Bogolyubov averaging method I need to solve the following problem using averaging method:
$$\frac{dx}{dt}=-2x+\varepsilon x^3+\varepsilon sin(\frac{2t}{\varepsilon}),\\
x(0)=1$$
Unfortunately, I am stuck at the very beginning. The Bogolyubov method suggests solution for equations of the following form:
$$\frac{dx}{dt} = \varepsilon F(x, t)$$
The main issue is that $\varepsilon$ is in denominator which does not allow to take it equal to $0$ (as Bogolyubov method assumes).
Could you please advice a method to solve this equation or a transformation that can adapt the source equation to standard Bogolyubov method?
Thanks in advance.
 A: The averaging method is in essence a technique of removing the very-small-$\epsilon$ behavior from the differential equation, and later replacing it in a context which is more amenable to perturbation theory.  The key point in this problem is that for small $\epsilon$, $\epsilon\sin(\frac{2t}\epsilon)$ is in fact well behaved.
We remove, temporarily, the $\epsilon$-related behavior by noting that as $\epsilon\to 0$, we can approximate $\sin(\frac{2t}\epsilon)\approx 2t$.  Solving
$$
\frac{d\bar{x}}{dt}=-2\bar{x}-2t
$$
is done by writing $\bar{x}(t) = u(t)+t$ and solving the now-separable equation for $u(t)$.  Imposing the boundary condition $\bar{x}(0) = 1$ we get
$$
x(t) = \frac32 e^{-2t}+t - \frac12
$$
Now let $x(t) = \bar{x}(t)+s(t) = s(t) + \frac32 e^{-2t}+t - \frac12$ and the original equation becomes 
$$
\frac{ds}{dt} = \epsilon \left( s- \frac32 e^{-2t}-t + \frac12 \right)^3
+\epsilon \left( \sin (\frac{2t}\epsilon)-\frac{2t}{\epsilon} \right)
$$ with $s(0)=0$, 
and the point is that when $\epsilon$ is small the last term remains small until
$t > \frac1{\epsilon}$,  so a perturbative solution will work until $t$ is that large.  (Actually, the cubed term becomes large when $t\sim \epsilon^{-1/3}$.)
