What is the solution for 2y^{''}+y=0 ?

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

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2y^{''}+y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

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