Solve the differential equation $x^2u'=0$ in the sense of distributions 
Solve the differential equation in the sense of distribution:
  $$x^{2}\frac{du}{dx}=0$$

This is from "Principles of Applied Mathematics" by Keener, problem 4.1.5.
The solution in the back of the text is 
$$u(x)=c_{1}+c_{2}H(x)+c_{3}\delta(x)$$
where $H(x)$ is the Heaviside function and $\delta(x)$ is the Dirac delta function.
I think that I understand what the solution means (the action of $u$ on test functions), but I do not understand how to arrive at such a solution.
 A: Following Keener's example 2, on page 143 of the year 2000 edition. 
We want to find a distribution $u$ such that :
$<x^2 u',\phi>=0$ for all $\phi \in D$, where $D$ is the set of test functions.
Which is equivalent to
$-<u,(x^2\phi)'>=0$ for all $\phi \in D$.
or $<u,\psi>=0$ for any test function $\psi$ such that $\psi=(x^2\phi)'$.
Now, it is not hard to see that:
Lemma: Given $\psi,\phi \in D$, then one can write $\psi=(x^2\phi(x))'$ if and only if $\displaystyle \int_{-\infty}^{+\infty} \psi(x) dx=0$, $\displaystyle \int_{0}^{+\infty} \psi(x) dx=0$, and $\psi(0)=0$. 
Assume the lemma, and pick $\phi_0(x),\phi_1(x),\phi_2(x) \in D$ such that: 
$\displaystyle \int_{0}^{+\infty} \phi_i(x) dx=0$ for $i\in\{1,2\}$ and $\displaystyle \int_{0}^{+\infty} \phi_i(x) dx=1$ for $i=0$.
$\displaystyle \int_{-\infty}^{+\infty} \phi_i(x) dx=0$ for $i\in\{0,2\}$ and 
$\displaystyle \int_{-\infty}^{+\infty} \phi_i(x) dx=1$ for $i=1$.
$\phi_i(0)=0$ for $i\in\{0,1\}$ and $\phi_i(0)=0$ for $i=2$.
Note that for any $\phi \in D$, we can write 
$\phi(x)=\phi_0(x) \displaystyle \int_{0}^{+\infty} \phi(s) ds + 
\phi_1(x) \displaystyle \int_{-\infty}^{+\infty} \phi(s) ds + \phi_2(x)\phi(0)+\psi(x)$.
Where $\psi(x)=\phi(x)-\left(\phi_0(x) \displaystyle \int_{0}^{+\infty} \phi(s) ds + 
\phi_1(x) \displaystyle \int_{-\infty}^{+\infty} \phi(s) ds + \phi_2(x)\phi(0)\right)$
Show that $\phi(x)$ satisfies the three conditions of the lemma. 
Then from $<u,\psi>=0$  where $\psi=(x^2\phi)'$ you get that 
$<u,\phi(x)>=<u,\phi_0(x) \displaystyle \int_{0}^{+\infty} \phi(s) ds > + <u,\phi_1(x) \displaystyle \int_{-\infty}^{+\infty} \phi(s) ds >+ <u,\phi_2(x)\phi(0)>$
Or equivalently 
$<u(x),\phi(x)> = <c_1 H(x) + c_2 \delta_0(x) + c_3 ,\phi(x)> $ for all $\phi(x) \in D$.
Therefore $u(x)=c_1 H(x) + c_2 \delta(x) + c_3$ in the distributional sense, as desired.
A: Let $v=du/dx$. The expression $x^2 v =0$ means that for every test function $\phi$, the distribution $v$ vanishes on $x^2 \phi$. It follows that $v$ vanishes on all test functions $\phi$ such that $\phi(0)=\phi'(0)$, because these can be written as $\phi=x^2\psi$ with $\psi$ another test function.  Hence 
$$
v(\phi) = c_1\phi(0)+c_2\phi'(0)
$$
which can be written as $v$ being a linear combination of $\delta$ and $\delta'$. Integration yields $u$ being a linear combination of $H$ and $\delta$, plus a constant.
