What is the solution for y^{''}+6y^'+9y=3 ?

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

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y^{''}+6y^'+9y=3

y^{^{''}} + 6 y^{^{'}} + 9 y = 3

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 25 y = 2 x cos (x)

y^{^{''}} + 4 y = sec^{3} (2 t)

12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = e^{x}

y^{^{''}} + 4 y = 16 t, y (0) = 3, y^{^{'}} (0) = - 6

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