What is the solution for 2y^{''}+25y=0 ?

The solution for 2y^{''}+25y=0 is y=c_{1}cos((5t)/(sqrt(2)))+c_{2}sin((5t)/(sqrt(2)))

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2y^{''}+25y=0

2 y^{^{''}} + 25 y = 0

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

Solution steps

2 y^{''} (t) + 25 y = 0

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})

y^{^{''}} + 0.46 y^{^{'}} + 128 y = 0

2 y^{^{''}} - 4 y = 0

f^{^{''}} (t) + 2 f^{^{'}} (t) + 5 f (t) = 0, f (0) = 3, f^{^{'}} (0) = - 7

y^{^{'}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

What is the solution for 2y^{''}+25y=0 ?
The solution for 2y^{''}+25y=0 is y=c_{1}cos((5t)/(sqrt(2)))+c_{2}sin((5t)/(sqrt(2)))