I encountered the following nonlinear 1st order ODE as a simplified case of this question.
$$\frac{dy}{dx}=\frac{1}{y^2}-\frac{1}{x^2}$$
I didn't find it in the literature and Wolfram Alpha couldn't solve it either.
However, I believe it might have a solution,.
Replace:
$$y= r \sin t, \qquad x=r \cos t$$
After simplifications, this leads to the following ODE:
$$(r^2 \sin^3 t \cos^2 t+\sin^2 t \cos t- \cos^3 t) dr+\left(r^3 \sin^2 t \cos^3 t+r( \sin t \cos^2 t-\sin^3 t) \right)dt=0$$
$$P(r,t) dr+Q(r,t)dt=0$$
Checking if the new ODE is exact or not, we find that it's not, but the partial derivatives look promising:
$$P_t =r^2 (3 \sin^2 t \cos^3 t-2 \sin^4 t \cos t) +5 \sin t \cos^2 t-\sin^3 t$$
$$Q_r =r^2 (3 \sin^2 t \cos^3 t) +\sin t \cos^2 t-\sin^3 t$$
I feel like there's a simple integrating factor we can use here, but I have no idea what it is.
Can we find an integrating factor which makes the last ODE exact? If we can, what is it?
Is there another way to obtain an explicit or implicit solution for the titular ODE in terms of known functions?
What about the more general case ($a,b$ - constants)? $$\frac{dy}{dx}=\frac{a}{y^2}-\frac{b}{x^2}$$
Edit
Multiplying by $\frac{1}{\cos^2 t}$ we obtain:
$$\left(r^2 \sin^3 t +\frac{\sin^2 t }{\cos t}- \cos t\right) dr+\left(r^3 \sin^2 t \cos t+r\left( \sin t -\frac{\sin^3 t}{\cos^2 t}\right) \right)dt=0$$
Now we have:
$$P_t=3 r^2 \sin^2 t \cos t+ \sin t +\frac{\sin^3 t }{\cos^2 t}$$
$$Q_r=3 r^2 \sin^2 t \cos t+ \sin t -\frac{\sin^3 t }{\cos^2 t}$$
It's almost exact. Did I make a mistake somewhere? Or do we need a more elaborate integrating factor?
Edit 2
Hyperbolic substitution:
$$y= r \sinh t, \qquad x=r \cosh t$$
leads to a similar equation, which is still not exact:
$$\left(r^2 \sinh^3 t-\frac{1}{\cosh t} \right) dr+\left(r^3 \sinh^2 t \cosh t-r \frac{\sinh t}{\cosh^2 t} \right) dt=0$$
$$P_t=3 r^2 \sinh^2 t \cosh t +\frac{\sinh t }{\cosh^2 t}$$
$$Q_r=3 r^2 \sinh^2 t \cosh t -\frac{\sinh t }{\cosh^2 t}$$
Hint:
With reference to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=180:
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{1}{x^2u^2}-\dfrac{1}{x^2}$
$x\dfrac{du}{dx}=\dfrac{1-u^2}{x^2u^2}-u$
$x\dfrac{du}{dx}=\dfrac{1-u^2-x^2u^3}{x^2u^2}$
Let $v=x^2$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=2x\dfrac{du}{dv}$
$\therefore2x^2\dfrac{du}{dv}=\dfrac{1-u^2-x^2u^3}{x^2u^2}$
$2v\dfrac{du}{dv}=\dfrac{1-u^2-u^3v}{u^2v}$
$(u^3v+u^2-1)\dfrac{dv}{du}=-2u^2v^2$
$\left(v+\dfrac{1}{u}-\dfrac{1}{u^3}\right)\dfrac{dv}{du}=-\dfrac{2v^2}{u}$
This belongs to an Abel equation of the second kind.