What is the solution for y^{''}+0/3 y^'+12/3 y=0 ?

The solution for y^{''}+0/3 y^'+12/3 y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)

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y^{''}+0/3 y^'+12/3 y=0

y^{^{''}} + \frac{0}{3} y^{^{'}} + \frac{12}{3} y = 0

y = c_{1} cos (2 t) + c_{2} sin (2 t)

Solution steps

y^{''} (t) + \frac{0}{3} y^{'} (t) + \frac{12}{3} y = 0

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t)

3 y^{^{'''}} - 2 y^{^{''}} - 37 y^{^{'}} - 12 y = 0

- y^{^{''}} + 4 y^{^{'}} + 5 y = 0

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y (0) = 1

(d^{3} + 9 d^{2} + 27 d + 27) y = 0

(D^{2} - D - 2) y = 0

What is the solution for y^{''}+0/3 y^'+12/3 y=0 ?
The solution for y^{''}+0/3 y^'+12/3 y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)