Solve $f(x+f(2y))=f(x)+f(y)+y$ 
Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ such that for each $x$ and $y$ in $\mathbb{R}^+$, $$f(x+f(2y))=f(x)+f(y)+y$$

Note:
$f(x)=x+b$ is a solution for all  $b\in\mathbb{R}^+$ but I can not prove it.

Another solution is
$f(x) =-x/2$.
Here,
at the end of my ramblings,
is how I found it.
Inspired by the comment
that a solution is
$x+b$,
let
$f(x) = x+g(x)$.
Then
$x+2y+g(2y)+g(x+2y+g(2y))
=x+g(x)+y+g(y)+y
$
or
$$g(2y)+g(x+2y+g(2y))
=g(x)+g(y).
$$
Letting $y=0$,
this is
$g(0)+g(x+g(0))
=g(x)+g(0)
$
or
$g(x) = g(x+g(0))$.
If $g(0) = 0$,
this tells nothing.
If $g(0) \ne 0$,
$g$ is periodic
with period $g(0)$.
In particular,
$g(0) 
= g(g(0))
= g(g(g(0)))
=...
$.
Setting $x=0$,
this is
$g(2y)+g(2y+g(2y))
=g(0)+g(y)
$.
This doesn't seem to help much.
If $x=2y$,
this is
$g(2y)+g(4y+g(2y))
=g(2y)+g(y)
$
or
$g(y)
=g(4y+g(2y))
$,
so
$g(y+g(0))
=g(4y+g(2y))
$.
If
$y+g(0)
=4y+g(2y)$,
then
$g(2y)
=g(0)-3y
$
or
$g(y)
=g(0)-3y/2
$.
This suggests trying
$f(x) = ax+b$.
Then
$a(x+2ay+b)+b
=a(x+y)+2b+y
$
or
$2a^2y+ab
=(a+1)y+b
$
or
$0
=(2a^2-a-1)y+b(a-1)
$.
Therefore
$0=(2a^2-a-1$
or
$a
=\dfrac{1\pm\sqrt{1+8}}{4}
=\dfrac{1\pm 3}{4}
=1, -\frac12
$.
If $a=1$
then any $b$ works.
If
$a=-\frac12$,
then
$b=0$,
so
$-x/2$
is a solution.
To check,
$f(x+f(2y))
=f(x-y)
=(y-x)/2
$
and
$f(x)+f(y)+y
=-x/2-y/2+y
=(y-x)/2
$.
That's all I have.
