A relationship between the Laguerre and the Hermite equations

Question

The Hermite and the Laguerre differential equations are respectivly given by $$(1)\quad y'' -2x y' + n y = 0$$ $$(2)\quad xy'' + (1 - x)y' + ny = 0$$ I look for a transformation that enables to go from $(1)$ and $(2)$ or the contrary?

Answer 1

The general solution of the Hermite equation is : $$ (1)\quad y'' -2xy' + ny = 0 \quad\to\quad y=c_1\: {_1F_1}\left(-\frac{n}{4}\:,\: \frac{1}{2}\:;\:x^2\right)+c_2 H_{n/2}(x)$$ ${_1F_1}$ is a confluent hypergeometric function of the first kind.

The general solution of the Laguerre equation is : $$(2)\quad xy''+(1-x)y'+ny=0 \quad\to\quad y=C_1 U(-n,1,x)+C_2 L_n(x)$$ $U$ is a confluent hypergeometric function of the second kind.

The Hermite functions and the Laguerre functions are not independent. The relationships are known : http://functions.wolfram.com/Polynomials/HermiteH/27/01/

On the other hand, the first and second confluent hypergeometric functions are independent.

As a consequence, one cannot expect a general relationship between the Hermite and Laguerre equations. One can only expect to transform one to another similar equation. By similar, I mean an equation on the same kind, but not exactly the same. To be more explicit :

Consider the Hermite equation (1). Change of variable :$\quad X=\frac{x^2}{4}$

$\frac{dy}{dx}=\frac{x}{2}\frac{dy}{dX} \quad\to\quad \frac{d^2y}{dx^2}=X\frac{d^2y}{dX^2}+\frac{1}{2}\frac{dy}{dX}$

$\left(X\frac{d^2y}{dX^2}+\frac{1}{2}\frac{dy}{dX}\right) -2x\left(\frac{1}{2}x\frac{dy}{dX}\right) + ny = 0$

$$X\frac{d^2y}{dX^2}+\left(\frac{1}{2}-4X\right)\frac{dy}{dX} + ny = 0$$ This equation is similar, but different from the Laguerre equation : $$x\frac{d^2y}{dx^2}+(1-x)\frac{dy}{dx}+ny=0$$