$xu'+u=1$ distribution equation

Question

While I know the answer to homogeneous equation $ xu'+u=0 $, that is $u=c_1\delta_0+\textbf{p.v.}(\frac{1}{x}) $ I'm struggling with the nonhomogeneous $$ xu'+u=1 $$ I've tried using the known facts about distributions in this way, $\phi(x)$ is some test function: $$ \int_\mathbb{R}\phi(x)=\langle 1, \phi(x)\rangle=\langle xu'+u, \phi(x) \rangle= \langle (xu)', \phi(x) \rangle = -\langle xu, \phi'(x) \rangle=-\langle u, x\phi'(x) \rangle $$ Then $$ \int_\mathbb{R} \phi=\int_\mathbb{R} (x\phi)' - x\phi'= (x\phi)|_{-\infty}^{+\infty}-\int_\mathbb{R}x\phi' \Rightarrow \int_\mathbb{R} \phi=-\int_\mathbb{R}x\phi' $$ As $\phi(x)=0$ at $\pm\infty$. This leads me to conclusion that $u=1$, but to be honest I'm not really good at distribution theory so I'd be thankful for comments whether my train of thought was correct and, if not, what's the correct way to solve this equation. Thanks.

Answer 1

The equation can be written as $(xu)' = 1.$ Taking antiderivatives gives $xu=x+A$ where $A$ is a constant. This can then be rewritten as $x(u-1) = A$ from which we conclude that $u-1 = \operatorname{pv}A/x + B\delta_0,$ where $B$ is a constant. Thus, $$u = 1 + A\operatorname{pv}\frac{1}{x} + B\delta_0.$$