What is the solution for y^'=(x^2-6)/(y^2),y(0)=2 ?

The solution for y^'=(x^2-6)/(y^2),y(0)=2 is y=\sqrt[3]{x^3-18x+8}

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y^'=(x^2-6)/(y^2),y(0)=2

y^{^{'}} = \frac{x ^{2} - 6}{y ^{2}}, y (0) = 2

Solution steps

y^{'} (x) = \frac{x ^{2} - 6}{y ^{2}}

Solve separable ODE:

\frac{d y}{d x} = \frac{x ^{2}}{x ^{2} ( 3 y ^{2} + 1 )}, y (1) = 4

\frac{d y}{d x} = x^{5} y^{2}

5 x y y^{^{'}} = 1

y^{^{'}} = \frac{2 y}{t - 1}

y^{^{'}} = 5 t^{3} y^{\frac{1}{5}}, y (0) = 0

What is the solution for y^'=(x^2-6)/(y^2),y(0)=2 ?
The solution for y^'=(x^2-6)/(y^2),y(0)=2 is y=\sqrt[3]{x^3-18x+8}