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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{'}} = x^{2} + 2 x e^{x} + e^{2 x}

y^{^{''}} + 6 y^{^{'}} + 8 y = 3 e^{5 x} + x^{2} - 1

y^{^{''}} + 9 y = sin (2 x)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + y = 2 sin (3 x)

y^{^{''}} - y^{^{'}} - 6 y = cos (2 t)

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