What is the solution for y^{''}+6y^'+9y=12e^t ?

The solution for y^{''}+6y^'+9y=12e^t is y=c_{1}e^{-3t}+c_{2}te^{-3t}+3/4 e^t

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y^{''}+6y^'+9y=12e^t

y^{^{''}} + 6 y^{^{'}} + 9 y = 12 e^{t}

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{3}{4} e^{t}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 9 y = 12 e^{t}

Solve linear ODE:

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{3}{4} e^{t}

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{3}{4} e^{t}

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 1, y (0) = 0, y^{^{'}} (0) = 0

D (x) = 82.5 - 2 \cdot 5 x

y^{^{''}} + 6 y = 8 t

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} e^{2 t}, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 6 y = 12 cos (3 x)

What is the solution for y^{''}+6y^'+9y=12e^t ?
The solution for y^{''}+6y^'+9y=12e^t is y=c_{1}e^{-3t}+c_{2}te^{-3t}+3/4 e^t