What is the solution for x^{''}+3x^'+6x=0 ?

The solution for x^{''}+3x^'+6x=0 is x=e^{-(3t)/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))

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x^{''}+3x^'+6x=0

x^{^{''}} + 3 x^{^{'}} + 6 x = 0

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

Solution steps

x^{''} (t) + 3 x^{'} (t) + 6 x = 0

Solve linear ODE:

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

\frac{d}{d x} (x^{2} - 6)

\int_{0}^{2} \frac{2 x}{( x ^{2} + 3 ) ^{2}} d x

(14, 448), (1, y)

\int 4

What is the solution for x^{''}+3x^'+6x=0 ?
The solution for x^{''}+3x^'+6x=0 is x=e^{-(3t)/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))