A special solution to the Hermite Differential Equation

Question

I know that the general form solution to the Hermite differential equation $$ y''-2xy'+2\lambda y=0$$ is $$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x),$$ where $M(\cdot,\cdot,\cdot)$ is a confluent hypergeometric function of the first kind, and $H(\cdot,\cdot)$ is a Hermite polynomial.

For a general value of $\lambda$ (negative and non-integer real valued), is there a special solution to the Hermite differential equation such that it's first order derivative goes to zero for $x\rightarrow-\infty$? In other words, is there a parametric characterization of $a_2$ as a function of $a_1$ and $\lambda$ such that $y'(x)\rightarrow 0$ as $x\rightarrow-\infty$?

From mathematica numerical calculations it seems that such a special case exists. However, I was not able to characterize it explicitly. I would appreciate any help on this. Thank you!

Answer 1

$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=-2a_1\lambda x M(1-\tfrac{1}{2}{\lambda},\tfrac{3}{2},x^2)+2 {a_2} {\lambda} H({\lambda}-1,x)$$ The asymptotics for $x\rightarrow-\infty$ and $\lambda$ negative non-integer is $$y'(x)\rightarrow\frac{2 e^{x^2}(-x)^{-\lambda} }{\sqrt{\pi }\, \Gamma \left(-\lambda/2\right)}\bigl(a_2 \Gamma \left(-\lambda/2\right) \Gamma (\lambda+1)\sin \pi \lambda -\pi a_1\bigr).$$ So this vanishes if $$a_2 \Gamma \left(-\lambda/2\right) \Gamma (\lambda+1)\sin \pi \lambda =\pi a_1.$$