Generalizing the solution to an ODE

Is there a way to solve the following ODE for general integral values of $$m$$

\begin{align} \frac{\partial A(x)}{ \partial x} = -A(x)^m + \frac1x \label{rec}\tag{1} \end{align}

I have some ways to approach this problem for a special case of $$m=2$$. For this case, if we substitute $$A(x) = \frac{u^\prime(x)}{u(x)}$$, we would get a differential equation of the form, $$u^{\prime\prime}(x) = \frac{u(x)}x$$ And, it is possible to write a solution for this equation in terms of Bessel functions. But I don't know how to generalize this for higher $$m$$.

Any help would be appreciated!

This is a Chini differential equation, but the Chini invariant depends on $$x$$. I don't think a closed-form solution is known in general. Maple does find the general solution for $$n=2$$:

$$A \left( x \right) =-{\frac {-c {{ I}_{0}\left(2\,\sqrt {x}\right)}+{{ K}_{0}\left(2\,\sqrt {x}\right)}}{\sqrt {x} \left( { \it c}\,{{ I}_{1}\left(2\,\sqrt {x}\right)}+{{ K}_{1}\left(2 \,\sqrt {x}\right)} \right) }}$$

but not for $$m=3$$.

Hint:

$$\dfrac{dA}{dx}=-A^m+\dfrac{1}{x}$$

$$-\dfrac{dA}{dx}=\dfrac{A^mx-1}{x}$$

$$(x-A^{-m})\dfrac{dx}{dA}=-A^{-m}x$$

This belongs to an Abel equation of the second kind.

Let $$u=x-A^{-m}$$ ,

Then $$x=u+A^{-m}$$

$$\dfrac{dx}{dA}=\dfrac{du}{dA}-mA^{-m-1}$$

$$\therefore u\left(\dfrac{du}{dA}-mA^{-m-1}\right)=-A^{-m}(u+A^{-m})$$

$$u\dfrac{du}{dA}-mA^{-m-1}u=-A^{-m}u-A^{-2m}$$

$$u\dfrac{du}{dA}=(mA^{-m-1}-A^{-m})u-A^{-2m}$$