What is the solution for 3y^{''}+2y^'+1y=0 ?

The solution for 3y^{''}+2y^'+1y=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^'+1y=0

3 y^{^{''}} + 2 y^{^{'}} + 1 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) + 1 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{y ^{^{''}}}{8} + 3 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} + 4 y = 0, y (0) = 6, y^{^{''}} (0) = - 4

4 y^{^{''}} - 12 y^{^{'}} + 73 y = 0

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

y^{^{'''}} + 3 y^{^{''}} - 7 y^{^{'}} + 3 y = 0

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