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

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

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

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

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

Solution steps

x^{''} (t) + 3 x^{'} (t) + 3 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}))

49 y^{^{''}} - 14 y^{^{'}} + y = 0

x^{^{''}} + 2 x^{^{'}} - 3 x = 0, x (0) = 2, x^{^{'}} (0) = 3

\frac{3}{4} \cdot x^{^{''}} + 36 x = 0, x^{^{'}} (0) = 0, x (0) = - 6

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} - y^{^{'}} - 72 y^{^{'}} = 0

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