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}))

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = x e^{x}

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 0, y (0) = 0, y^{^{'}} (0) = 2

625 y^{^{''''}} = y

y^{^{''}} + 14 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = 14

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

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))