Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$x^2y'' + xy' + (x^2-v^2)y = 0$$
Divide everything by $x^2$ to get $y''$ by itself.
$$y'' + {1\over x}y' + {(x^2-v^2)\over x^2}y = 0$$
Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(x)dx}$ where c is a constant and $p(x)$ is $1\over x$ in this case. Now we solve the integral:
$$ce^{-\int{1\over x}dx} = ce^{-ln(x)+C} = ce^{ln(x^{-1})+C} = cx^{-1}e^C$$
But $ce^C$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:
$$W = {c\over x}$$
