Consider the ordinary differential equation $\dfrac{\text{d}y}{\text{d}x}=\dfrac{y-3}{x^2+y^2},$ with $y(0)=1$. My question is about determining the graph of $y$. Here is most of the information that I have gathered, arguing somewhat heuristically:
$1$. For all $x>0$, $y$ is decreasing. Also, $\dfrac{\text{d}^2y}{\text{d}x^2} = -10$ at $x=0$, so $y$ is concave downwards at this point.
$2$. $\displaystyle \lim_{x \rightarrow \infty} \dfrac{\text{d}y}{\text{d}x}=0$, hence $y$ is bounded below and has a horizontal asymptote. Similarly, $\displaystyle \lim_{x \rightarrow -\infty} \dfrac{\text{d}y}{\text{d}x}=0.$
$3$. By formulating a quadratic in terms of $y$ using the given D.E. and considering its discriminant, I can show (using contradiction) that $y$ must intersect the $x$-axis and hence will eventually be negative. This implies the lower bound is negative, and the horizontal asymptote lies below the $x$-axis. Hence the second derivative must be zero for some positive $x$, and $y$ changes concavity there.
$4$. $y$ attains its maximum for some (maximal) $x_0<0$, and $y(x_0) = 3$. So either $y$ is the constant function $y(x)=3$ for all $x<x_0$, or we have a local maximum that is also a unique stationary point, since $\dfrac{\text{d}y}{\text{d}x}=0 \iff y=3.$ For the latter, $y$ intersects the $x$ axis for some negative $x$, and there is another horizontal asymptote (the two horizontal asymptotes could be equal).
Would it be possible to determine the following?
- $ -2 \leq \dfrac{\text{d}y}{\text{d}x} < 0$ for all positive $x$.
- Coordinates of the point(s) where the second derivative equals zero ($y$ changes concavity).
- Concavity of the graph for negative $x$.
- The coordinates of the point(s) where $y$ intersects the $x$-axis, or give a bound.
- The equation of the horizontal asymptote(s), or give a bound.
- Existence of the turning point (maximum, assuming $y$ does not become constant for sufficiently small $x$), and its $x$-coordinate, or give a bound for it/them.
For reference, here are the possible shapes of the graph that I have in mind. Any help (including any other interesting key features that I might have missed out) will be much appreciated.
Decompose the equation into a system $$ \dot x(t)=x(t)^2+y(t)^2,\\ \dot y(t)=y(t)-3,\\ x(0)=x_0,~~y(0)=y_0. $$ The second equation has a simple exponential as solution $$ y(t)=3+(y_0-3)e^t. $$ With that, the first equation is of type Riccati. Setting $x=-\frac{\dot u}{u}$, $u(0)=1$, $\dot u(0)=-x_0$, transforms this into $$ \ddot u(t) + y(t)^2 u(t)=0. $$ In both time directions $y(t)$ is eventually bounded away from zero, so that the Sturm-Picone comparison theorem guarantees the existence of roots of $u$. As these are poles for $x$, the domain of $x$ is a bounded interval, its image is $\Bbb R$ and the image of $y$ is bounded on both sides.
Translated back into a $x$-$y$ relation, this means indeed that the solution $y(x)$ has horizontal asymptotes. It is not obvious if $x(t)$ is monotonous