How can I solve this equation involving a distribution?

Question

I'm trying to find all distributions $u \in \mathcal{D'}(\mathbb{R})$ such that $$(x^2+1)u = \delta$$ and I'm having some trouble. I can see that since $x \mapsto \frac{1}{x^2+1}$ is smooth, we have a particular solution given by $u = \frac{1}{x^2+1} \delta$. So then I look for solutions to the homogeneous equation $(x^2+1)u =0$ to add to this, and this is what I can't do. I know that if $x^2 u = 0$ then $x = c_0 \delta + c_1 \partial \delta$ for some constants $c_0$ and $c_1$ but I don't know how to apply this in this case. Any help appreciated!

Answer 1

Since $f\delta=f(0)\delta$, the solution $u = \frac{1}{x^2+1} \delta$ reduces to $u=\delta$.

There are no other solutions. The homogeneous equation $(x^2+1)u=0$ is equivalent with $u=0$ since $x^2+1\neq 0$ for all $x$.