To solve $ \dfrac {dy}{dx}=\dfrac 1{x^2+y^2}$

Question

How do we solve $ \dfrac {dy}{dx}=\dfrac 1{x^2+y^2}$ ? In general for which positive real $b$ do the equation

$\dfrac {dy}{dx}=\dfrac 1{(x^2+y^2)^b}$ admit an analytic solution ?

Answer 1

Let us first note that $x$ as a function of $y$ satisfies the equation $x'_y=x^2+y^2$. This is a special case of the Riccati equation which can be solved by substituting $x=-(\ln u)'_y$ and observing that $u$ satisfies a linear ODE reducible to a particular Bessel equation. The result is

$$x(y)=-\left[\ln\sqrt{y}\,\left(J_{1/4}\bigl({y^2}/{2}\bigr)+\nu J_{-1/4}\bigl({y^2}/{2}\bigr)\right)\right]'_y,$$ where $\nu$ is an arbitrary integration constant. See this answer for more details.

I believe this is the only case that admits analytic solution - and we surely can't have it for generic $b$.