I'm in trouble with this exercise:
If T $\in D’$ is a solution of the equation $xT'=-T$ then I have to prove that there exists a constant $C \in \mathbb{R}$ such that the distribution $P=T-C \cdot PV(1/x)$ is zero in $\mathbb{R}$ without the origin. After that, I have to prove that $P$ is supported in zero and, using this fact, to find all solutions to the above equation.
Firstly I proved that if $T$ solves the above equation then $T$ solves also $(xT)'=0$ but I don't know if this could be usefull. But know I'm stuck. Some ideas?
Here $PV(1/x)$ is the principal value of $(1/x)$ and it is such that $\langle PV(1/x),\phi\rangle := \lim_{\epsilon \to 0^{+}} \int_{\mathbb{R} \setminus (-\epsilon,\epsilon)} (1/x)\phi(x)dx$
where $\phi$ is a test function
Your first step is correct; indeed $(xT')=0$. It implies that $xT = C$ for a constant $A$ (this is an exercise covered in all textbooks and on Math.SE).
Now consider $S=T - C\cdot PV(1/x)$ and take $\phi\in D$ such that $supp\, \phi$ is separated from zero. What can you say about $\frac{\phi(x)}{x}$? What happens when we apply $$\langle S ,\phi\rangle =\left \langle S, x \cdot \frac{\phi(x)}{x}\right\rangle?$$