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

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

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y^{''}+12y=0

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

y = c_{1} cos (23 t) + c_{2} sin (23 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (23 t) + c_{2} sin (23 t)

y = c_{1} cos (23 t) + c_{2} sin (23 t)

25 y^{^{''}} - 64 y = 0

y^{^{'''}} - 4 y^{^{''}} + 9 y^{^{'}} = 0

y^{^{''}} + 16 y = 0, y (\frac{π}{12}) = 4, y^{^{'}} (\frac{π}{12}) = - 8

y^{^{''}} - 7 y^{^{'}} - 8 y = 0

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

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