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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = 3

y^{^{''}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 2

u^{2} y^{^{''}} - 4 u y^{^{'}} + 6 y = 0

9 y^{^{''}} + 30 y = 0, y (0) = 8, y^{^{'}} (0) = 8

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