Solution of complex differential equation 
Finding solution of differential equation $\displaystyle \frac{xdy-ydx}{ydy-xdx}=\sqrt{\frac{x^2-y^2+1}{x^2-y^2}},$


Where $y=f(x)$ and $f(0)=0$ and $f(1)=1$

I have write $\displaystyle \int \frac{xdy-ydx}{x^2}=-\int \frac{1}{2x^2}(xdx-ydy)\sqrt{\frac{x^2-y^2+1}{x^2-y^2}}$
$\displaystyle \int d\bigg(\frac{y}{x}\bigg)=-\int \frac{1}{x^2}d(x^2-y^2)\sqrt{\frac{x^2-y^2+1}{x^2-y^2}}$
But I did not understand what I do with $x^2$ term in denominator
so that right side become integrable.
Please have a look.

$$\displaystyle \frac{xdy-ydx}{ydy-xdx}=\sqrt{\frac{x^2-y^2+1}{x^2-y^2}},$$
$$2x^2d\dfrac yx=-\sqrt{\frac{x^2-y^2+1}{x^2-y^2}}d(x^2-y^2)$$
Substitute $x^2-y^2=v$ and $u=\dfrac yx$:
$$2x^2du=-\sqrt{\frac{v+1}{v}}dv$$
For the $x^2$ factor we have:
$$v=x^2-y^2=x^2(1-u^2)$$
$$ \implies {x^2}=\dfrac v{1-u^2}$$
The DE becomes:
$$\dfrac {2v} {u^2-1}du=\sqrt{\frac{v+1}{v}}dv$$
The DE is separable but not sure if it's easy to integrate. Try to substitute $w=\dfrac 1 v$. Or maybe it's better to substitute $w^2=1+\dfrac 1v$ and $2wdw=-\dfrac {d
v}{v^2}$.
