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

The solution for y^'=((y^2-1))/(2x),y(1)=-2 is y=-1/(3x-1)-(3x)/(3x-1)

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y^'=((y^2-1))/(2x),y(1)=-2

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

y = - \frac{1}{3 x - 1} - \frac{3 x}{3 x - 1}

Solution steps

y^{'} (x) = \frac{( y ^{2} - 1 )}{2 x}

Solve linear ODE:

y = - \frac{1}{3 x - 1} - \frac{3 x}{3 x - 1}

y = - \frac{1}{3 x - 1} - \frac{3 x}{3 x - 1}

What is the solution for y^'=((y^2-1))/(2x),y(1)=-2 ?
The solution for y^'=((y^2-1))/(2x),y(1)=-2 is y=-1/(3x-1)-(3x)/(3x-1)