How to solve this ordinary differential equation? I am just trying to find general solution 
$$\frac{dy}{dx} = 1 + \sqrt{1 - xy}$$
 A: Here is partial answer. I hope this question doesn't get closed and someone can give a better answer.
I looked at the numerical solution to the initial value problem $$\frac{dy}{dx} = 1 + \sqrt{1-xy}, y(0) = 0$$ the solution exist for only finite time. 
The solution must satisfy $$x < y < 2x$$ and these line cut the boundary $xy = 1$ at $(\pm 1, \pm 1), (\pm \frac{1}{\sqrt 2}, \pm \sqrt 2)$ so the interval of existence cannot be bigger than $(-1, 1)$ and no smaller than $(-1/\sqrt 2, 1/\sqrt 2).$
The Runge-Kutta numerical solution I tried stops at $x = 0.75128$.
A: Let $u=-\sqrt{1-xy}$ ,
Then $y=\dfrac{1-u^2}{x}$
$\dfrac{dy}{dx}=\dfrac{u^2-1}{x^2}-\dfrac{2u}{x}\dfrac{du}{dx}$
$\therefore\dfrac{u^2-1}{x^2}-\dfrac{2u}{x}\dfrac{du}{dx}=1-u$
$\dfrac{2u}{x}\dfrac{du}{dx}=u-1+\dfrac{u^2-1}{x^2}$
Approach $1$:
$\dfrac{2u}{x}\dfrac{du}{dx}=\dfrac{(u-1)x^2+(u+1)(u-1)}{x^2}$
$(x^2+u+1)\dfrac{dx}{du}=\dfrac{2ux}{u-1}$
Let $v=x^2$ ,
Then $\dfrac{dv}{du}=2x\dfrac{dx}{du}$
$\therefore\dfrac{x^2+u+1}{2x}\dfrac{dv}{du}=\dfrac{2ux}{u-1}$
$(x^2+u+1)\dfrac{dv}{du}=\dfrac{4ux^2}{u-1}$
$(v+u+1)\dfrac{dv}{du}=\dfrac{4uv}{u-1}$
This belongs to an Abel equation of the second kind.
Let $w=v+u+1$ ,
Then $v=w-u-1$
$\dfrac{dv}{du}=\dfrac{dw}{du}-1$
$\therefore w\left(\dfrac{dw}{du}-1\right)=\dfrac{4u(w-u-1)}{u-1}$
$w\dfrac{dw}{du}-w=\dfrac{4uw}{u-1}-\dfrac{4u(u+1)}{u-1}$
$w\dfrac{dw}{du}=\dfrac{(5u-1)w}{u-1}-\dfrac{4u(u+1)}{u-1}$
Approach $2$:
$\dfrac{2u}{x}\dfrac{du}{dx}=u-1+\dfrac{u^2-1}{x^2}$
$u\dfrac{du}{dx}=\dfrac{u^2}{2x}+\dfrac{xu}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$
Let $u=\sqrt{x}v$ ,
Then $\dfrac{du}{dx}=\sqrt{x}\dfrac{dv}{dx}+\dfrac{v}{2\sqrt{x}}$
$\therefore\sqrt{x}v\left(\sqrt{x}\dfrac{dv}{dx}+\dfrac{v}{2\sqrt{x}}\right)=\dfrac{xv^2}{2x}+\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$
$xv\dfrac{dv}{dx}+\dfrac{v^2}{2}=\dfrac{v^2}{2}+\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$
$xv\dfrac{dv}{dx}=\dfrac{x\sqrt{x}v}{2}-\dfrac{x}{2}-\dfrac{1}{2x}$
$v\dfrac{dv}{dx}=\dfrac{\sqrt{x}v}{2}-\dfrac{1}{2}-\dfrac{1}{2x^2}$
Let $t=\dfrac{x^\frac{3}{2}}{3}$ ,
Then $\dfrac{dv}{dx}=\dfrac{dv}{dt}\dfrac{dt}{dx}=\dfrac{\sqrt{x}}{2}\dfrac{dv}{dt}$
$\therefore\dfrac{\sqrt{x}v}{2}\dfrac{dv}{dt}=\dfrac{\sqrt{x}v}{2}-\dfrac{1}{2}-\dfrac{1}{2x^2}$
$v\dfrac{dv}{dt}=v-\dfrac{1}{\sqrt{x}}-\dfrac{1}{x^\frac{5}{2}}$
$v\dfrac{dv}{dt}=v-\dfrac{1}{\sqrt[3]{3t}}-\dfrac{1}{(3t)^\frac{5}{3}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf
