What is the solution for 3y^{''}+4y=0 ?

The solution for 3y^{''}+4y=0 is y=c_{1}cos((2t)/(sqrt(3)))+c_{2}sin((2t)/(sqrt(3)))

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3y^{''}+4y=0

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

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

Solution steps

3 y^{''} (t) + 4 y = 0

Solve linear ODE:

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

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

\frac{x ^{^{''}}}{x} = a

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

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 3

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

15 y^{^{''''}} + 92 y^{^{''}} + 64 y = 0

What is the solution for 3y^{''}+4y=0 ?
The solution for 3y^{''}+4y=0 is y=c_{1}cos((2t)/(sqrt(3)))+c_{2}sin((2t)/(sqrt(3)))