Solution of first order nonlinear differential equation without inverse function The following differential equation is part of a path integral problem in quantum mechanics:
$$(\partial_{r} y)^2 = y^4 (-1+p y -q^2 y^2)^{3/2}$$
with $p>0$ and $q>0$ and $p,q \in \mathbb{R}$
Additionally it is required, that
$$\lim_{r\to \infty} y(r) = 0$$
To solve the equation we transform to
$$\int y^{-2} (-1+p y -q^2 y^2)^{-3/4} dy = \pm (r+C_1)$$
With mathematica we can solve the integral to
$$\frac{1}{8 x \left(p x-q^2 x^2-1\right)^{3/4}} \left(\left(-\sqrt{2}\right) p x \left(\frac{\sqrt{p^2-4 q^2}-p+2 q^2 x}{q^2 x}\right)^{3/4} \left(-\frac{\sqrt{p^2-4 q^2}+p-2 q^2 x}{q^2 x}\right)^{3/4} F_1\left(\frac{3}{2};\frac{3}{4},\frac{3}{4};\frac{5}{2};\frac{p+\sqrt{p^2-4 q^2}}{2 q^2 x},\frac{p-\sqrt{p^2-4 q^2}}{2 q^2 x}\right)-4 \left(\sqrt[4]{2} x \left(\sqrt{p^2-4 q^2}-p+2 q^2 x\right) \left(\frac{-2 q^2 \left(x \sqrt{p^2-4 q^2}+2\right)+p \sqrt{p^2-4 q^2}+p^2}{p^2-4 q^2}\right)^{3/4} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\frac{2 x q^2-p+\sqrt{p^2-4 q^2}}{2 \sqrt{p^2-4 q^2}}\right)-2 p x+2 q^2 x^2+2\right)\right)$$
We can see here Hypergeometric functions. The first is an Appell function. It is obviously not possible to find ad hoc the inverse function to determine $y(r)$.
I have the assumption, that it is possible to find a transformation of the integral, that the solution can be used to evaluate $y(r)$
A second option could be, that the limit of $y(r)$ can be used to find an equation for the values $p$ and $q$, which simplifies the Hypergeometric functions to arithmetic expressions which are good to handle to find $y(r)$
Does anyone has a tipp for me?

I have an example for a simplification:
$$\int y^{-2} (-1+p y -q^2 y^2)^{-3/4} dy = \int y^{-2} \left(1-\frac{p^2}{4 q^2}+\left(q y-\frac{p}{2 q}\right)^2\right)^{-{3/4}} dy $$
now we substitute
$x = q y-\frac{p}{2 q}$ with $dy = \frac{1}{q} dx$ which leads to
$$\int y^{-2} (-1+p y -q^2 y^2)^{-3/4} dy = q \int \left( x + \frac{p}{2 q} \right)^{-2} \left(1-\frac{p^2}{4 q^2}+x^2\right)^{-{3/4}} dx $$
Now the integral has the solution
$$-\frac{8 \sqrt{2} q \left(\frac{2 x-\sqrt{\frac{p^2}{q^2}-4}}{p q+2 x}\right)^{3/4} \left(\frac{\sqrt{\frac{p^2}{q^2}-4}+2 x}{p q+2 x}\right)^{3/4} F_1\left(\frac{5}{2};\frac{3}{4},\frac{3}{4};\frac{7}{2};\frac{p q-\sqrt{\frac{p^2}{q^2}-4}}{p q+2 x},\frac{p q+\sqrt{\frac{p^2}{q^2}-4}}{p q+2 x}\right)}{5 (p q+2 x) \left(-\frac{p^2}{q^2}+4 x^2+4\right)^{3/4}}$$
The simplification is, that it depends now only on one Appell hypergeometric function
