Deriving a differential equation for the Lambert W function. I would like to know if I am correct in the following:
Let $f(x) = e^{x}x$, and $g(x) = w(x)$, where $w(x)$ is the Lambert W function.
By the rule that inverse function integral relation, which states that:
$$\int_{a}^{b} f(x) \,dx = bf(b) - \int_{f(a)}^{f(b)} g(x) \,dx$$
$$\int_{a}^{b} e^{x}x \,dx = e^{b}b - \int_{f(a)}^{f(b)} W(x) \,dx$$
One can show that:
$$\int e^{x}x \,dx = e^{x}(x - 1) + C$$
Let $W(x)$ be the antiderivative of the Lambert W function, $w(x)$, then:
$$W(be^{b}) - W(ae^{a}) = e^{a}(a - 1) + e^{b}$$
$$\int_{ae^{a}}^{be^{b}} w(x) \,dx = e^{a}(a - 1) + e^{b}$$
Let $a = 0$, then:
$$\int_{0}^{be^{b}} w(x) \,dx = e^{b} - 1$$
Therefore:
$$\int_{0}^{x} w(t) \,dt = \frac{x}{W(x)} - 1$$
$$w(x) = \frac{w(x) - xw'(x)}{w^{2}(x)}$$
$$w^{3}(x) = w(x) - xw'(x)$$
$$w^{3}(x) - w(x) + xw'(x) = 0$$
I am unsure as to how to progress, or if these steps are even correct.
I would appreciate any help.
Thank you.
 A: Here is what I tried: The function $x\mapsto w(x)$ in question satisfies
$$x=w(x)e^{w(x)},\qquad w(0)=0\ .\tag{1}$$
Differentiating with respect to $x$ we obtain
$$1=w'(x)e^{w(x)}\bigl(1+w(x)\bigr)\ .$$
We now substitute $e^{w(x)}={x\over w(x)}$ from $(1)$ and get the followsing differential equation:
$$w'(x)={w(x)\over x\bigl(1+w(x)\bigr)}\ .$$
A: I think there are some errors. The theorem you cite says that holds $$
\int_{a}^{b}f\left(x\right)dx=bf\left(b\right)-af\left(a\right)-\int_{f\left(a\right)}^{f\left(b\right)}f^{-1}\left(x\right)dx
 $$ then if we put $f\left(x\right)=xe^{x}
 $ and $f^{-1}\left(x\right)=w\left(x\right)
 $ we have $$\int_{a}^{b}xe^{x}dx=b^{2}e^{b}-a^{2}e^{a}-\int_{ae^{a}}^{be^{b}}w\left(x\right)dx
 $$ then $$\int_{ae^{a}}^{be^{b}}w\left(x\right)dx=b^{2}e^{b}-a^{2}e^{a}-e^{b}\left(b-1\right)+e^{a}\left(a-1\right)
 $$ now if we take $a=0
 $ we get $$\int_{0}^{be^{b}}w\left(x\right)dx=b^{2}e^{b}-e^{b}\left(b-1\right)-1
 $$ so if we put $z=be^{b}=w\left(z\right)e^{w\left(z\right)}
 $ we have $$W\left(z\right)=\int_{0}^{z}w\left(x\right)dx=w^{2}\left(z\right)e^{w\left(z\right)}-e^{w\left(z\right)}\left(w\left(z\right)-1\right)-1
 $$ now if you take the derivative, you get a differential equation.
