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

The solution for y^{''}+6y^'+9y=t+e^{3t} is y=c_{1}e^{-3t}+c_{2}te^{-3t}+t/9-2/27+1/36 e^{3t}

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

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

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{t}{9} - \frac{2}{27} + \frac{1}{36} e^{3 t}

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{t}{9} - \frac{2}{27} + \frac{1}{36} e^{3 t}

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{t}{9} - \frac{2}{27} + \frac{1}{36} e^{3 t}

y^{^{''}} + 16 y = 5 cos (4 t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 9 y = 4, y^{1} (0) = e^{- 3 x}

y^{^{''''}} + 9 y^{^{''}} = 20 e^{- t}

y^{^{''''}} - y = 9 t + cos (t)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 4 e^{2 x}

What is the solution for y^{''}+6y^'+9y=t+e^{3t} ?
The solution for y^{''}+6y^'+9y=t+e^{3t} is y=c_{1}e^{-3t}+c_{2}te^{-3t}+t/9-2/27+1/36 e^{3t}