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x
                 5.8-3. Equations of the Form y(x)+  K(x, t)G t, y(t) dt = F (x)
                                                  a
                             x


               18.   y(x)+    f t, y(t) dt = g(x).
                            a
                     The solution of this integral equation is determined by the solution of the first-order ordinary
                     differential equation


                                                y + f(x, y) – g (x)=0
                                                 x          x
                     under the initial condition y(a)= g(a). For the exact solutions of the first-order differential
                     equations with various f(x, y) and g(x), see E. Kamke (1977), A. D. Polyanin and V. F. Zaitsev
                     (1995), and V. F. Zaitsev and A. D. Polyanin (1994).
                               x

               19.   y(x)+    (x – t)f t, y(t) dt = g(x).
                            a
                     Differentiating the equation with respect to x yields
                                                   x


                                             y +     f t, y(t) dt = g (x).                  (1)


                                              x                   x
                                                  a
                     In turn, differentiating this equation with respect to x yields the second-order nonlinear
                     ordinary differential equation
                                               y     + f(x, y) – g (x) = 0.                 (2)

                                                xx          xx
                     By setting x = a in the original equation and equation (1), we obtain the initial conditions for
                     y = y(x):

                                             y(a)= g(a),   y (a)= g (a).                    (3)

                                                                  x
                                                            x
                        Equation (2) under conditions (3) defines the solution of the original integral equation. For
                     the exact solutions of the second-order differential equation (2) with various f(x, y) and g(x),
                     see A. D. Polyanin and V. F. Zaitsev (1995), and V. F. Zaitsev and A. D. Polyanin (1994).
                               x
                                   n


               20.   y(x)+    (x – t) f t, y(t) dt = g(x),  n =1, 2, ...
                            a
                     Differentiating the equation n+1 times with respect to x, we obtain an (n+1)st-order nonlinear
                     ordinary differential equation for y = y(x):
                                            y (n+1)  + n! f(x, y) – g (n+1) (x)=0.
                                             x               x
                     This equation under the initial conditions

                                   y(a)= g(a),  y (a)= g (a),  ... ,  y (n) (a)= g x (n) (a),

                                                       x
                                                x
                                                                   x
                     defines the solution of the original integral equation.
                               x


               21.   y(x)+    e λ(x–t) f t, y(t) dt = g(x).
                            a
                     Differentiating the equation with respect to x yields
                                                        x
                                                         λ(x–t)
                                     y + f x, y(x) + λ  e    f t, y(t) dt = g (x).


                                                                          x
                                      x
                                                      a
                     Eliminating the integral term herefrom with the aid of the original equation, we obtain the
                     first-order nonlinear ordinary differential equation
                                          y + f(x, y) – λy + λg(x) – g (x)=0.


                                           x                     x
                     The unknown function y = y(x) must satisfy the initial condition y(a)= g(a). For the exact
                     solutions of the first-order differential equations with various f(x, y) and g(x), see E. Kamke
                     (1977), A. D. Polyanin and V. F. Zaitsev (1995), and V. F. Zaitsev and A. D. Polyanin (1994).
                 © 1998 by CRC Press LLC




               © 1998 by CRC Press LLC
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