What is the solution for (dy)/(dx)=(x^2-1)/(y^2),y(0)=2 ?

The solution for (dy)/(dx)=(x^2-1)/(y^2),y(0)=2 is y=\sqrt[3]{x^3-3x+8}

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(dy)/(dx)=(x^2-1)/(y^2),y(0)=2

\frac{d y}{d x} = \frac{x ^{2} - 1}{y ^{2}}, y (0) = 2

y = 3 x^{3} - 3 x + 8

Solution steps

\frac{d y}{d x} = \frac{x ^{2} - 1}{y ^{2}}

Solve separable ODE:

y = 3 x^{3} - 3 x + 8

y = 3 x^{3} - 3 x + 8

\frac{d x}{d y} = \frac{1 + x ^{2}}{y}

cos (y) \frac{d y}{d x} = \frac{- x sin ( y )}{1 + x ^{2}}

2 x^{^{'}} + x = 10

y y^{^{'}} = x e^{- y^{2}}

sin (3 x) d x + 2 y \cdot cos^{3} (3 x) d y = 0

What is the solution for (dy)/(dx)=(x^2-1)/(y^2),y(0)=2 ?
The solution for (dy)/(dx)=(x^2-1)/(y^2),y(0)=2 is y=\sqrt[3]{x^3-3x+8}