4206, - 1.486

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