Long term property of solutions of a nonlinear first order ODE Let $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ with $|f(t, x)| \leq \phi (t) |x| \ \forall (t, x) \in \mathbb{R} \times \mathbb{R}$, where $\int_{- \infty}^{\infty} \phi < \infty$.
Consider the differential equation $\dot x = - x + f(t, x)$. Show that every solution goes to zero as $t \rightarrow \infty$.
My idea is to first show that $x(t)$ converges as $t \rightarrow \infty$ by showing that it is Cauchy and then try to get the result, but I cannot even show the former. Beyond that, I really don't have much thought. Any ideas?
 A: The equation $\dot x=-x$ has solutions of the form $Ce^{-t}$. To get the nonlinear part $f$, use the variation of parameter: write $x=e^{-t}y$ where $y$ is another unknown function. In terms of $y$, the ODE becomes
$$\dot y=e^t f(t,e^{-t}y) \tag1$$
which according to your assumption implies 
$$|\dot y| \le \phi(t) |y| \tag2$$
Now this looks like first-order linear equation, traditionally solved with integrating factor $e^{\Phi}$, where $\Phi$ is an antiderivative of $\phi$. Well, (2) is actually an inequality, but it goes the way that helps us: bounding the growth of $y$. So we should expect   $|y| =O( e^{\Phi})$. Here is a sketch of the proof that $e^{-\Phi}y$ is bounded from above. 


*

*if $y(t)\le  0$, then $e^{-\Phi}y\le 0$ (duh)

*if $y(t)>0$, then 
$\dfrac{d}{dt}(e^{-\Phi}y)=-\phi e^{-\Phi}y + e^{-\Phi}\dot y\le 0$, by (2)


The bound from below works exactly the same: after all, (2) stays the same if we swap $y$ for $-y$. 
So, $e^{-\Phi}y$ is bounded. But $\Phi$ has a finite limit at infinity, due to $\phi$ being integrable. Thus, $y$ is bounded, and $x=e^{-t}y =O(e^{-t})$.
