To solve it analytically, you may need to use the beta function due to the structure of the RHS. You could replace the RHS with a high order taylor series and then solve it keeping track of the error term. Otherwise, you can always use a numerical solver.
The Mathematica file doesn't seem to solve the given equation, however: The DSolve command asks for the solution of the equation (y''[x])^(-1)-y[x)=0, which isn't the same as the original equation. This last equation is autonomous, for starters, while the original equation is not-it's that of an oscillator with ``time-dependent'' frequency; its solution is, also, the putative eigenfunction, of eigenvalue zero, of the ``Schrödinger-like operator'' in the potential x^(3/4)/(1-x)^(1/2). This potential has a ``hard wall'' at x=0 and goes to infinity at x=1 and is bounded from below, therefore one knows that the spectrum is discrete. However it's not obvious that there does exist an eigenfunction, of eigenvalue zero.
It's not clear just what the transformation from x to v accomplishes-it's not used.
While (1-x)^(-1/2) does have a Taylor expansion around x=0, x^(3/4) does not. And while x^(3/4) does have a Taylor expansion about x=1, (1-x)^(-1/2) does not.
It is, of course, possible to solve the equation numerically, but boundary conditions must be imposed.
It depends it you need analytical or numerical approximation of the solution. For analytical solution, you can use Mathematica or xmaxima or any free solfware.
For numerical solution, do a simple change of variable to convert to a system of first order ODE and use Matlab, or Octave, python or write a code by yourself for that.'
Well it's always possible to set y'[x] = z[x], thus z'[x] = -(x^(3/4))*y[x]/(1-x)^(1/2). To solve it numerically requires boundary conditions and, as mentioned, it's not obvious that an analytical solution, in terms of ``elementary'' functions is available, unless it's possible to prove that the potential x^(3/4)/(1-x)^(1/2), defined in the interval [0,1), belongs to the class of exactly solvable potentials. However it's clear that its form implies that the spectrum is discrete and bounded from below.