What is the solution for y^{''}+6y^'+9=0 ?

The solution for y^{''}+6y^'+9=0 is y=c_{1}+c_{2}e^{-6t}-(3t)/2

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y^{''}+6y^'+9=0

y^{^{''}} + 6 y^{^{'}} + 9 = 0

y = c_{1} + c_{2} e^{- 6 t} - \frac{3 t}{2}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 9 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 6 t} - \frac{3 t}{2}

y = c_{1} + c_{2} e^{- 6 t} - \frac{3 t}{2}

\frac{d}{d x} (e^{- 0.5 x} (4 - x) - 2)

(1 - x + x^{2}) e^{x}

\frac{\partial}{\partial x} (x^{5} - 6 x + 4 y^{3} - y^{2})

(4.6) (3.1)

f (9) = 3.6 x + 500

What is the solution for y^{''}+6y^'+9=0 ?
The solution for y^{''}+6y^'+9=0 is y=c_{1}+c_{2}e^{-6t}-(3t)/2