What is the solution for y^{''}+4y^'+y=-9 ?

The solution for y^{''}+4y^'+y=-9 is y=c_{1}e^{(-2+sqrt(3))t}+c_{2}e^{(-2-sqrt(3))t}-9

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y^{''}+4y^'+y=-9

y^{^{''}} + 4 y^{^{'}} + y = - 9

y = c_{1} e^{(- 2 + 3) t} + c_{2} e^{(- 2 - 3) t} - 9

Solution steps

y^{''} (t) + 4 y^{'} (t) + y = - 9

Solve linear ODE:

y = c_{1} e^{(- 2 + 3) t} + c_{2} e^{(- 2 - 3) t} - 9

y = c_{1} e^{(- 2 + 3) t} + c_{2} e^{(- 2 - 3) t} - 9

y^{^{'''}} + 4 y^{^{'}} = sin (x) \cdot cos (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 2, y^{^{'}} (0) = - 1

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = 6

y^{^{''}} + y = 10

4 y^{^{''}} + y = e^{- 2 t} cos (\frac{t}{2}) + 6 t sin (\frac{t}{2})

What is the solution for y^{''}+4y^'+y=-9 ?
The solution for y^{''}+4y^'+y=-9 is y=c_{1}e^{(-2+sqrt(3))t}+c_{2}e^{(-2-sqrt(3))t}-9