What is the solution for 3y^{''}-2y^'+6y=0 ?

The solution for 3y^{''}-2y^'+6y=0 is y=e^{t/3}(c_{1}cos((sqrt(17)t)/3)+c_{2}sin((sqrt(17)t)/3))

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3y^{''}-2y^'+6y=0

3 y^{^{''}} - 2 y^{^{'}} + 6 y = 0

y = e^{\frac{t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

Solution steps

3 y^{''} (t) - 2 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{\frac{t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

y = e^{\frac{t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

y^{^{'''}} + 36 y^{^{'}} = 0, y (0) = 1

y^{^{''}} - y^{^{'}} = - 10 y

25 y^{^{''}} + 4 y = 0

y^{^{''}} + 5 y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 5 y^{^{'}} - 36 y = 0, y^{0} = a, y^{^{'}} 0 = 32

What is the solution for 3y^{''}-2y^'+6y=0 ?
The solution for 3y^{''}-2y^'+6y=0 is y=e^{t/3}(c_{1}cos((sqrt(17)t)/3)+c_{2}sin((sqrt(17)t)/3))