if I'm not mistaken $\partial H(\lambda,x)/\partial x\rightarrow 2^{\lambda} \lambda x^{\lambda-1}$ for $x\rightarrow \infty$, so the $a_2$ term vanishes for any $\lambda<1$; hence if the condition you are looking for exists, it can only involve $\lambda$, and then the condition seems to be $\lambda=-2,-4,-6,\ldots$ (for any other value of $\lambda$ the derivative blows up $\propto e^{x^2/2}$).
$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=\frac{{a_1}}{x} \left(\left({\lambda}+x^2\right) M(-\frac{{\lambda}}{2},\frac{1}{2},x^2)-({\lambda}-2) M(1-\frac{{\lambda}}{2},\frac{1}{2},x^2)\right)+2 {a_2} {\lambda} H({\lambda}-1,x)$$ for $\lambda=-2$, $y'(x)=-a_1 e^{-\frac{x^2}{2}} x \left(x^2-2\right)-4 a_2 H(-3,x)\rightarrow 0$ for $x\rightarrow\infty$, and similarly for $\lambda$ any negative even integer.
Carlo Beenakker
- 188.1k
- 18
- 448
- 651