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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = 5 e^{4 t} + 6 t e^{4 t} + 3 t^{2} e^{4 t}

y^{^{''}} + 7 y^{^{'}} + 10 y = - 4 sin (3 t)

y^{^{''}} + 42 y = 42 cos (42 t)

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5 x}, y (0) = 1, y (0) = - 2

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{x} sin (x)

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