How to solve $\frac{dy}{dx}=-\frac{y}{x^2+y^2}$?

Question

How does one solve the differential equation $$\frac{dy}{dx}=-\frac{y}{x^2+y^2}$$ I've tried to convert to polar coordinates:

I let $$x=r(\varphi)\cdot\cos(\varphi)$$ $$y=r(\varphi)\cdot\sin(\varphi)$$ Then, $$\frac{dx}{d\varphi}=r'(\varphi)\cdot\cos(\varphi)+r(\varphi)\cdot\sin(\varphi)$$ $$\frac{dy}{d\varphi}=r'(\varphi)\cdot\sin(\varphi)+r(\varphi)\cdot\cos(\varphi)$$ such that $$\frac{dy}{dx}=\frac{r'(\varphi)\cdot\sin(\varphi)+r(\varphi)\cdot\cos(\varphi)}{r'(\varphi)\cdot\cos(\varphi)+r(\varphi)\cdot\sin(\varphi)}$$ Substitution into the differential equation yields $$\frac{r'(\varphi)\cdot\sin(\varphi)+r(\varphi)\cdot\cos(\varphi)}{r'(\varphi)\cdot\cos(\varphi)+r(\varphi)\cdot\sin(\varphi)}=-\frac{\sin(\varphi)}{r(\varphi)}$$ It does not seem to get any easier now does it.

Answer 1

It seems to me that if we consider it as $$x'=\frac{x^2+y^2}{-y}=\dfrac{1}{y}x^2-y,~~(y\neq0)$$ then it can be regarded as a Riccati ODE.

Answer 2

Hint.

$$ y y'+\frac{y^2}{x^2+y^2}=0\Rightarrow \frac 12u'+ \frac{u}{x^2+u}=0 $$

with $u = y^2$