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.
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.