Solve the following differential equaction in the sense of distribution

Question

I have a following problem in functional analysis $$x^2\frac{du}{dx}=0$$ and I know I should solve it like this $$\langle x^2u',\phi \rangle=\langle0,\phi\rangle \Rightarrow \langle u, (x^2\phi)' \rangle = 0$$, set $\psi=(x^2\phi)'$, then $$\langle u,\psi \rangle=0$$ but then how can I solve $\psi$ in sense of distributions? How to prove it is a test function?

Answer 1

$\langle u, (x^2\phi)' \rangle = \langle a+bH+c\delta, (x^2\phi)'\rangle= a\langle 1, (x^2\phi)'\rangle+b\langle H, (x^2\phi)'\rangle+c\langle \delta, (x^2\phi)'\rangle$, where $\langle 1, (x^2\phi)'\rangle=\int_{-\infty}^{\infty}(x^2\phi)'dx=x^2\phi\bigg|_{-\infty}^{\infty}=0$,

$\langle H, (x^2\phi)'\rangle=\int_{-\infty}^{\infty}H(x)(x^2\phi)'dx=\int_{0}^{\infty}(x^2\phi)'dx=x^2\phi\bigg|_{0}^{\infty}=0$, and

$\langle\delta, (x^2\phi)'\rangle=\int_{-\infty}^{\infty}\delta(x)(x^2\phi)'dx=(x^2\phi)'\bigg|_{x=0}=0$. In computations we have used the properties of $\phi$ and $\delta$.