What is the solution for y^'=(x^2-5)/(y^2),y(0)=3 ?

The solution for y^'=(x^2-5)/(y^2),y(0)=3 is y=\sqrt[3]{x^3-15x+27}

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y^'=(x^2-5)/(y^2),y(0)=3

y^{^{'}} = \frac{x ^{2} - 5}{y ^{2}}, y (0) = 3

Solution steps

y^{'} (x) = \frac{x ^{2} - 5}{y ^{2}}

Solve separable ODE:

x^{^{'}} = (x)^{22} (ln (t) + (10) + t e^{t})

y^{^{'}} = x y - x - y + 1

y^{^{'}} + \frac{1}{2} y = \frac{5}{2}

f (x) (g (x)) f^{^{'}} (x) = 1 + (f (x))^{2}

y^{- 1} d y - y e^{c o s (x)} sin (x) d x = 0

What is the solution for y^'=(x^2-5)/(y^2),y(0)=3 ?
The solution for y^'=(x^2-5)/(y^2),y(0)=3 is y=\sqrt[3]{x^3-15x+27}