Distribution equation

Question

I am trying to solve $xT'+T = 0$ where $T$ is a distribution

I already know (it is asked in the previous point) that $(xT)'= T + xT'$

my thoughts : $ (xT)'(f) = - (xT)(f')$ and from this point I immediately think of $T$ equals the dirac measure at $0$ since $\int xδ(f'(x)) = 0*f'(0) = 0$, or an arbitrary constant $c$ times the dirac measure at $0$

is it correct ?

Answer 1

$(xT)' = 0$ means $xT = c$, then let $S = pv(1/x)$,

$x S = 1$ thus $x(T-c S) = 0$ which means $T-c S = b \delta$