The differential equation $y' = \frac{\ln(x^2+y^2)}{x^2 + y^2}$ In my university, the integral calculus teacher gave me this differential equation to solve.
$$ y' = \frac{\ln(x^2+y^2)}{x^2 + y^2} $$
I don't have any clue of what form the solution of this differential equation has.
 A: I can't comment because I don't have enough points but this is what I came up with.
If we assume $y=r\sin(\theta)$ and $x=r\cos(\theta)$ where $r=\sqrt{x^2+y^2}$ and $\theta = a\tan(y/x)$ then
$$
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}x}+\frac{\mathrm{d}y}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}x}=[\text{calculations}]=0.
$$
Inserting this in the given ODE we get $r^2=x^2+y^2=1$. But this feels wrong. Please comment if you guys know where I made a mistake.
A: A differential geometric approach for numerical integration is used as no analytical solution seems possible.
Using polar coordinates differentiation with respect to four variables arc length, two polar coordinates and connecting parameter $\psi$ as shown in the exaggerated differential elements triangle
$$s,\;dr/ds=\cos\psi, r d\theta/ds= \sin \psi,dy/dx= \tan \phi \tag 1$$

$$ \phi=\tan^{-1}\dfrac{\log r^2}{r^2} \tag1$$
Differentiating with respect to arc, the curvature
$$ =\dfrac{ d\phi}{ds} =2\; \dfrac{\cos \psi}{r}\cdot\dfrac{1-\log r^2}{r^2+(\log (r^2)/r)^2} \tag2 $$
From the above figure we have $ \phi= \psi+\theta$ so that this equals
$$=\dfrac{d\theta+ d\psi}{ds} =\psi'+ \dfrac{\sin \psi}{r}=2 \dfrac{\cos \psi}{r}\cdot\dfrac{1-\log r^2}{r^2+(\log (r^2)/r)^2} \tag3$$
Inflection points occur at radius $\sqrt{e}$ at zero curvature.
Integration has initial conditions in the center of interval at start point or "Nose" point (for lack of a classical term known to me) as $\psi=0,r=r_i$ and integrates numerically on either side of this start point.
Nose point or Power Tangent point
The "nose" can be defined for any continuous curve in the plane in one of the three following ways:
$$ \psi=0\quad; \dfrac{dy}{dx}=\dfrac{y}{x}; \quad \dfrac{dy}{dx}=\tan\theta \tag4 $$
The powers of a circle
$$ T^2= (x-h)^2+(y-k)^2-R^2$$
are obtained by setting $(x=y=0)$ to be $T^2= h^2+k^2-R^2$
According as the circle is outside, passes through origin or inside of the origin powers correspond to
$$T^2>0,\;T^2=0,\;T^2<0\;;$$
Plot of Curves

Tangency at the nose of curves at $(r_i=0,0.25,0.75,1.0,1.25,1.50\;)$are marked.
For these curves we have all powers $r_i^2=T^2$ positive.
A: I would say to simply draw the vector field.  Inside the unit circle, $y'$ is negative, so the flow vectors are down and to the right.  On the unit circle, $y'=0$, so the flow vectors are horizontal there.  Outside the unit circle, the flow vectors are up and to the right (though they become close to horizontal far from the unit circle).  So the solutions are essentially constant functions (increasing very slowly) for very positive and very negative initial values of $y$.  For intermediate initial values of $y,$ we will see an upward slope approaching the unit circle from left to right, a downward dive passing through the unit circle, and then an upward slope again after leaving the unit circle.  There is one solution, symmetric under a $180^\circ$ rotation, where $y\rightarrow 0$ as $x\rightarrow\pm\infty$.

Update.  The attached figure shows approximately what the solutions look like.  I misstated the behavior of the symmetric solution; it does not necessarily approach $y=0$ as $x\rightarrow\pm\infty$, but rather has two asymptotes, $y=\pm y_\infty$.

