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I have found that for the Hermite differential equation, $$ y'' -2xy' +2\lambda y = 0 $$

Solutions exist in multiple forms:

$\bullet$ Popularly, the Hermite polynomials are solutions for $\lambda = 0,1,2,3...$: $$ H_\lambda(x) = (-1)^\lambda e^{x^2}\frac{d^\lambda}{dx^\lambda}e^{-x^2} $$ using the Rodrigues formula.

$\bullet$ Another way I found is by solving the d.e. with a series solution, which yields two linearly independent solutions: $$ y_1(x) = 1 + \sum_{k=1}^{\infty}{(-1)^k \frac{2^k \lambda(\lambda - 2)(\lambda - 4)...(\lambda-2k+2)}{(2k)!}x^{2k}} \\ y_2(x) = x + \sum_{k=1}^{\infty}{(-1)^k \frac{2^k (\lambda-1)(\lambda - 3)...(\lambda-2k+1)}{(2k+1)!}x^{2k+1}} $$ and the total solution is $y(x) = a_0y_1(x) + a_1y_2(x)$.

$\bullet$ I also used Sturm-Liouville theory to change the d.e. to this eigenvalue equation: $$ -\frac{d}{dx}[e^{-x^2}y'] = 2\lambda e^{-x^2}y $$

but haven't been able to derive any solution for $y(x)$ outside of the above.

My goal is to generate two linearly independent solutions that satisfy this boundary condition: $$\lim_{x \rightarrow \infty} y(x) = 0$$

My hint was to use the Sturm-Liouville method and I was told the boundary condition would become obvious. At present I still don't know how to solve the Sturm-Louiville equation that I've verified in correct form above.

Guidance is tremendously appreciated!

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As you mentioned, your equation can be written in the Sturm-Liouville normal form $$ -\frac{d}{dx}e^{-x^2}\frac{d}{dx}y=\lambda e^{-x^2}y. $$ That puts the problem into the context of $L^2_{e^{-x^2}}(\mathbb{R})$. For $\lambda = 0$, the solutions are $$ e^{-x^2}y'= C \\ y' = Ce^{x^2} \\ y = C\int_{0}^{x}e^{t^2}dt+D. $$ Only one of the two solutions is in $L^2_{e^{-x^2}}(\mathbb{R})$ near $x=\infty$ or near $x=-\infty$. Therefore, the equation is in the limit point case at both endpoints $x=\pm\infty$. That means that, for every $\lambda\in\mathbb{C}$, there is only one solution $y$ of the eigenvalue equation that is in $L^2_{e^{-x^2}}[0,\infty)$. However, if both linearly independent solutions were to satisfy your condition, then they would definitely be in $L^2_{e^{-x^2}}[0,\infty)$. So what you want is impossible.

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  • $\begingroup$ My professor at the time thought differently. Not sure what really was sought for. The issue was never really clarified. Thanks for your effort and insight. $\endgroup$
    – khaverim
    Mar 7, 2016 at 17:07

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