Proving Hermite polynomials satisfy Hermite's equation

[weezy]

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:
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What is the purpose of $e^{-x^2/2}$ factor?
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How can I show $H_n(x)$ is a solution to the equation $y''-2xy'+2ny=0$
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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?
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[Stephen Tashi]

All I did was make a substitution: $H_n(x) = e^{x^2} u, u=D^ne^{-x^2}$

From that substitution, I'm curious how you got other terms with an "$n$" in them to cancel out the "$2ny$" term in the differential equation.

The substitution $H_n(x) = (-1)^n e^{x^2/2} u$ with $u = e^{x^2/2} D^n e^{-x^2}$ would allow us to express $u''$ as a function of $u$ since $y = u$ satisfies the first differential equation you gave: $y''- x^2 y = -(2n+1)y$ , which implies $u'' = x^2 u - (2n+1)u$.

I don't understand why is the extra factor $e^{-x^2/2}$ there?

A simple minded way to motivate a factor is that we want the Hermite polynomials to be polynomials in x. The result of $D^n e^{-x^2}$ is going to have the factor $e^{-x^2}$ in it, which needs to be "canceled out" in order to produce a polynomial.