Solve $xu'+u=0$ as distribution If $u \in \mathcal{D}'(\mathbb{R})$ where $\mathcal{D}'(\mathbb{R})$ is the space of all tempered distributions defined on $C_c^{\infty}(\mathbb{R})$ and $xu'+u=0$. Show that $u = A \mathbf{p.v.}\frac{1}{x} + B\delta$, where
\begin{equation}
\mathbf{p.v.}\frac{1}{x}(f(x)) = \lim_{\epsilon \to 0^+} \int_{|x|\ge\epsilon} \frac{f(x)}{x} \mathbf{d}x,
\quad f \in \mathcal{D}(\mathbb{R})
\end{equation}
I have already shown that $\mathbf{p.v.}\frac{1}{x}$ is  a tempered distribution. For $f \in \mathcal{D}(\mathbb{R})$, we have
\begin{align}
0 &= \langle xu', f \rangle + \langle u, f \rangle \\
&= \langle u', xf \rangle + \langle u, f \rangle \\
&= -\langle u, f+ xf' \rangle + \langle u, f \rangle \\
&= -\langle u, xf' \rangle \\
&= -\langle xu, f'\rangle \\
&= \langle (xu)', f\rangle
\end{align}
Since this holds for any $f$, it must be true that $xu=0$.
But I don't know where to go next.
 A: $\newcommand{\tf}{\mathcal D}$
Hint: If $u\in\tf'$ and $u'=0$ then $u$ is constant: There exists $c\in\mathbb C$ such that $u=c$.
Regarding what that means and then why it's true: First, saying $u'=0$ means $\langle u',\phi \rangle=0$ for all $\phi\in\tf$, which by the definition of $u'$ means $$\langle u,\phi' \rangle=0\quad(\phi\in\tf).$$
In general, if $u\in\tf'$ and $f$ is a locally integrable function then saying $u=f$ means $$\langle u,\phi \rangle=\int f\phi\quad(\phi\in\tf).$$
SO what's being asserted is this:


If $u\in\tf'$ and $\langle u,\phi' \rangle=0$ for all $\phi\in\tf'$ then there exists $c\in\mathbb C$ such that $\langle u,\phi \rangle= c\int\phi$ for all $\phi\in\tf'$.


Proof:


Exercise Suppose $\phi\in\tf$. There exists $\psi\in\tf$ with $\phi=\psi'$ if and only if $\int\phi=0$.


Now suppose $u'=0$. Choose $\phi_0\in\tf$ with $\int\phi_0=1$ and let $$c=\langle u,\phi_0 \rangle.$$ Suppose $\phi\in\tf$. Let $\alpha=\int\phi$. The exercise show that there exists $\psi\in\tf$ with $$\phi-\alpha\phi_0=\psi'.$$Hence $$(u,\phi-\alpha\phi_0 \rangle=0,$$or $$\langle u,\phi \rangle=c\int\phi.$$
Now to finish your problem: You have $(xu)'=0$, so $xu=c$. If you let $v=u-c\frac1x$ it follows that $xv=0$, and now an argument much like what's above shows that $v$ is a multiple of $\delta$. (Hint for that last bit: If $\phi\in\tf$ and $\phi(0)=0$ then there exists $\psi\in\tf$ with $\phi=x\psi$.)
