how to show that all solutions tend to zero? Here is our nonlinear first order ode:
\begin{equation*}
y'(t) +2y(t)+y^3(t)=e^{-t} .
\end{equation*}
We want to show that all solutions tend to zero as $t$ goes to infinity.
Attempt:
Multiply both side by $y$. Then we will have:
\begin{equation*}
(1/2)(y^2)'=-2y^2-y^4+ye^{-t}, \\
(1/2)(y^2)'\le ye^{-t}.
\end{equation*}
After this, I think Grönwall's inequality may help but we are not sure of $y\le y^2$. If this was the case:
\begin{equation*}
(1/2)(y^2)'\le y^2e^{-t}
\end{equation*}
then by Grönwall: $y=0$. However $y=0$ also does not satisfy the ode anyway.
 A: I would rewrite it to
$$ y' = -2(y-\tfrac12 e^{-t})-y^3 $$
Intuitively the first term pulls $y$ towards $\frac12 e^{-t}$ and the second term pulls it towards $0$. So eventually it should be expected to end up between those targets (and thus converge towards $0$).
More formally, if you have a starting condition, you can bound the solution from below by the solution to $ y' = -(2y+y^3) $ (which will easily converge to $0$).
From above is a little more involved, with a bit of footwork it can be done. Let $z(t)$ be a solution to $z'= -z^3$ such that for some $t_0>1$ where $y$ is defined, we have $z(t_0) > \max(y(t_0),e^{-t_0})$. Then it is impossible for $z(t)$ to cross below $\frac12e^{-t}$ because at the intersection we would have
$$ \frac{d}{dt}(z-\tfrac12e^{-t}) = -(\tfrac12e^{-t})^3+\tfrac12te^{-t} = \tfrac12e^{-t}(t-\tfrac14e^{-2t}) $$
which is positive when $t>1$.
This we always (for $t>1$) have $z(t)>\frac12e^{-t}$. But that means it is impossible for $y(t)$ to cross above $z(t)$, because in the region above $\frac12e^{-t}$ we would have $y' < z'$ at the point of intersection.
So $y$ is bounded from above by $z$ everywhere to the right of $t_0$, and $z$ goes towards $0$ (not quickly, but it does).
