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})

2500 y^{^{''}} + 729 y = 0, y (0) = 2, y^{^{'}} (0) = 3

x^{^{''}} + 8 x^{^{'}} + 10 x = 0, x (0) = 10, x^{^{'}} (0) = 0

2500 y^{^{''}} + 729 y = 0, y (0) = 2, y^{^{'}} (0) = 5

35 y^{^{''''}} + 52 y^{^{''}} + 9 y = 0

(D^{4} + D^{2}) y = 0

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