I am stuck trying to solve the following differential equation in terms of distribution (theory).
$$x\frac{du}{dx} - \lambda u = 0.$$
This is just for $\mathbb{R}$ and $\lambda \in \mathbb{C}$.
I know the regular distribution corresponding to the function $f(x) = Ax^{\lambda}$ is a solution but the solution says that $c\delta$ is also a solution for some $c \in \mathbb{C}$ and I am not sure I can see how to derive this.
On
The equation is $$xu'-\lambda u = 0,$$ hence $$(xu)'-(\lambda+1) u = 0$$
If $\lambda =-1$, then the equation becomes $(xu)' = 0$, which then becomes $xu=c$ ($c$ is a constant). The latter equation has a solution $u = c\cdot pv (1/x) + c_1 \delta_0$ (this equation should be discussed in your textbook; it was also discussed on this site).
If on the other hand $\lambda \ne -1$, you can employ the usual techniques to solve this equation (eulerian chaange of variables $x=e^t$, for example).
You have to rewrite the equation itself in terms of distributions. You get then $x\frac{du}{dx}(\phi(x))=\lambda u(\phi(x))$. If $u$ is a distribution then $x\frac{du}{dx}(\phi(x))= -u((x\phi(x))^{'})$. So You have then $-u(x\phi^{'}(x)+\phi(x))=\lambda u(\phi(x))$. This equation is satisfied for $u=\delta$ distribution and $\lambda=-1$. See also here: http://mathworld.wolfram.com/DeltaFunction.html