What is the solution for 4y^{''}+75y=0 ?

The solution for 4y^{''}+75y=0 is y=c_{1}cos((5sqrt(3)t)/2)+c_{2}sin((5sqrt(3)t)/2)

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4y^{''}+75y=0

4 y^{^{''}} + 75 y = 0

y = c_{1} cos (\frac{5 3 t}{2}) + c_{2} sin (\frac{5 3 t}{2})

Solution steps

4 y^{''} (t) + 75 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{5 3 t}{2}) + c_{2} sin (\frac{5 3 t}{2})

y = c_{1} cos (\frac{5 3 t}{2}) + c_{2} sin (\frac{5 3 t}{2})

x^{^{''}} + 5 x^{^{'}} + 9 x = 0

25 y^{^{''}} - 30 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{\frac{3 x}{5}}

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = - 4

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = - 3

5 y^{^{''}} + 5 y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

What is the solution for 4y^{''}+75y=0 ?
The solution for 4y^{''}+75y=0 is y=c_{1}cos((5sqrt(3)t)/2)+c_{2}sin((5sqrt(3)t)/2)