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

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

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

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

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{e ^{- 3 t} t ^{3}}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{e ^{- 3 t} t ^{3}}{6}

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{e ^{- 3 t} t ^{3}}{6}

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = cos (2 x)

y^{^{''}} + y = 2 + t, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 6 y^{^{'}} + 8 y = 4 e^{e - 2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = - 80 sin (3 x) - 23 cos (3 x)

\frac{d ^{2}}{d t ^{2}} (θ (t)) + 4 \cdot θ (t) = csc^{2} (2 t)

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