What is the solution for 12x^{''}+6x^'+18x=0 ?

The solution for 12x^{''}+6x^'+18x=0 is x=e^{-t/4}(c_{1}cos((sqrt(23)t)/4)+c_{2}sin((sqrt(23)t)/4))

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12x^{''}+6x^'+18x=0

12 x^{^{''}} + 6 x^{^{'}} + 18 x = 0

x = e^{- \frac{t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

Solution steps

12 x^{''} (t) + 6 x^{'} (t) + 18 x = 0

Solve linear ODE:

x = e^{- \frac{t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

x = e^{- \frac{t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

b \cdot y = (a - v^{2}) y^{^{''}}

y^{^{''}} = 9 y^{^{'}} - 20 y

y^{^{''}} - 4 y^{^{''}} - 5 y = 0

0.25 x^{^{''}} + 2 x^{^{'}} + 2 x = 0

y^{^{''}} - 5 y^{^{'}} - 14 y = 0, y (0) = α, y^{^{'}} (0) = - 14

What is the solution for 12x^{''}+6x^'+18x=0 ?
The solution for 12x^{''}+6x^'+18x=0 is x=e^{-t/4}(c_{1}cos((sqrt(23)t)/4)+c_{2}sin((sqrt(23)t)/4))