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

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

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3y^{''}+y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

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