$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$

Question

Solve the differential equation - $$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$$

I tried various methods such as putting $y=vx$, squarring both sides and then manipulating the expression but couldnt solve it. Any hint will be helpful. Thanks

Answer 1

$$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$$ Let $X=x^2$ and $Y=y^2$ $$X\frac{dY}{dX} = \sqrt{ X- Y - XY -1}$$ Let $u(X)=\sqrt{ X- Y - XY -1}$ $$Y=\frac{X-1-u^2}{X+1}$$ $$X\;Y' =X\left(\frac{1}{X+1}-\frac{2u\,u'}{X+1}-\frac{X}{(X+1)^2}+\frac{1}{(X+1)^2} +\frac{u^2}{(X+1)^2}\right)=u$$ $$u\,u'+\frac{X+1}{2X}u-\frac{u^2}{2(X+1)}=\frac{1}{X+1}$$ Let $v(X)=\frac{1}{u(X)}$ $$v'=-\frac{1}{X+1}v^3+\frac{X+1}{2X}v^2-\frac{1}{2(X+1)}v$$ This is an Abel's ODE of the first kind. Many of them are not analytically solvable in terms of standard elementary and/or special functions. This is probably the case. For deeper investigation, see : https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf

Answer 2

Hint:

Let $t=x^2$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=2x\dfrac{dy}{dt}$

$\therefore2x^2y\dfrac{dy}{dt}=\sqrt{x^2-y^2-x^2y^2-1}$

$2ty\dfrac{dy}{dt}=\sqrt{t-y^2-ty^2-1}$

Let $u=y^2$ ,

Then $\dfrac{du}{dt}=2y\dfrac{dy}{dt}$

$\therefore t\dfrac{du}{dt}=\sqrt{t-u-tu-1}$

$t\dfrac{du}{dt}=\sqrt{t-1-(t+1)u}$