homework and exercises - Wave speed of a hanging rope

Let us consider a homogeneous rope hanging from the ceiling. I will call the vertical direction $x$ and the horizontal displacement $y$. When we apply the second Newton's Law to a portion of mass $\Delta m$ and proceed in the same way we do for a horizontal string we get $$\mu\frac{\partial^2 y(x,t)}{\partial t^2}=\frac{\partial}{\partial x}\left[T(x)\frac{\partial y(x,t)}{\partial x}\right].$$ The difference now is that the tension is not constant. Defining $x=0$ at the free end of the rope and orienting it upwards we have the tension $T=\mu gx$. Therefore, $$\frac{\partial^2 y}{\partial t^2}=gx\frac{\partial^2y}{dx^2}-g\frac{\partial y}{\partial x}.$$

So, my issue is that I have seen in a couple of physics books (for instance) that the wave speed of this rope is simply $$v=\sqrt{\frac{T}{\mu}}=\sqrt{gx}.$$ In my opinion these books are either:

i) Cheating the students: they know it is wrong but assume it is right just to show some nice features (the speed would increase as the wave goes upwards).

ii) Had misconception assuming that the usual wave equation are still valid.

iii) Is doing some obscure approximation, which is another cheat if you do not reveal it.

My questions are:

1) Is there an expression for the wave speed for the "wave equation" above?

2) Is there an approximation leading to the wave speed given above?