I am solving a BVP, as the example (1) in attachment, where all the variables only have one condition on the boundary.
So I tried to do the change of variable s = 2 - t for p(t) and q(t), and I obtained , as example (2) in attachment, but that didn't work because solution of (1) and (2) do not are the same.
The solution of (1) and (2) will never be the same because your system (2) is not equivalent to the original system (1).
For example, you have defined s(t) = p(2 - t) but then, in your second equation, you have simply substituted p(t) by s(t) which is not correct because p(t) and s(t) are not the same.
Sorry but the attachment had a rather basic mistake. When I was writing it in latex, I forgot to multiply second member, of second and third equations of (2), by (-1). I already correct the error.
If you are interested in finding numerical solutions then apply shooting method to original system (1). I understand they are a coupled equations of second order. Assume p(0) = alpha1 and q(0) = beta1 and try to find solution at t=2 by Runge-Kutta method. Now you take another set of values p(0) = alpha2 and q(0) = beta2 and find solution for t=2. Interpolate so that you get values of alpha and beta where p(2) = 0 and q(2) =0. Find final solutions by Runge- Kutta method by these alpha beta.
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