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x

                                                   2
               46.   y(x)+ A    y(t)y(x – t) dt =(AB x + B)e λx .
                              0
                                       λx
                     A solution: y(x)= Be .
                            λ      x             1
               47.   y(x)+       y(t)y(x – t) dt =  2 β sinh(λx).
                           2β  0
                     A solution: y(x)= βI 1 (λx), where I 1 (x) is the modified Bessel function.

                           λ      x
               48.   y(x) –      y(t)y(x – t) dt =  1 2  β sin(λx).
                           2β  0
                     A solution: y(x)= βJ 1 (λx), where J 1 (x) is the Bessel function.

                              x

                                                       λ
               49.   y(x)+ A    x –λ–1 y(t)y(x – t) dt = Bx .
                              0
                                                      λ
                                       λ
                     Solutions: y 1 (x)= β 1 x and y 2 (x)= β 2 x , where β 1,2 are the roots of the quadratic equation
                                                                          2
                                                           1             Γ (λ +1)
                                      2
                                                             λ
                                                                   λ
                                  AIβ + β – B =0,    I =    z (1 – z) dz =       .
                                                          0              Γ(2λ +2)
               5.2. Equations With Quadratic Nonlinearity That Contain
                      Arbitrary Functions
                                              x
                 5.2-1. Equations of the Form  G(···) dt = F (x)
                                            a
                       x

                                2
               1.       K(x, t)y (t) dt = f(x).
                      a
                                         2
                     The substitution w(x)= y (x) leads to the linear equation
                                                 x

                                                  K(x, t)w(t) dt = f(x).
                                                a
                         x
               2.       K(t)y(x)y(t) dt = f(x).
                      a
                     Solutions:
                                                          x         
 –1/2

                                          y(x)= ±f(x) 2    K(t)f(t) dt  .
                                                         a
                       x
                            t
                                                 λ
               3.       f      y(t)y(x – t) dt = Ax .
                           x
                      0
                     Solutions:
                                                              1
                                            A   λ–1                λ–1     λ–1
                                  y(x)= ±     x 2 ,    I =    f(z)z 2 (1 – z) 2 dz.
                                            I
                                                            0
                       x
                            t
               4.       f      y(t)y(x – t) dt = Ae λx .
                           x
                      0
                     Solutions:
                                                     λx           1
                                                 A e                f(z) dz
                                        y(x)= ±     √ ,     I =    √       .
                                                 I   x              z(1 – z)
                                                                0
                 © 1998 by CRC Press LLC








               © 1998 by CRC Press LLC
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