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b
               4.    y(x) – λ  [Ag(x)+ Bg(t)]y(t) dt = f(x).
                             a
                     The characteristic values of the equation:


                                                              2 2
                                           (A + B)g 1 ±  (A – B) g +4AB(b – a)g 2
                                                               1
                                     λ 1,2 =                                 ,
                                                          2
                                                    2AB[g – (b – a)g 2 ]
                                                          1
                     where

                                                b               b
                                                                  2
                                          g 1 =  g(x) dx,  g 2 =  g (x) dx.
                                               a               a
                     1 . Solution with λ ≠ λ 1,2 :
                      ◦
                                             y(x)= f(x)+ λ[A 1 g(x)+ A 2 ],
                     where the constants A 1 and A 2 are given by
                              Af 1 –λAB[f 1 g 1 –(b–a)f 2 ]          Bf 2 –λAB(f 2 g 1 –f 1 g 2 )
                      A 1 =    2                        ,  A 2 =    2                        ,
                                          2
                                                                               2
                          AB[g –(b–a)g 2 ]λ –(A+B)g 1 λ+1      AB[g –(b–a)g 2 ]λ –(A+B)g 1 λ+1
                               1
                                                                    1
                                                b               b
                                         f 1 =  f(x) dx,  f 2 =  f(x)g(x) dx.
                                              a               a
                     2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
                      ◦
                                                                      1 – λ 1 Ag 1
                                    y(x)= f(x)+ Cy 1 (x),  y 1 (x)= g(x)+      ,
                                                                      λ 1 A(b – a)
                     where C is an arbitrary constant and y 1 (x) is an eigenfunction of the equation corresponding
                     to the characteristic value λ 1 .
                     3 . The solution with λ = λ 2 ≠ λ 1 and f 1 = f 2 = 0 is given by the formulas of item 2 in
                                                                                           ◦
                      ◦
                     which one must replace λ 1 and y 1 (x)by λ 2 and y 2 (x), respectively.
                                                                                         2
                     4 . Solution with λ = λ 1,2 = λ ∗ and f 1 = f 2 = 0, where the characteristic value λ ∗ =
                      ◦
                                                                                      (A + B)g 1
                     is double:
                                                                      (A – B)g 1
                                    y(x)= f(x)+ Cy ∗ (x),  y ∗ (x)= g(x) –     .
                                                                      2A(b – a)
                     Here C is an arbitrary constant and y ∗ (x) is an eigenfunction of the equation corresponding
                     to λ ∗ .

                              b

               5.    y(x) – λ  [g(x)+ h(t)]y(t) dt = f(x).
                             a
                     The characteristic values of the equation:


                                                               2
                                               s 1 + s 3 ±  (s 1 – s 3 ) +4(b – a)s 2
                                         λ 1,2 =                         ,
                                                     2[s 1 s 3 – (b – a)s 2 ]
                     where

                                       b               b                  b
                                 s 1 =  g(x) dx,  s 2 =  g(x)h(x) dx,  s 3 =  h(x) dx.
                                      a               a                  a


                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
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