Machine Computations for Formulas (19.93)-(19.95)

The version number of Mathematica used to evaluate this notebook:

$VersionNumber

(The technique is similar to that for page 170.)

p  =  5

Do[PowerSum[n, j_] = Sum[ k^n, {k, 0, j}], {n, 0, 2*p + 1}]

Gam[-1, j_] = 0

Gam[0, j_] = 1

summandGam[i_, j_] := 
      Expand[D[ Gam[i, j], z] + (i + j) alpha[z]*
            Gam[i, j] + (i + j) (m - i - j) bet[z]*
            Gam[i - 1, j]]

Do[Gam[i, j_] = 
        Module[{t, k}, t = Expand[summandGam[i - 1, k]]; 
          Sum[Coefficient[t, k, n]*PowerSum[n, j], {n, 0, 2*i - 1}]], {i, 1, 
        p}]

Gam[1, j]

Gam[2, j] // Simplify

eta[m_, i_] := Gam[i, m - i]

Machine Computations for the Context of Section 19.9.2

In terms of a given (19.97), we set

alpha[z_] = c[1][z]/Binomial[m, 2]

bet[
        z_] = (1/
            Binomial[m + 1, 3])*( c[2][
              z] - ((m - 2)/3) c[1]'[
                z] - ((3  m - 
                      1) (m - 2)/(6  m (m - 1)))*c[1][
                  z]^2)

Bscript[m_, i_] := c[i][z] - eta[m, i]

Then, for the formula on page 212 in line 18, we have

eta[m, 1]

And, for the formula on page 212 in line 19, we have

eta[m, 2] // Simplify

Example 19.25 on page 212

For (19.105), we have

m = 4

c[1][z_] = -6 z/(z - 1)

c[2][z_] = (11  z^2 + 10  z - 6)/(z - 1)^2

c[3][
        z_] = -(6  z^3 + 30  z^2 - 13  z - 8)/(z - 
              1)^3

c[4][z_] = 3 (6  z^2 + 8  z + 1)/(z - 1)^3

In terms of the preceding formulas, we find that this yields

alpha[z]

bet[z] // Simplify

eta[4, 1]

eta[4, 2] // Simplify

eta[4, 3]  // Simplify

eta[4, 4] // Simplify

Bscript[4, 1]

Bscript[4, 2] // Simplify

Bscript[4, 3] // Simplify

Bscript[4, 4] // Simplify

eqn = u''[z] + alpha[z]*u'[z] + bet[z]*u[z];

DSolve[eqn == 0, u[z], z]

Thus, the solutions of  u''(z) + ((-z)/(z - 1))u'(z) + (1/(z - 1))u(z) = 0  are given by

u0[z_] = C1 *z + C2*Exp[z]

As a check, we have

eqn /. { 
          u[z] -> u0[z], u'[z] -> u0'[z], u''[
              z] -> u0''[z]} // Simplify

Thus, the solutions of (19.107) are given by

y0[z_] = 
      C1*z^3 + C2*z^2*Exp[z] + C3*z*Exp[2  z] + C4*Exp[3*z]

As a direct check that the preceding functions are solutions of (19.105), we find for

expr = y''''[
          z] + c[1][z]*y'''[z] + c[2][z]*y''[
            z] + c[3][z]*y'[z] + c[4][z]*y[z]

that

expr /. {y[z] -> 
            y0[z],  y'[z] -> y0'[z], y''[z] -> y0''[
              z], y'''[z] -> y0'''[
              z],  y''''[z] -> y0''''[z]} // 
      Simplify

However, for Version 3.0 of Mathematica, the  DSolve  command does not solve (19.107) directly.  Instead,  it returns the expression unevaluated.

$Version

DSolve[ expr == 0,  y[z], z ]

Of coursae, Version 3.0 has no difficulty with

eq =  y''[z] + 3  y'[z] + 2  y[z]

DSolve[eq == 0, y[z], z]

We have found that Version 5.0 and 5.1 give a desirable solution for (19.105).   
For instance, we have

$VersionNumber

DSolve[ y''''[
            z]  - ((6 z)/(-1 + z)) y'''[
              z] + ((-6 + 10 z + 
                  11 (z^2))/(((-1 + z))^2)) y''[
              z] + ((8 + 13 z - 30 (z^2) - 
                  6 (z^3))/(((-1 + z))^3)) y'[
              z] + ((3 (1 + 8 z + 6 (z^2)))/(((-1 + z))^3)) 
            y[z] == 0, y[z], z]