Analyze $y' = y\cdot \cos(x+y)$ Anyone familiar with Mathematica will realize this ODE is the first example of the NDSolve command's documentation, a screenshot of which is shown below

I am wondering if it possible to get some insight of this ODE analytically. For example, I tried the standard perturbation approach. First introduce a small parameter $\epsilon$ so that the ODE becomes
$$
y' = y \cdot \cos(x + \epsilon y)
$$
The initial conditions is still $y(0) = 1$. Substitute $y = y_0 + \epsilon y_1$ and expand, the 0th order system is
$$
y_0'=y_0 \cos(x), y_0(0) = 1
$$
The solution is $y_0(x) = e^{\sin(x)}$. The 1st order system is
$$
y_1' = y_1 \cos(x)  -\sin(x) e^{2 \sin(x)}, y_1(0) = 0
$$
Unfortunately, there is no analytical solution for $y_1$ and I got stuck. Are there more advanced analytical methods that can extract something from this equation? For example, to predict that there is a decreasing trend of the function? Or even get the rate of this trend? Thanks!
Update (Aug. 14)
A minor improvement is to use the ansatz $y = f(x)e^{\sin(x)}$, substitute to have
$$
f' + f\cos(x) = f\cos(x + f e^{\sin(x)}); f(0) = 1
$$
If we lump $e^{\sin(x)}$ to be the average over $2\pi$, which is about 1.27; and we further use $f\sim 1$ in the $\cos$ function to linearize it, we have:
$$
f' + f\cos(x) = f\cos(x + 1.27); f(0) = 1
$$
Solve $f$ and $f e^{\sin(x)}$ is plotted (blue) below together with the "exact" numerical solution (yellow).

The analytical solution roughly captures the solution behavior up to $x \sim \pi$, and becomes useless for larger $x$. I would appreciate it if anyone can bring up a method that captures the "long-term" solution behavior.
Update 2 (Aug. 16) 
Having no clue of analytical methods, I decide to play with the numbers. Putting the numerical solution in log-log plot suggests overall the function decreases as $1/x$. For large $x$, $y \ll x$ in the cosine function, and $y' = y \cos(x)$ gives $y \propto e^{\sin(x)}$. Unfortunately, the decreasing trend is NOT captured at all! 
Anyway, piecing all the hints together, and with some eyeballing, I got to "fit" a function $2 e^{\sin(x)} /x$. It is plotted below (black, oscillating curve). Shown together are two envelope curves $5/x$ and $0.6/x$ (black, smooth curves), and the numerical curve (red, oscillating curve).
I do feel the leading order behavior is captured--I may still be wrong, but how to derive it analytically?

 A: Hint:
$y'=y\cos(x+y)$
$\dfrac{dy}{dx}=y\cos(x+y)$
$\dfrac{dx}{dy}=\dfrac{1}{y\cos(x+y)}$
Let $u=x+y$ ,
Then $x=u-y$
$\dfrac{dx}{dy}=\dfrac{du}{dy}-1$
$\therefore\dfrac{du}{dy}-1=\dfrac{1}{y\cos u}$
$\dfrac{du}{dy}=\dfrac{1}{y\cos u}+1$
Let $u=v-\dfrac{\pi}{2}$ ,
Then $\dfrac{dv}{dy}=\dfrac{1}{y\cos\left(v-\dfrac{\pi}{2}\right)}+1$
$\dfrac{dv}{dy}=\dfrac{1}{y\sin u}+1$
Let $w=\tan\dfrac{v}{2}$ ,
Then $v=2\tan^{-1}w$
$\dfrac{dv}{dy}=\dfrac{2}{w^2+1}\dfrac{dw}{dy}$
$\therefore\dfrac{2}{w^2+1}\dfrac{dw}{dy}=\dfrac{w^2+1}{2wy}+1$
$\left(y+\dfrac{w^2+1}{2w}\right)\dfrac{dy}{dw}=\dfrac{2y}{w^2+1}$
This belongs to an Abel equation of the second kind.
Let $z=y+\dfrac{w^2+1}{2w}$ ,
Then $y=z-\dfrac{w}{2}-\dfrac{1}{2w}$
$\dfrac{dy}{dw}=\dfrac{dz}{dw}-\dfrac{1}{2}+\dfrac{1}{2w^2}$
$\therefore z\left(\dfrac{dz}{dw}-\dfrac{1}{2}+\dfrac{1}{2w^2}\right)=\dfrac{2}{w^2+1}\left(z-\dfrac{w^2+1}{2w}\right)$
$z\dfrac{dz}{dw}-\dfrac{(w^2-1)z}{2w^2}=\dfrac{2z}{w^2+1}-\dfrac{1}{w}$
$z\dfrac{dz}{dw}=\dfrac{(w^4+4w^2-1)z}{w^2(w^2+1)}-\dfrac{1}{w}$
