Does the solution of $y' = (x^2 + y^2) e^{-(x^2+y^2)}$ have a limit for $x \to \infty$? An old exam problem I am trying to solve is as follows:
Given the cauchy problem $y' = (x^2 + y^2) e^{-(x^2+y^2)}, y(x_0) = y_0$, do the following:

*

*Show that there is a unique solution for all $x \in \mathbb{R}$

*Does the limit $ \lim_{x \to \infty} y(x) $ exist? Hint: evaluate the limit $ \lim_{x \to \infty} (x^2 + y^2) e^{-(x^2+y^2)} e^x$
For part (1), I managed to bound the functions $f(x,y) = (x^2 + y^2) e^{-(x^2+y^2)}$ and $f_y(x,y)$ under some fixed value. Then, I concluded, that we have a solution at some interval $[x_0 - \varepsilon, x_0 + \varepsilon]$ and we can extend this to $\mathbb{R}$ by moving to the right and left and applying the same result.
I had quite a hard time bounding the function $f_y(x,y)$, so first of all, I would be glad if someone shows me a quick and elegant way to do so.
Secondly, and more importantly, I don't know how to approach part (2). I suppose I should somehow bound the integral $\int  (x^2 + y^2) e^{-(x^2+y^2)}$ by $(x^2 + y^2) e^{-(x^2+y^2)} e^x$ and then show that this limit tends to $0$. But I have been so far unsuccesful with showing either of these two claims to be true. So any help here would be much appreciated!
 A: Part (1)
The right-hand side is locally Lipschitz continuous with respect to $y$, so that every solution is locally unique.
Since $0 \le u e^{-u} \le 1/e$ for all $u \ge 0$, $|y'(x)| \le 1/e$ and the inequality
$$
  y_0 - \frac 1e |x-x_0| \le y(x)  \le y_0 + \frac 1e |x-x_0|
$$
holds on the maximal interval of existence, which implies that the solution exists on all of $\Bbb R$. (The solution can not “blow up,“ compare e.g. Maximal interval of existence of an ODE $y'=f(y), y(0)=a.$)
Part (2)
$u \mapsto u e^{-u}$ is decreasing for $u \ge 1$, so that $(x^2 + y^2) e^{-(x^2+y^2)} \le x^2 e^{-x^2}$ for $x \ge 1$ and arbitrary $y\in \Bbb R$. It follows that for $1 \le x_1 < x_2$
$$
 0 \le y(x_2) - y(x_1) = \int_{x_1}^{x_2} y'(t) \, dt 
\le \int_{x_1}^\infty t^2 e^{-t^2} \, dt
$$
and that is arbitrarily small for sufficiently large $x_1$ because the integral on the right converges.
It follows that $y(x_n)$ is a Cauchy sequence (and therefore convergent) for every sequence $(x_n)$ converging to $+\infty$. This implies that $\lim_{x \to \infty} y(x)$ exists.
Or, slightly simpler: $\lim_{x \to \infty} y(x)$ exists because $y$ is increasing and bounded above: For $x > 1$ is
$$
 y(x) - y(1) = \int_{1}^{x} y'(t) \, dt 
\le \int_{1}^\infty t^2 e^{-t^2} \, dt \, .
$$
