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246   6. Applications of Determinants in Mathematical Physics

            Equation (6.2.25) was solved by Ohta, Kajiwara, Matsukidaira, and
          Satsuma in 1993. A brief note on the solutions is given in Section 6.11.



          6.3   The Dale Equation

          Theorem. The Dale equation, namely
                                        y     y +4n
                                      1 2  	       2
                                  2
                             (y ) = y                 ,
                                        x     1+ x
          where n is a positive integer, is satisfied by the function
                                               11
                                  y =4(c − 1)xA ,
                                               n
          where A 11  is a scaled cofactor of the Hankelian A n = |a ij | n in which
                 n
                                     x i+j−1  +(−1) i+j c
                               a ij =
                                         i + j − 1
          and c is an arbitrary constant. The solution is clearly defined when n ≥ 2
          but can be made valid when n =1 by adopting the convention A 11 =1 so
          that A 11  =(x + c) −1 .
          Proof. Using Hankelian notation (Section 4.8),
                             A = |φ m | n ,  0 ≤ m ≤ 2n − 2,
          where
                                     x m+1  +(−1) c
                                                 m
                                φ m =               .                (6.3.1)
                                          m +1
            Let
                                     P = |ψ m | n ,
          where
                                ψ m =(1 + x) −m−1 φ m .

          Then,
                                   ψ    = mFψ m−1
                                    m
          (the Appell equation), where
                                   F =(1 + x) −2 .                   (6.3.2)

          Hence, by Theorem 4.33 in Section 4.9.1 on Hankelians with Appell
          elements,

                                  P = ψ P 11
                                         0
                                       (1 − c)P 11
                                     =           .                   (6.3.3)
                                        (1 + x) 2
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