What is the solution for y^{''}+16/9 y=0 ?

The solution for y^{''}+16/9 y=0 is y=c_{1}cos((4t)/3)+c_{2}sin((4t)/3)

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y^{''}+16/9 y=0

y^{^{''}} + \frac{16}{9} y = 0

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

Solution steps

y^{''} (t) + \frac{16}{9} 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})

u, \frac{d ^{2} u}{d x ^{2}} = u

16 y^{^{''}} + 8 y^{^{'}} + 5 y = 0

y^{^{'''}} - 2 y^{^{'}} - y^{^{'}} + 2 y = 0

3 y^{^{''}} + 11 y^{^{'}} - 4 y = 0, y (0) = 4, y^{^{'}} (0) = - \frac{5}{3}

y^{^{''}} + 20 y = 0

What is the solution for y^{''}+16/9 y=0 ?
The solution for y^{''}+16/9 y=0 is y=c_{1}cos((4t)/3)+c_{2}sin((4t)/3)