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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. My first thought was to find out what ##D^n e^{f(x)}## and I thought if ##f(x)=x^2## I would get a finite polynomial but that didn't happen.
All my questions can be summarised as follows:

What is the purpose of ##e^{x^2/2}## factor?

How can I show ##H_n(x)## is a solution to the equation ##y''2xy'+2ny=0 ##

What is the relation between the differential equation of Hermite functions to that of Hermite polynomials? Are the two equations/solutions equivalent in some way?

$$y_n''  x^2y_n=(2n+1)y_n$$ 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. My first thought was to find out what ##D^n e^{f(x)}## and I thought if ##f(x)=x^2## I would get a finite polynomial but that didn't happen.
All my questions can be summarised as follows:

What is the purpose of ##e^{x^2/2}## factor?

How can I show ##H_n(x)## is a solution to the equation ##y''2xy'+2ny=0 ##

What is the relation between the differential equation of Hermite functions to that of Hermite polynomials? Are the two equations/solutions equivalent in some way?
