Why can't I solve this homogenous second order differential equation?

Question

I've been banging my head on the wall for quite some time trying to come up with a solution to the following:

$$\frac {\partial^2 y(x)} {\partial x^2} + (A-B*V(x)) y(x) = 0 $$

$$V(x) = (36 + (2 - x)^2)^{-1/2}$$

With A and B constants, and $y$ solely a function of $x$.

If it helps, in my area of concern $0 \leq x \leq 4$, you can treat V as: $$V(x) = (-1/432)*(x - 2)^2 + 1/6$$

With no real loss of accuracy (that I care about). Generally, I know this is equvalent to the form:

$$y''(x) + p(x)y' + q(x)y = 0$$

With $p(x) = 0$. I can find plenty of examples of constant coefficients and solutions for the form of $q(x)=0$.

Can anybody recommend an anzatz/method/approximation solution that might help me solve this?

Answer 1

Solutions to the equation with quadratic potential can be expressed via the parabolic cylinder functions. The result being $$ y(x)=C_1 D_{-\frac{i \sqrt{3} (6A+1)}{\sqrt{B}} -\frac{1}{2}} \left(\frac{\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt[4]{B} (x-2)}{3^{3/4}}\right)+ $$ $$ C_2 D_{\frac{i \sqrt{3}(6A+1)} {\sqrt{B}}-\frac{1}{2}}\left(-\frac{\left(\frac{1}{2}-\frac{i}{2} \right) \sqrt[4]{B} (x-2)}{3^{3/4}}\right). $$

Answer 2

Just one note, if you care about the accuracy, the validity of approximation depends on the actual values of $A$ and $B$. Perhaps the best way is to expand both $V(x)$ and $y(x)$ in to sums of series and truncate at the right order according to the requirement of accuracy.

Answer 3

Suppose $B\neq0$ :

$\dfrac{d^2y}{dx^2}+\left(A-B(36+(2-x)^2)^{-\frac{1}{2}}\right)y=0$

$\dfrac{d^2y}{dx^2}+\biggl(A-\dfrac{B}{\sqrt{(x-2)^2+36}}\biggr)y=0$

$\sqrt{(x-2)^2+36}\dfrac{d^2y}{dx^2}+\left(A\sqrt{(x-2)^2+36}-B\right)y=0$

Let $r=x-2$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=\dfrac{dy}{dr}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)=\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}=\dfrac{d^2y}{dr^2}$

$\therefore\sqrt{r^2+36}\dfrac{d^2y}{dr^2}+\left(A\sqrt{r^2+36}-B\right)y=0$

Let $s=\sqrt{r^2+36}$ ,

Then $\dfrac{dy}{dr}=\dfrac{dy}{ds}\dfrac{ds}{dr}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{dy}{ds}$

$\dfrac{d^2y}{dr^2}=\dfrac{d}{dr}\biggl(\dfrac{r}{\sqrt{r^2+36}}\dfrac{dy}{ds}\biggr)=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d}{dr}\biggl(\dfrac{dy}{ds}\biggr)+\dfrac{36}{(r^2+36)^\frac{3}{2}}\dfrac{dy}{ds}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d}{ds}\biggl(\dfrac{dy}{ds}\biggr)\dfrac{ds}{dr}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d^2y}{ds^2}\dfrac{r}{\sqrt{r^2+36}}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{r^2}{r^2+36}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{s^2-36}{s^2}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}$

$\therefore s\biggl(\dfrac{s^2-36}{s^2}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}\biggr)+(As-B)y=0$

$\dfrac{(s+6)(s-6)}{s}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^2}\dfrac{dy}{ds}+(As-B)y=0$

$\dfrac{d^2y}{ds^2}+\dfrac{36}{s(s+6)(s-6)}\dfrac{dy}{ds}+\dfrac{s(As-B)}{(s+6)(s-6)}y=0$

$\dfrac{d^2y}{ds^2}+\left(\dfrac{1}{2(s-6)}+\dfrac{1}{2(s+6)}-\dfrac{1}{s}\right)\dfrac{dy}{ds}+\left(\dfrac{6A-B}{2(s-6)}-\dfrac{6A+B}{2(s+6)}+A\right)y=0$