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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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