What is the solution for 2x^{''}+6x^'+9x=0 ?

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

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2x^{''}+6x^'+9x=0

2 x^{^{''}} + 6 x^{^{'}} + 9 x = 0

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Solution steps

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

Solve linear ODE:

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

(2 x y + x) d x + (x^{2} + y) d y = 0

\int 9.8 x d x

\frac{\partial}{\partial t} (0)

\int - 4 x^{\frac{3}{2}} + 3 x^{\frac{2}{3}} - 4 x^{2} d x

\int \frac{2}{x 4 - ( ln ( x ) ) ^{2}} d x

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