how to solve this variant of Hermite differential equation The Hermite differential equation is defined as :
$y''-2xy' + \lambda y = 0$
I have a differential equation that is a slight variation:
$y''-(x+\alpha)y' + \lambda y = 0$
where $\alpha$ is a constant. Has this equation been solved already? Does it have any name?

You can write the solution of your differential equation in terms of the solution to
$$\frac{d^2y}{dx^2}-2x\frac{dy}{dx} + \lambda y = 0$$
I'll specify different variables in your equation to make the transformation easier to see:
$$\frac{d^2y}{dw^2}-(w+\alpha)\frac{dy}{dw} + \lambda' y = 0$$
Now set $x = \frac{w + \alpha}{\sqrt{2}}$, i.e. $w = \sqrt{2}x - \alpha$.
Since $\frac{dy}{dx} = \frac{dy}{dw}\frac{dw}{dx} = \sqrt{2}\frac{dy}{dw}$ 
and $\frac{d^2y}{dx^2} = 2\frac{d^2y}{dw^2}$ we can rewrite your equation as:
$$\frac{1}{2}\frac{d^2y}{dx^2}-x\frac{dy}{dx} + \lambda' y = 0$$
Multiplying through by $2$ we arrive at the original equation with a different constant multiple of $y$.
$$\frac{d^2y}{dx^2}-2x\frac{dy}{dx} + 2\lambda' y = 0$$
i.e. your solution is given by the original equation with $\lambda = 2 \lambda'$ and the substitution $x = \frac{w + \alpha}{\sqrt{2}}$.
