Solving $(x+y) \exp(x+y) = x \exp(x)$ for $y$.

While thinking about the Lambert $W$ function I had to consider

Solving $(x+y) \exp(x+y) = x \exp(x)$ for $y$.

This is what I arrived at:

(for $x$ and $y$ not zero)

$(x+y) \exp(x+y) = x \exp(x)$

$x\exp(x+y) + y \exp(x+y) = x \exp(x)$

$\exp(y) + y/x \exp(y) = 1$

$y/x \exp(y) = 1 - \exp(y)$

$y/x = (1-\exp(y))/\exp(y)$

$x/y = \exp(y)/(1-\exp(y))$

$x = y\exp(y)/(1-\exp(y))$

$1/x = 1/y\exp(y) -1/y$

And then I got stuck.

Can we solve for $y$ by using Lambert $W$ function?

Or how about an expression with integrals?

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To solve $exp(a)=exp(a+b)$ we can do for $a$ not $0$ : $f_1(a) = ln(-exp(a))$ Then $f_1(f_1(a)) = a + 2 \pi i$ which is a good answer. Likewise let $f(a) = W(-a e^a)$ , then $f(f(a)) = a + b$ wich is probably a solution to $(a+b) e^{a+b} = a e^a$ for $a$ and $b$ not $0$. Maybe I should post this as the answer. However I still do not know if $f(f(a))$ can be simplified or written as an integral ? –  mick Sep 20 '12 at 21:24

The solution of $(x+y) \exp(x+y) = x \exp(x)$ is given in terms of the Lambert W function
Let $z=x+y$, then we have
$$z {\rm e}^{z} = x {\rm e}^{x} \Rightarrow z = { W} \left( x{{\rm e}^{x}} \right) \Rightarrow y = -x + { W} \left( x{{\rm e}^{x}} \right) \,.$$
Added: Based on the comment by Robert, here are the graphs of $y = -x + { W_0} \left( x{{\rm e}^{x}} \right)$ and $y = -x + { W_{-1}} \left( x{{\rm e}^{x}} \right)$
Hmm but if $W(xe^x) = x$ , then $y = -x + x = 0$ ? –  mick Sep 20 '12 at 21:04
There are different branches of the Lambert W function. In Maple's numbering, $W_0(x e^x) - x = 0$ for $x \ge -1$, while $W_{-1}(x e^x) - x = 0$ for $x \ge -1$. For $x \le -1$, $W_0(x e^x) - y$ is the nonzero real solution of $(x+y) e^{x+y} = x e^x$. For $-1 \le x < 0$, $W_{-1}(x e^x) - x$ is the nonzero real solution. –  Robert Israel Dec 25 '12 at 5:14