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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{1}^{5} \frac{ln ( x )}{x} d x

\int_{- \infty}^{0} \frac{1}{5 - 9 x} d x

3 e^{4 t}

4 y^{^{''}} - 4 y^{^{'}} - 3 y = cos (2 x)

\frac{d}{d a} (tan (a) + cot (a))

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