My book (by Mary L Boas) introduces first the Hermite differential equation for Hermite functions: $$y_n'' - x^2y_n=-(2n+1)y_n$$ and we find solutions like $$y_n=e^{x^2/2}D^n e^{-x^2}$$ where $D^n=\frac{d^n}{dx^n}$
Now she says that multiplying $y_n$ by $(-1)^ne^{x^2/2}$ gives us what is known as Hermite polynomials: $$H_n(x)=(-1)^ne^{x^2}D^n e^{-x^2}$$which satisfies another equation: $$y''-2xy'+2ny=0 $$ So far so good until I try to prove if $H_n(x)$ does indeed satisfy the equation above. In her problem set she asks to check if $e^{-x^2/2}H_n(x)$ satisfies the Hermite polynomial equation and here I don't understand why is the extra factor $e^{-x^2/2}$ there?
Also, when I try to just check if $H_n(x)$ satisfies the equation I fail to get past this step which is: $$H_n'(x)=(-1)^ne^{x^2}[D^{n+1}e^{-x^2}+2x \cdot D^ne^{-x^2}] $$ I don't know if this can be further simplified because when I try to calculate $H_n''(x)$ it gets even more complicated and I am unable to prove that $H_n(x)$ is a solution to the equation.
All my questions can be summarised as follows:
Premultiplication of a Hermite polymomial $H_n(x)$ by $e^{-x^2/2}$ converts it into a Hermite function (aside from the $(-1)^n$ factor), which you expressed as $y_n=e^{x^2/2}D^n e^{-x^2}.$ As you noted, the Hermite functions satisfy the differential equation $$y_n'' - x^2y_n=-(2n+1)y_n,$$ so this premultiplication allows us to write down a solution to this differential equation as $e^{-x^2/2}H_n(x)$.
So, simply substitute the expression $e^{-x^2/2}H_n(x)$ into the same differential equation. After some straightforward simplification, you’ll find that the Hermite polynomial $H_n(x)$ satisfies the differential equation $y''-2xy'+2ny=0$.