Consider the following IVP non-linear ODE about $y(x)$:
$$
y'(x)=\frac{1}{e^{y(x)}+x^2}=:f(x,y), y(x_0)=y_0
$$
I try to show that the existence and uniqueness of $y(x)$ for $x\ge x_0$ and the existence of $\lim_{x\to \infty} y(x)$.
I am stuck on the uniqueness of $y(x)$ and the existence of $\lim_{x\to \infty} y(x)$.
My proof is as follows.
(1)Show that the locally existence of the solution to IVP.
I try to check the condition in Picard–Lindelöf theorem. Here we consider a closed rectangle $D\subset R\times R$ with $(x_0, y_0)\in int D$ and write $D=\{(x,y): |x-x_0|<A, |y-y_0|<B\}$. Here $f(x,y)$ is continuous in $x$. Also, for $(x,y_1), (x,y_2)\in D$,
$$
|f(x,y_1)-f(x,y_2)|=\frac{|e^{y_2}-e^{y_1}|}{(x^2+e^{y_1})(x^2+e^{y_2})}<\frac{|e^{y_2}-e^{y_1}|}{x^4}\le K|y_1-y_2|
$$
where we use that $|e^{y_2}-e^{y_1}|\le M|y_2-y_1|$ by mean value theorem and $e^y>0$ and $x>x_0-A$.
Hence, there exists $\epsilon>0$, such that the initial value problem has a unique solution $y(x)$ on $[x_0-\epsilon, x_0+\epsilon]$.
(2) Show that the solution can be extended globally.
I try to use Theorem 3 in https://q.utoronto.ca/courses/107052/files/4402384/download?wrap=1:
Let $(a,b)$ be a maximal interval of existence of a solution of $x'=f(x)$, then either $a=a_0$ or $\lim_{t\to a^{+}}|x(t)|=\infty$.
My proof for existence of $y(x)$ on $x\ge x_0$ is as follows.
Let $(a,b)$ be the maximal interval of existence for the solution. If $y(x)$ does not exist globally for $x\ge x_0$, then $y(x)$ only exists on a interval $(x_0,b)$ with $b<\infty$.
Note that since $y'>0$ and $y$ is increasing , then $e^y >e^{y_0}$ for $x\ge x_0$,
$$
y'(x)\le \frac{1}{x^2+e^{y_0}}
$$
Then integrate on the both side $$
y(x)\le y_0+\int_{x_0}^x \frac{1}{s^2}ds=y_0+e^{-y_0/2}[arctan(xe^{-y_0/2})-arctan(x_0e^{-y_0/2})
$$
By Theorem 3, we know that
$$
\lim_{x\to b^-} |y(x)|=\infty.
$$
But note that
$$
\lim_{x\to b^-} y(x)<y_0+e^{-y_0/2}[arctan(b e^{-y_0/2})-arctan(x_0e^{-y_0/2})<\infty
$$
So $$\lim_{x\to b^-} |y(x)|<\infty$$, which is a contradiction.
Hence, $y(x)$ exists globally.
(3) Existence of limit of $y(x)$
Since
$$
y(x)\le y_0+e^{-y_0/2}[\frac{\pi}{2}-arctan(x_0e^{-y_0/2})]
$$
Then we take $x\to \infty$,
$$
\lim_{x\to\infty} <\infty
$$
So the limit exists.
Best Answer
At the current state of the question text, you have handled the local existence and uniqueness. In general it is sufficient to notice that the ODE function $f$ is continuously differentiable, as then $\partial_yf(x,y)$ is continuous, and its maximum over a bounded set is a constant for the local Lipschitz condition.
It is a seemingly trivial fact, but requires a non-constructive proof that a unique solution with a maximal domain exists. One characteristic property is that it will leave any compact set inside the domain of the ODE.
As the function is growing and as observed bounded above by $y_0+x_0^{-1}$, one can construct compact rectangles $[x_0,x_1]\times[y_0,y_0+x_0^{-1}]$ that the solution can only leave to the vertical sides, independent of how large $x_1$ is. Thus the maximal solution is defined at least over $[x_0,+\infty)$.
Note that your proof for the bound requires $x_0>0$. The bound can be improved to avoid that singularity by using $e^{y_0}$ as lower bound of the exponential instead of zero.
You got an always positive derivative, $y(x)$ is strictly monotonous, thus one can switch the independent variable to $y$. Horizontal asymptotes of $y(x)$ switch to vertical asymptotes or pole singularities of $x(y)$. $$ x'(y)=x(y)^2+e^y. $$ This is a Riccati equation, where such singularities are routine. One can apply standard upper and lower bounds to enclose the singularities to both sides. (Which circles back to the bounds as in the OP text that are probably expected in a direct treatment of the task.)
One can also apply a standard substitution $x(y)=-\frac{u'(y)}{u(y)}$ to get $$ 0=u''(y)+e^yu(y). $$ This is an elliptic second order equation, the frequency factor $e^y$ is always positive. Now one can apply Sturm-Picone comparison theory to get an idea of the oscillation.
Both of these transformed problems might provide more intuitive improvements to the bounds and estimates for the location of the asymptotes.