How to solve DE $y'=1/(x^2y)(y^2-1)^{3/2}$ Man, I'm having troubles with this differential equation. I just can't do any math if I'm tired...
What I have done:
$$\frac{y'\cdot y}{(y^2-1)^{3/2}}=\frac{1}{x^2}$$
Now I integrated both sides from $a$ to $x$ and I've got 
$$-\frac{1}{(y^2-1)^{1/2}}+ \frac{1}{(y(a)^2-1)^{1/2}} = -\frac{1}{x} + \frac{1}{a}$$ 
But this seems wrong after checking the result on WolframAlpha. Please help!

I don't understand the $_a^x$ part. Why would you do that? You can evaluate the antiderivative directly(we may have the same solution but here it goes anyway);
$$\int \frac{y' y}{(y^2-1)^{3/2}} \mathrm{d}x =\frac12 \int \frac{1}{u^{3/2}} \mathrm{d}u=-\frac{1}{\sqrt{y^2-1}}+c$$
And $$\int \frac{1}{x^2} \mathrm{d}x = -\frac{1}{x}$$
Then the DE solution becomes;
$$y(x) = \frac{\sqrt{c_1^2 x^2-2 c_1 x+x^2+1}}{1-c_1 x}$$
or 
$$y(x)=\frac{\sqrt{c_1^2 x^2-2 c_1 x+x^2+1}}{c_1 x-1}$$
