What is the solution for y^{''}+10y^'+25y=t^{-2}e^{-5t} ?

The solution for y^{''}+10y^'+25y=t^{-2}e^{-5t} is y=c_{1}e^{-5t}+c_{2}te^{-5t}-e^{-5t}ln(t)-e^{-5t}

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y^{''}+10y^'+25y=t^{-2}e^{-5t}

y^{^{''}} + 10 y^{^{'}} + 25 y = t^{- 2} e^{- 5 t}

y = c_{1} e^{- 5 t} + c_{2} t e^{- 5 t} - e^{- 5 t} ln (t) - e^{- 5 t}

Solution steps

y^{''} (t) + 10 y^{'} (t) + 25 y = t^{- 2} e^{- 5 t}

Solve linear ODE:

y = c_{1} e^{- 5 t} + c_{2} t e^{- 5 t} - e^{- 5 t} ln (t) - e^{- 5 t}

y = c_{1} e^{- 5 t} + c_{2} t e^{- 5 t} - e^{- 5 t} ln (t) - e^{- 5 t}

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

\frac{d}{d x} (x (4 x - 5)^{3})

\int_{0}^{2} 4 x e^{x} d x

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

n = 1 \sum \infty 4 \cdot (\frac{4}{7})^{n}

What is the solution for y^{''}+10y^'+25y=t^{-2}e^{-5t} ?
The solution for y^{''}+10y^'+25y=t^{-2}e^{-5t} is y=c_{1}e^{-5t}+c_{2}te^{-5t}-e^{-5t}ln(t)-e^{-5t}