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

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

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

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

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

Solution steps

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

a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = 0

2 y^{^{''}} + 16 y = 0

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 6 x = 0

y^{^{''}} - 15 y^{^{'}} + 54 y = 0

2 y^{^{''}} + 4 y^{^{'}} + 2 y = 0

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