Method for solving distributional ode

Question

Let $u$ be a distribution on the real line. I know that the equation $x(u-u')= \delta_0 -\delta_2$ ($\delta_a$ is the dirac delta at the point $a$) can be solved by putting $u-u'=v$ and then solving $xv=\delta_0$ and $xv=-\delta_2$. However I dont understand why this works. If somebody could explain to me whats going on here Id be very grateful!

Answer 1

Let $v_1$ satisfy $xv_1=\delta_0$ and $v_2$ satisfy $xv_2=\delta_2$. Then $x(v_1-v_2) = xv_1-xv_2 = \delta_0-\delta_2.$