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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{x ^{^{''}} - 3 x ^{^{'}}}{2} = 2 x - \frac{( x ^{^{'}} - 3 x )}{2}

y^{^{''}} + y^{^{'}} - 12 y = 0, y (0) = - 6, y^{^{'}} (0) = 17

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 8, y^{^{'}} (0) = 6

36 y^{^{''}} + 132 y^{^{'}} + 121 y = 0, y (0) = 5, y^{^{'}} (0) = 5

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

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