What is the solution for y^{''}+6y^'+9y=t^2*e^{-3t} ?

The solution for y^{''}+6y^'+9y=t^2*e^{-3t} is y=c_{1}e^{-3t}+c_{2}te^{-3t}+(e^{-3t}t^4)/(12)

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y^{''}+6y^'+9y=t^2*e^{-3t}

y^{^{''}} + 6 y^{^{'}} + 9 y = t^{2} \cdot e^{- 3 t}

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{e ^{- 3 t} t ^{4}}{12}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 9 y = t^{2} \cdot e^{- 3 t}

Solve linear ODE:

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{e ^{- 3 t} t ^{4}}{12}

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{e ^{- 3 t} t ^{4}}{12}

3 y^{^{''}} + 75 y = 10 cos (5 t)

y^{^{''''}} - y = 8 e^{x}

(D^{2} + 9) y = cos (2 x)

y^{^{''}} + y = t, y (0) = 4, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 3 y^{^{'}} - 2 y = 2 t^{2} + 4 t + 4

What is the solution for y^{''}+6y^'+9y=t^2*e^{-3t} ?
The solution for y^{''}+6y^'+9y=t^2*e^{-3t} is y=c_{1}e^{-3t}+c_{2}te^{-3t}+(e^{-3t}t^4)/(12)