What is the solution for y^'=(x^2-8)/(y^2),y(0)=2 ?

The solution for y^'=(x^2-8)/(y^2),y(0)=2 is y=\sqrt[3]{x^3-24x+8}

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y^'=(x^2-8)/(y^2),y(0)=2

y^{^{'}} = \frac{x ^{2} - 8}{y ^{2}}, y (0) = 2

Solution steps

y^{'} (x) = \frac{x ^{2} - 8}{y ^{2}}

Solve separable ODE:

y^{^{'}} = - (sin (x)) (cos^{2} (y))

y^{^{'}} = - \frac{9 x}{y}

6 (y^{2} + 2 x y \frac{d y}{d x}) + 18 x = - 4 y^{3} \frac{d y}{d x}

y^{^{'}} (x) = \frac{y}{x ( y - 3 )}, y (2) = 6

\frac{d}{d x} (x^{2} sec (y)) = 8 x

What is the solution for y^'=(x^2-8)/(y^2),y(0)=2 ?
The solution for y^'=(x^2-8)/(y^2),y(0)=2 is y=\sqrt[3]{x^3-24x+8}