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

The solution for 3y^{''}+2y^'+2y=0 is y=e^{-t/3}(c_{1}cos((sqrt(5)t)/3)+c_{2}sin((sqrt(5)t)/3))

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

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

y = e^{- \frac{t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

y = e^{- \frac{t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0

\frac{1}{8} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = 14

4 y^{^{''}} + 24 y^{^{'}} + 37 y = 0

x^{^{''}} - 6 x^{^{'}} + 9 x = 0

What is the solution for 3y^{''}+2y^'+2y=0 ?
The solution for 3y^{''}+2y^'+2y=0 is y=e^{-t/3}(c_{1}cos((sqrt(5)t)/3)+c_{2}sin((sqrt(5)t)/3))