I’m practicing some problems for applied mathematics, and couldn’t quite solve the following:

In differential equations of equilibrium, from u(0) = a, du/dx(0) = b, u(1) = c, du/dx(1) = d we get the Hermite polynomial u = a(x-1)²(2x+1) + b(x-1)²x + cx²(3-2x) + dx²(x-1). But how do I get the cubic which substitutes u = a(x-1)²(2x+1) + b(x-1)²x + cx²(3-2x) + dx²(x-1) when we move the right end of the interval from x=1 to x=h?

And how do I get the new form of 2(s0) + 1(s1) = 3(u1) – 3(u0) for the case described above, and given that d²u/dx² = 0 at x = 0 (in which u are the heights and s the slopes)

If you know the solution, or could guide me in the right direction, I would appreciate it a lot! Thanks in advance..