Conjecture: Suppose we have a function $f(x)$ and some real number $\alpha$ such that $f(\alpha)=\alpha$ and $f'(\alpha)<1$. Then there exists a solution $y(t)$ to the differential equation $f(y)-y=y'$ such that $\lim_{t\to\infty}y(t)=\alpha$.
I've tested some functions and the conjecture always seems to hold:
For $f(x)=x^2, \alpha=0$ and the corresponding solution is $\dfrac{1}{ce^t+1}$, which goes to $\alpha=0$.
If $f(x)=\sin(x)+x, \alpha = (2k+1)\pi$, then $y(t)=2\cot^{-1}(ce^{-t})+2k\pi\to(2k+1)\pi$ and so on...
I would appreciate any kind of help, discussion or source for further reading. Thanks!
The conjecture is obviously correct, as $y(t)=\alpha$ satisfies the ODE. What you probably wanted to conjecture is that the steady state at $y=\alpha$ is locally attractive. The easiest to do this is by showing that it is locally exponentially stable by linearizing the ODE around $y=\alpha$:
$\frac{d}{dt}z=(-1+f'(\alpha))z+O(z^2)$,
with $z=y-\alpha$. Since $f'(\alpha)<1$, the only pole of the system, $(-1+f'(\alpha))$, is negative, the steady state thus locally exponential stable, and thus it is also locally attractive.