Here, $y$ is a dependent variable of the independent variable $x$. Now Mark Beadles already answered that part, the part I'm going to answer is the "*I don't know what that means" part. The notion of parameter here means that for each value of $\nu$, we are given the differential equation. Solving this equation for $y$ as a function of $x$ and $\nu$ does not mean that $y$ is a function of $\nu$, but rather that given $\nu$, $y(x)$ solve the differential equation with $\nu$ fixed. For instance, the well-known equation $ \frac{dy}{dx} = \lambda x, $ where $\lambda$ is a parameter, has the solution $y(x) = Ce^{\lambda x}$, with $C$ a real constant. The fact that $\lambda$ is in the expression of $y$ does not mean that $y$ is a function of $\lambda$ ; when you consider the differential equation, you fix $\lambda$ beforehand, and then you solve the differential equation to find $y(x)$. In other words, you're not solving for $y(x,\lambda)$, but you're finding $y(x)$ for a family of differential equations parametrized by the parameter $\lambda$.
Hope that helps,