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◦
                     3 . The substitution y(x)= e –βx w(x) leads to an equation of the form 6.2.7:
                                            b

                                             e –β(λ+1)t f(t)w(t)w(x + λt) dt = A.
                                           a
                       b

               10.      f(t)y(t)y(x – t) dt = A.
                      a
                     1 . Solutions*
                      ◦

                                        y 1 (x)=  A/I 0 ,  y 2 (x)= –  A/I 0 ,
                                        y 3 (x)= q(I 0 x – I 1 ),  y 4 (x)= –q(I 0 x – I 1 ),
                     where

                                           b                  A
                                            m
                                   I m =   t f(t) dt,  q =   2   2  ,  m =0, 1, 2.
                                         a                 I 0 I – I I 2
                                                             1
                                                                 0
                     2 . The integral equation has some other (more complicated) solutions of the polynomial
                      ◦
                                     k
                     form y(x)=  n    λ k x , where the constants λ k can be found from the corresponding system
                               k=0
                     of algebraic equations. For n = 3, such a solution is presented in 6.1.18.
                       b

               11.      f(t)y(t)y(x – t) dt = Ax + B.
                      a
                     A solution: y(x)= λx + µ, where the constants λ and µ are determined from the following
                     system of two second-order algebraic equations:
                                                                    b
                                    2
                                                                     m
                                                   2
                                              2
                           I 0 λµ + I 1 λ = A,  I 0 µ – I 2 λ = B,  I m =  t f(t) dt,  m = 0, 1, 2.  (1)
                                                                   a
                     Multiplying the first equation by B and the second by –A and adding the results, we obtain
                     the quadratic equation
                                           2
                                       AI 0 z – BI 0 z – AI 2 – BI 1 =0,  z = µ/λ.          (2)
                     In general, to each root of equation (2) two solutions of system (1) correspond. Therefore,
                     the original integral equation can have at most four solutions of this form. If the discriminant
                     of equation (2) is negative, then the integral equation has no such solutions.
                        The integral equation has some other (more complicated) solutions of the polynomial
                                     k
                     form y(x)=  n    λ k x , where the constants λ k can be found from the corresponding system
                               k=0
                     of algebraic equations.
                       b                   n

                                                  k
               12.      f(t)y(t)y(x – t) dt =  A k x .
                      a
                                           k=0
                     This equation has solutions of the form
                                                         n
                                                               k
                                                   y(x)=    λ k x ,                         (1)
                                                         k=0
                     where the constants λ k are determined from the system of algebraic equations obtained by
                     substituting solution (1) into the original integral equation and matching the coefficients of
                     like powers of x.
                 * The arguments of the equations containing y(x–t) in the integrand can vary within the following intervals: (a) –∞<t<∞,
               –∞ < x < ∞ for a = –∞ and b = ∞ or (b) a ≤ t ≤ b, –∞ ≤ x < ∞, for arbitrary a and b such that –∞ < a < b < ∞.
               Case (b) is a special case of (a) if f(t) is nonzero only on the interval a ≤ t ≤ b.




                 © 1998 by CRC Press LLC









                © 1998 by CRC Press LLC
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