What is the solution for 9y^{''}+6y^'+4y=0 ?

The solution for 9y^{''}+6y^'+4y=0 is y=e^{-t/3}(c_{1}cos(t/(sqrt(3)))+c_{2}sin(t/(sqrt(3))))

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9y^{''}+6y^'+4y=0

9 y^{^{''}} + 6 y^{^{'}} + 4 y = 0

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

Solution steps

9 y^{''} (t) + 6 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

What is the solution for 9y^{''}+6y^'+4y=0 ?
The solution for 9y^{''}+6y^'+4y=0 is y=e^{-t/3}(c_{1}cos(t/(sqrt(3)))+c_{2}sin(t/(sqrt(3))))