2nd order distribution differential equation

Question

Denote by $f(x)$ a quadratic polynomial with coeffitients $a, b, c$: $$ f(x) = ax^2 + bx+ c$$ And let's look at distribution differential equation $$f \left(\dfrac{d}{dx}\right) y =\text{p.v.} \dfrac{1}{x}$$ Or in another form $$a\dfrac{d^2y}{dx^2} + b\dfrac{dy}{dx} + cy = \text{p.v.} \dfrac{1}{x}$$ Where the distribution $\text{p.v.} \dfrac{1}{x}$ act's in the following way $$(\text{p.v.} \dfrac{1}{x} , \psi ) = \int_{\mathbb{R}} \dfrac{\psi(x) - \psi(0)}{x} dx$$ Is there any way to solve this equations? I have tryid it in simple case, where $f(x)$ hase 2 real roots, but i failed, any help and links will be appriciated.

Answer 1

Let $u$ be some distribution, e.g. $\operatorname{pv}\frac{1}{x}$ and consider the differential equation ("d.e.") $$ ay'' + by' + cy = u. $$

Let $r=r_1,r_2$ be the two (possible equal) roots of $ar^2+br+c.$ Then we have $$ ay'' + by' + cy = (D-r_1)(D-r_2)y, $$ where $D$ is the differentiation operator, $Df=f'.$

Thus the d.e. can be written $$ (D-r_1)(D-r_2)y = u. $$

Now multiply the equation with $e^{-r_1 x}$: $$ e^{-r_1 x} (D-r_1)(D-r_2)y = e^{-r_1 x} u. $$

The left hand side can then be rewritten: $$ D(e^{-r_1 x}(D-r_2)y) = e^{-r_1 x} u. $$

Taking antiderivative gives $$ e^{-r_1 x}(D-r_2)y = \int e^{-r_1 x} u \, dx $$ so $$ (D-r_2)y = e^{r_1 x} \int e^{-r_1 x} u \, dx. $$

There is always a distribution $U$ such that $U'=e^{-r_1 x} u$ but it can be impossible to write in an explicit form. I think that is the case when $u=\operatorname{pv}\frac{1}{x}.$

Anyway, to solve for $y,$ just repeat what we did: multiply with $e^{-r_2 x}$ and rewrite $e^{-r_2 x}(D-r_2)$ as $D(e^{-r_2 x}\cdots),$ take antiderivative and multiply with $e^{-r_2 x}.$