What exactly is $\frac {d^n} {dx^n}$ in rodrigues formula as used for hermite equations.

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I have wrecked my brain trying to understand the example and solution below but I came up empty.

(This was translated from another language so it might not be entirely correct, please do correct me if I erred anywhere.)

Example 1.6: Place $n=2$ in Hermite Equation and use the Rodrigues formula to find the answer.

Solution: By placing $n=2$ in Hermite equation we have $$\frac {d^2y} {dx^2} - x\frac {dy}{dx} + 4y= 0$$

The Rodriguez formula for Hermite equations is:

$$H(x) = (-1)^n e^{x^2} \frac {d^n} {dx^n} (e^{-x^2})$$

By placing $n = 2$ we have: $H(x) = (-1)^2 e^{x^2}\frac {d^2}{dx^2} (e^{-x^2})$ This equation is the second Hermite equation. By pluging in we have $4x^2-2$ which is the answer of this equation.

Okay to start off: What on earth is $\frac {d^n} {dx^n}$? What am I supposed to replace it with? I don't understand how they arrived at the answer at all. (Sorry for the messy english, it is my third langauge.)

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$$ \frac {d^n} {dx^n} $$ Is the nth derivative of whatever it acts on. Thus, $$ \frac {d^n} {dx^n} e^{-x^2} $$ is the nth derivative of $e^{-x^2}$. In your case you are asked to consider the case $n=2$ i.e. $$ \frac {d^2} {dx^2} e^{-x^2} $$ For more on Hermite polynomials see here

and for more on the differential equation see here