To practice ordinary differential calculus, I set myself a few problems to solve. One of those problems is "Solve $y(x)^{y'(x)} = |x|^{|x|}$ for $y(x)$!" with $x \in \mathbb{R} \backslash \left\{ 0 \right\}$ and $y(x) \in \mathbb{C} \backslash \left\{ 0 \right\}$.
So I started: $$ \begin{align*} y(x)^{y'(x)} &= |x|^{|x|} \quad\mid\quad \ln\left( ~~ \right)\\ \ln\left( y(x)^{y'(x)} \right) &= \ln\left( |x|^{|x|} \right)\\ y'(x) \cdot \ln\left( y(x) \right) &= |x| \cdot \ln\left( |x| \right) \quad\mid\quad \int ~\operatorname{d}x\\ \int y'(x) \cdot \ln\left( y(x) \right) ~\operatorname{d}x &= \int |x| \cdot \ln\left( |x| \right) ~\operatorname{d}x + c_{1}\\ -y(x) + \ln\left( y(x)^{y(x)} \right) &= \int |x| \cdot \ln\left( |x| \right) ~\operatorname{d}x + c_{1}\\ \end{align*} $$
Since the equation reminds me of power towers, I tried to solve the equation with the lambert W-function: $$ \begin{align*} -y(x) + \ln\left( y(x)^{y(x)} \right) &= \int |x| \cdot \ln\left( |x| \right) ~\operatorname{d}x + c_{1}\\ -y(x) + \ln\left( y(x) \right) \cdot y(x) &= \int |x| \cdot \ln\left( |x| \right) ~\operatorname{d}x + c_{1}\\ -y(x) + \ln\left( y(x) \right) \cdot e^{\ln(y(x))} &= \int |x| \cdot \ln\left( |x| \right) ~\operatorname{d}x + c_{1}\\ \end{align*} $$
Now comes the problem: I don't know how to continue. Is there a nice way to continue?
Wolfram|Alpha tells me that there is solution to this formula via using the lambert W-function: $$y(x) = \frac{\int |x| \cdot \ln\left( |x| \right) ~\operatorname{d}x + c_{1}}{\operatorname{W}\left( \frac{\int |x| \cdot \ln\left( |x| \right) ~\operatorname{d}x + c_{1}}{e} \right)} \text{ or } y(x) = \frac{\int^{x}_{1} |\xi| \cdot \ln\left( |\xi| \right) ~\operatorname{d}\xi + c_{1}}{\operatorname{W}\left( \frac{\int^{x}_{1} |\xi| \cdot \ln\left( |\xi| \right) ~\operatorname{d}\xi + c_{1}}{e} \right)}$$
Just how?
$$ \begin{align} -y(x) + \ln\left( y(x) \right) y(x)&=E \\ y(\ln y -1)&=E\\ y(\ln y -\ln e)&=E\\ y(\ln (y/e)=e\dfrac y e \ln \dfrac ye&=E\\ e^{\ln \dfrac y e} \ln \dfrac ye&=\dfrac E e\\ \end{align} $$ Take the Lambert function on both sides: $$\ln \dfrac y e=W \left (\dfrac E e\right)$$ $$y =ee^{W \left (\dfrac E e\right)}$$ and use $e^{W(y)}=\dfrac y {W(y)}$: $$y(x) =\dfrac E {W \left (\dfrac E e\right)}$$ Where $E= \int_1^x t \ln t dt+C$.