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 for a nonzero $a_2$, 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 $\propto e^{x^2/2}$).

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$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$
$$y'(x)=\frac{{a_1}}{x} \left(\left({\lambda}+x^2\right) M(-\frac{{\lambda}}{2},\frac{1}{2},x^2)-({\lambda}-2) M(1-\frac{{\lambda}}{2},\frac{1}{2},x^2)\right)+2 {a_2} {\lambda} H({\lambda}-1,x)$$
$$\rightarrow \frac{a_1  e^{x^2/2} x^{{\lambda}+1}}{\Gamma(1+\lambda/2)}+a_2 2^{\lambda} \lambda x^{\lambda-1}$$
$$\text{so this vanishes for large $x$ when $a_1=0$ and $\lambda<1$ or when $a_1\neq 0$ and $\lambda$ an even negative integer.}$$
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