What is the solution for 4y^{''}=-25y ?

The solution for 4y^{''}=-25y is y=c_{1}cos((5t)/2)+c_{2}sin((5t)/2)

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4y^{''}=-25y

4 y^{^{''}} = - 25 y

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2})

Solution steps

4 y^{''} (t) = - 25 y

Solve linear ODE:

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2})

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2})

What is the solution for 4y^{''}=-25y ?
The solution for 4y^{''}=-25y is y=c_{1}cos((5t)/2)+c_{2}sin((5t)/2)