Just a partial answer: on the link provided there is the description of the method for an analytical resolution. For numerical solution in matlab, there are classical ODE solvers like ode45, but for first order ODE. Yours is 2nd order, so it is needed to transform it in a first order ODE of 2 variables, defining the first derivative as this new variable (order reduction).
But beware that classical ODE have initial conditions, i.e. y(0) and y'(0) prescribed. Yours is an initial and final value prescribed ODE, i.e. y(0) and y(1) prescribed, so classical routines does not solve this. One way would be to iterate on the resolution to determine y'(0) so that you end up with y(1) = 1, for instance with the shooting method (https://en.wikipedia.org/wiki/Shooting_method). An other way could be to reformulate the ODE as a PDE with boundary conditions, and solve it with a variational method (finite element, finite volume...).
After searching on internet, seems that I found that there is a function in matlab suited for your problem (with a first order ODE, so you will still need to reduce the order), called bvp4c.
Here is the worked solution to your ODE. Please note that it is NOT nonlinear, but linear, as it can be re-written in the form f(x)y''(x) = A. A Nonlinear form of this equation would be f(y)y''(x) = A.