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{d ^{2} y}{d x ^{2}} - 5 y = 0

y^{^{''''}} + 8 y^{^{'''}} + 16 y^{^{''}} = 0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

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))