Differential Equation and Limit of Solution Let $g: \mathbb{R} \to \mathbb{R}$ be a continuous function and consider
\begin{align}
x'(t) = - \frac{2t}{1+t^2}\cdot x(t) + g(t).
\end{align}
Let $x(t)$ be a solution of this equation. How can I show that $\lim_{t \to \infty} x(t) = 0$ if $\lim_{t \to \infty} t\cdot g(t) = 0$?
I guess my most promising attempt is to reduced the problem to the equivalent problem of showing
\begin{align}
\lim_{t \to \infty} t\cdot x'(t) = -2 \cdot \lim_{t \to \infty} x(t), 
\end{align}
which follows when you multiply the first equation with $t$ and take the limit of both sides. Does anyone see how I can continue the argument?
 A: The differential equation is equivalent to
$$ ((1+t^2) x)' = g(t)(1+t^2),$$
when integrated, we have
$$x(t) = \frac{1}{1+t^2}\left( \int_0^t g(s)(1+s^2) ds - x(0)\right).$$
The term $\frac{x(0)}{1+t^2}$ goes to zero as $t\to +\infty$. So you are asking if
$$\frac{1}{1+t^2}\int_0^t g(s)(1+s^2) ds$$
tends to zero as $t\to +\infty$ given the condition on $g$.
Now $tg(t) \to 0$ when $t\to \infty$, so for all $\epsilon >0$, there is $M>0$ so that
$$ |sg(s)|< \epsilon \frac{s^2}{1+s^2}$$
for all $s\ge M$. Thus when $t \ge M$,
\begin{align}
\left| \frac{1}{1+t^2} \int_0^t g(s) (1+s^2) ds\right| &\le \frac{1}{1+t^2} \int_0^M |g(s)| (1+s^2) ds+\frac{1}{1+t^2} \int_M^t |g(s)| (1+s^2) ds \\
& \le \frac{1}{1+t^2} \int_0^M |g(s)| (1+s^2) ds + \frac{\epsilon}{1+t^2} \int_M^t s ds \\
&< \frac{1}{1+t^2} \int_0^M |g(s)| (1+s^2) ds  + \frac{\epsilon}{2}.
\end{align}
By choosing $M_1 \ge M$ large enough, the first term on the right hand side is $\le \frac{\epsilon}{2}$. Thus
$$\left| \frac{1}{1+t^2} \int_0^t g(s) (1+s^2) ds\right|  <\epsilon$$
whenever $t \ge M_1$. So $x(t) \to 0$ as $t\to +\infty$.
