Nonlinear ODE and $_2\text{F}_1$ Consider the differential equation $y'(x) = B \, y(x)^{-b} - A \, y(x)^a$ to which WA returns the following:
$$x - c_1 = \frac{y(x)^{b+1}}{B\,(b+1)} \; _2\text{F}_1 \left( 1, \frac{b+1}{a+b}; 1 + \frac{b+1}{a+b}; \frac{A}{B} \, y(x)^{a+b} \right)$$
Is there a way to turn this into an explicit expression for $y(x)$ or to approximate the solution? Thanks for your help.
 A: $$\frac{dy}{dx} = B \, y(x)^{-b} - A \, y(x)^a$$
This is a separable ODE.
$$x=\int\frac {dy}{B\,y^{-b} - A \, y^a} $$
$$x=\frac{1}{B(b+1)}y^{b+1}\;_2F_1\left(1\:,\:\frac{b+1}{a+b}\:;\:\frac{a+2b+1}{a+b}\:;\:\frac{A}{B}y^{a+b}\right)+\text{constant}$$
$y(x)$ is the inverse function of the above.
There is no available closed form for the inverse function $y(x)$.
Simplification might occur for particular values of the parameters $a,b,A,B\:$  but not in the general case.
Approximate formulas could be derived but arduous and valid only on limited ranges depending on $a,b,A,B$ . One could use the series expension of the hypergeometric function and the method of reversion of series : https://mathworld.wolfram.com/SeriesReversion.html . Since this involves complicated numerical calculus it should be much simpler to directly solve the original ODE thanks to numerical calculus.
A: Firstly, the question should have your ideas on how to solve the problem.
The answer is yes, but it needs a limit since your hypergeometric function is a Lerch Transcendent. Here is the “closed form” solution using the Wolfram Language’s Incomplete Beta function $\text B_z(a,b)$ and Inverse Beta Regularized $\text I^{-1}_z(a,b)$:
$$x - c= \frac{y^{b+1}}{B\,(b+1)} \; _2\text{F}_1 \left( 1, \frac{b+1}{a+b}; 1 + \frac{b+1}{a+b}; \frac{A}{B} \, y^{a+b} \right)=  \frac{y^{b+1}}{B\,(b+1)} \frac{b+1}{a+b}\left( \frac{A}{B} \, y^{a+b} \right)^{-\frac{b+1}{a+b}}\text B_{ \frac{A}{B} \, y^{a+b} }\left(\frac{b+1}{a+b},0\right)\mathop=^{a,b\in\Bbb R^+}   \frac{\text B_{ \frac{A}{B} \, y^{a+b} }\left(\frac{b+1}{a+b},0\right)}{\sqrt[a+b]{A^{b+1}B^{a-1}}(a+b)}$$
Now invert:
$$\boxed{y’=By^{-b}-Ay^a\implies y=\lim_{t\to0}\sqrt[a+b]{\frac BA\text I^{-1}_{\left(\frac AB\right)^\frac{b+1}{a+b}(a+b)Bt(x-c)}\left(\frac{b+1}{a+b},t\right)}}$$
which works when you use the decimal answer from the left bolded link and compare use it like in the right bolded link.
Take $t\to0$ to adjust the accuracy
