Consider the following equation $Lf= -f'' -f' + x^2f$ =0$
Can we construct orthogonal polynomials from the solution of this differential operator?
The general shape for classical orthogonal polynomial is $Q(x) = f_n'' (x)+ L(x) f_n'+ \lambda_n f_n=0$
Hermite polynomial has $x$ squared in its equation but there is no first derivative term
$$f''+f'-x^2f=0\tag 1$$ The Hermite differential equation is :
$$y''-2xy'+\lambda y=0 \tag 2$$ In order to reduce Eq.$(1)$ into Eq.$(2)$ the usual change of function is : $$f(x)=e^{u(x)}y(x)$$ $f'=e^u(u'y+y'u)$ and $f''=e^u(u'^2y+2u'y'+u''y+y'')$ that we put into Eq.$(1)$. After simplification : $$y''+(2u'+1)y'+(u''+u'^2+u'-x^2)y=0$$ We get Eq.$(2)$ with $2u'+1=-2x$ : $$u=-\frac12x(x+1)$$ and with $u''+u'^2+u'-x^2=\lambda \quad\implies\quad 1+(x+\frac12)^2-(x+\frac12)-x^2=\frac34$ $$\lambda=\frac34$$ $$y(x)=H_{\lambda/2}(x)$$ See Eq.(10) in https://mathworld.wolfram.com/HermiteDifferentialEquation.html for two independent solutions of $(2)$. $$f(x)=e^{-\frac12x(x+1)}\left(c_1H_{3/8}(x)+c_2\:_1F_1(-\frac{3}{16}\:;\: \frac12\:;x^2)\right)$$ $\:_1F_1$ is a confluent hypergeometric function.