Hermite Differential Equation

$${d^2y\over dx^2} - 2x{dy\over dx} + \lambda y = 0.$$ (1)

This differential equation has an irregular singularity at $\infty$. It can be solved using the series method

$$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - \sum_{n=1}^\infty 2na_nx^n + \sum_{n=0}^\infty \lambda a_nx^n = 0$$ (2)

$$(2a_2+\lambda a_4) + \sum_{n=1}^\infty [(n+2)(n+1)a_{n+2}-2na_n+\lambda a_n]x^n = 0.$$ (3)

Therefore,

$$a_2 = - {\lambda a_0\over 2}$$ (4)

and

$$a_{n+2} = {2n-\lambda\over (n+2)(n+1)} a_n$$ (5)

for $n = 1$, 2, .... Since (4) is just a special case of (5),

$$a_{n+2} = {2n-\lambda\over (n+2)(n+1)} a_n$$ (6)

for $n=0$, 1, .... The linearly independent solutions are then

$$y_1 = a_0\left[{1 - {\lambda\over 2!} \,x^2 - {(4-\lambda )\lambda\over 4!}\, x^4 - {(8-\lambda)(4-\lambda)\lambda\over 6!} \,x^6 - \ldots}\right]$$ (7)

$$y_2 = a_1\left[{x + {(2-\lambda)\over 3!} \,x^3 + {(6-\lambda)(2-\lambda)\over 5!}\, x^5 + \ldots}\right].$$ (8)

If $\lambda\equiv 4n = 0$, 4, 8, ..., then $y_1$ terminates with the Power $x^\lambda$, and $y_1$ (normalized so that the Coefficient of $x^n$ is $2^n$) is the regular solution to the equation, known as the Hermite Polynomial. If $\lambda \equiv 4n+2 = 2$, 6, 10, ..., then $y_2$ terminates with the Power $x^\lambda$, and $y_2$ (normalized so that the Coefficient of $x^n$ is $2^n$) is the regular solution to the equation, known as the Hermite Polynomial.

If $\lambda=0$, then Hermite's differential equation becomes

$$y''-2xy'=0,$$ (9)

which is of the form $P_2(x)y''+P_1(x)y'=0$ and so has solution

$$y = c_1\int{dx\over \mathop{\rm exp}\nolimits \left({\int{P_1\over P_2}\,dx}\right)} +c_2$$

$$= c_1\int{dx\over \mathop{\rm exp}\nolimits \int -2x\,dx}+c_2$$

$$= c_1 \int {dx\over e^{-x^2}}+c_2 = c_1\int e^{x^2}\,dx+c_2.$$ (10)