What is the solution for 3y^{''}+y^'+2y=0 ?

The solution for 3y^{''}+y^'+2y=0 is y=e^{-t/6}(c_{1}cos((sqrt(23)t)/6)+c_{2}sin((sqrt(23)t)/6))

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3y^{''}+y^'+2y=0

3 y^{^{''}} + y^{^{'}} + 2 y = 0

y = e^{- \frac{t}{6}} (c_{1} cos (\frac{23 t}{6}) + c_{2} sin (\frac{23 t}{6}))

Solution steps

3 y^{''} (t) + y^{'} (t) + 2 y = 0

Solve linear ODE:

y = e^{- \frac{t}{6}} (c_{1} cos (\frac{23 t}{6}) + c_{2} sin (\frac{23 t}{6}))

y = e^{- \frac{t}{6}} (c_{1} cos (\frac{23 t}{6}) + c_{2} sin (\frac{23 t}{6}))

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What is the solution for 3y^{''}+y^'+2y=0 ?
The solution for 3y^{''}+y^'+2y=0 is y=e^{-t/6}(c_{1}cos((sqrt(23)t)/6)+c_{2}sin((sqrt(23)t)/6))