what can be said about the solution of the following differential equation $y(x)=f(x)+\log(1+y'(x))$ with $\lim_{x \to 0}f(x)=\infty,f(\infty)=0$ and $f(x)$ is contiuoues and decreasing. In particular, can we prove that the solution $y(x)$ is convex in general?
from $e^{y(x)-f(x)}=1+y'(x)$ we get $1+y'(x) > 0$ and $y'(x) > -1$. Then it follow that $y(x) \leq f(x)$ 
$1+y'(x) = e^{y(x) - f(x)} \leq 1$ so also we have $y'(x) \leq 0$, so generally:
$$-1 < y'(x) \leq 0$$
Another observation: Take the derivative:
$y'(x)=f'(x)+\frac{y''(x)}{1+y'(x)}$ so: $$(y'(x)-f'(x))(1+y'(x))= y''(x)$$
and if $y'(x)-f'(x) \geq 0$ the function $y(x)$ is convex. When $f'(x) < -1$ this is always true, but what happend for $-1 \leq f'(x) \leq 0$?

If you put $z(x)=x+y(x)$ and $g(x)=x+f(x)$, the equation becomes $$z'(x)=e^{z(x)-g(x)},$$
which is separable, with general solution
$$z=-\ln\bigl(C-h(x)\bigr),\qquad h(x)=\int_0^x e^{-g(t)}\,dt.$$
Then
$$z'(x)=\frac{e^{-g(x)}}{C-h(x)},\quad
  z''(x)=\frac{e^{-2g(x)}-(C-h(x))e^{-g(x)}g'(x)}{(C-h(x))^2},$$
so the sign of $z''(x)$ is that of $e^{-g(x)}+\bigl(h(x)-C\bigr)g'(x)$.
It follows from the assumptions that $h(\infty)$ exists and is finite, and you need $C\ge h(\infty)$ for the solution to exist for all $x$, and so the second term above is negative, if $g'(x)>0$. In typical cases $g'(x)\sim 1$ for large $x$, so $z$ (and hence $y$) becomes concave for large $x$, at least if $C>h(\infty)$
