What is the solution for y^{''}+16/3 y=0 ?

The solution for y^{''}+16/3 y=0 is y=c_{1}cos((4t)/(sqrt(3)))+c_{2}sin((4t)/(sqrt(3)))

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y^{''}+16/3 y=0

y^{^{''}} + \frac{16}{3} y = 0

y = c_{1} cos (\frac{4 t}{3}) + c_{2} sin (\frac{4 t}{3})

Solution steps

y^{''} (t) + \frac{16}{3} y = 0

Solve linear ODE:

y = c_{1} cos (\frac{4 t}{3}) + c_{2} sin (\frac{4 t}{3})

y = c_{1} cos (\frac{4 t}{3}) + c_{2} sin (\frac{4 t}{3})

4 y^{^{''}} + 22 y^{^{'}} + 18 y = 0, y (0) = 3, y^{^{'}} (0) = 0

2 y^{^{''}} - 5 y - 3 y = 0

y^{^{''}} + 10.5 y = 0

\frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} - 12 y = 0

y^{^{''}} - 4 y = 0, y (0) = 0

What is the solution for y^{''}+16/3 y=0 ?
The solution for y^{''}+16/3 y=0 is y=c_{1}cos((4t)/(sqrt(3)))+c_{2}sin((4t)/(sqrt(3)))