I am posing a variant of this question.
Data: a polynomial $p(x)$, $x\in\mathbb{R}$, with complex coefficients, and function $\varphi:\mathbb{R}\to\mathbb{C}$, smooth, with rapid decrease (say, faster than any polynomial).
Goal: to solve the distribution equation $$ p(x)\,T\,=\,\varphi(x)\,. $$
In the original problem $\varphi\equiv 1$.
Now, pretty exhaustive discussions can be found for $p(x)T=0$ and $p(x)T=1$. General strategy (I'm being sketchy): to look for the roots to $p(x)=0$ and their degeneracy, they determine the general solution $T$ as a combinations of $\delta$-distributions (and their derivatives) and $PV$-distributions (and their derivatives). So, for instance, $$ x\,T\,=\, 1\qquad \Rightarrow\qquad T= c\delta(x)+PV\frac{1}{x} $$ or $$ \begin{split} &(x-x_1)(x+x_2)\,T\,=\, 1 \\ & \Rightarrow\; T= c_1\delta(x-x_1)+c_2\delta(x-x_2)+\frac{1}{x_2-x_1}\Big(PV\frac{1}{x-x_1}-PV\frac{1}{x-x_2}\Big) \end{split} $$ ($x_1\neq x_2$).
I am trying to understand how the above strategy gets modified with a nice $\varphi$ instead of a constant. To be concrete, let me pose the question for the equation $$\tag{*} (x-x_1)(x-x_2)\,T\,=\,\varphi(x) $$ with $x_1\neq x_2$. The roots of the polynomial are simple and distinct, so the homogeneous equation is solved by a linear combination of $\delta(x-x_1)$ and $\delta(x-x_2)$. Next, I have to fix a special solution $T_0$. I would be tempted to say $$ T_0\,=\,\varphi(x)\frac{1}{x_2-x_1}\Big(PV\frac{1}{x-x_1}-PV\frac{1}{x-x_2}\Big)\,. $$
Is that correct or wrong? and what is the rigorous reasoning one should make?
Thank you!
Yes, it's just to multiply a solution to $pT=1$ with $\varphi$ to get a solution to $pT=\varphi$:
Let $pT_0 = 1.$ Then, $\varphi = \varphi 1 = \varphi (pT_0) = p(\varphi T_0),$ so $T=\varphi T_0$ is a solution to $pT=\varphi.$
Assume that $S$ is another solution, i.e. $pS=\varphi.$ Then $p(S-T)=\varphi-\varphi=0,$ so $S-T$ is a solution to the homogeneous equation.