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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.
                     4 . Solution with λ = λ 1,2 = λ ∗ and f 1 = f 2 = 0, where the characteristic value λ ∗ =
                      ◦
                             2
                                       (A ≠ 0) is double:
                     A(b – a)+(B + C)h 1
                                                 y(x)= f(x)+ Cy ∗ (x),

                     where C is an arbitrary constant and y ∗ (x) is an eigenfunction of the equation corresponding

                     to λ ∗ . Two cases are possible.
                        (a) If A(b – a)+ Bh 1 ≠ Ch 1 , then

                                                        A(b – a)+(B – C)h 1
                                     y ∗ (x)= A + Bxh(x) –                h(x).
                                                               2h 0
                                                                  2
                                                                      2
                        (b) If A(b – a)+ Bh 1 = Ch 1 , then, in view of h 0 [A(b – a )+2Bh 2 ] = 0, we have
                                  
                                                                          2
                                                                              2
                                   A + Bxh(x)             for h 0 ≠ 0 and A(b – a )= –2Bh 2 ,
                                                                              2
                                                                          2
                           y ∗ (x)=  h(x)                  for h 0 = 0 and A(b – a ) ≠ –2Bh 2 ,
                                                                         2   2
                                    C 1 [A + Bxh(x)] + C 2 h(x)  for h 0 = 0 and A(b – a )= –2Bh 2 ,


                     where C 1 and C 2 are arbitrary constants.


                              b

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

                                                                               2
                                                  2
                                  s 1 + s 3 +  (s 1 – s 3 ) +4s 2 2  s 1 + s 3 –  (s 1 – s 3 ) +4s 2 2
                              λ 1 =                     ,  λ 2 =                    ,
                                                                             2
                                                2
                                        2(s 1 s 3 – s )              2(s 1 s 3 – s )
                                                2                            2
                     where
                                      b                b                  b

                                                                            2
                                        2
                                s 1 =  g (x) dx,  s 2 =  g(x)h(x) dx,  s 3 =  h (x) dx.
                                     a                a                  a
                     1 . Solution with λ ≠ λ 1,2 :
                      ◦
                                           y(x)= f(x)+ λ[A 1 g(x)+ A 2 h(x)],
                     where the constants A 1 and A 2 are given by
                                     f 1 – λ(f 1 s 3 – f 2 s 2 )   f 2 – λ(f 2 s 1 – f 1 s 2 )
                            A 1 =      2                ,  A 2 =      2               ,
                                          2
                                                                         2
                                 (s 1 s 3 – s )λ – (s 1 + s 3 )λ +1  (s 1 s 3 – s )λ – (s 1 + s 3 )λ +1
                                                                      2
                                       2
                                              b                   b
                                       f 1 =  f(x)g(x) dx,  f 2 =  f(x)h(x) dx.
                                            a                   a
                      ◦
                     2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
                                                                     1 – λ 1 s 1
                                   y(x)= f(x)+ Cy 1 (x),  y 1 (x)= g(x)+    h(x),
                                                                       λ 1 s 2
                     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.
                      ◦
                     4 . Solution with λ = λ 1,2 = λ ∗ and f 1 = f 2 = 0, where the characteristic value λ ∗ =1/s 1 is
                     double:
                                             y(x)= f(x)+ C 1 g(x)+ C 2 h(x),


                     where C 1 and C 2 are arbitrary constants.


                 © 1998 by CRC Press LLC
               © 1998 by CRC Press LLC
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