What is the solution for y^{''}+10y^'+25y=3e^{-5t} ?

The solution for y^{''}+10y^'+25y=3e^{-5t} is y=c_{1}e^{-5t}+c_{2}te^{-5t}+(3e^{-5t}t^2)/2

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y^{''}+10y^'+25y=3e^{-5t}

y^{^{''}} + 10 y^{^{'}} + 25 y = 3 e^{- 5 t}

y = c_{1} e^{- 5 t} + c_{2} t e^{- 5 t} + \frac{3 e ^{- 5 t} t ^{2}}{2}

Solution steps

y^{''} (t) + 10 y^{'} (t) + 25 y = 3 e^{- 5 t}

Solve linear ODE:

y = c_{1} e^{- 5 t} + c_{2} t e^{- 5 t} + \frac{3 e ^{- 5 t} t ^{2}}{2}

y = c_{1} e^{- 5 t} + c_{2} t e^{- 5 t} + \frac{3 e ^{- 5 t} t ^{2}}{2}

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What is the solution for y^{''}+10y^'+25y=3e^{-5t} ?
The solution for y^{''}+10y^'+25y=3e^{-5t} is y=c_{1}e^{-5t}+c_{2}te^{-5t}+(3e^{-5t}t^2)/2