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∞

               19.       K(ax + t)y(t) dt = Ae λx .
                      –∞
                     Solution:
                                        A       λ             ∞            λ
                                  y(x)=   exp – x ,      B =    K(z)exp – z dz.
                                        B       a                         a
                                                             –∞
                       ∞

               20.       K(ax + t)y(t) dt = f(x).
                      –∞
                     The transformation τ = –t, z = ax, y(t)= Y (τ) leads to an equation of the form 3.8.15:
                                               ∞

                                                 K(z – τ)Y (τ) dt = f(z/a).
                                              –∞
                       ∞

                                        µt
                          βt
               21.       [e K(ax + t)+ e M(ax – t)]y(t) dt = Ae λx .
                      –∞
                     Solution:
                                        I k (q)e px  – I m (p)e qx  λ          λ
                               y(x)= A                     ,   p = –  – β,  q =  – µ,
                                       I k (p)I k (q) – I m (p)I m (q)  a      a
                     where
                                         ∞                         ∞

                                 I k (q)=  K(z)e (β+q)z  dz,  I m (q)=  M(z)e –(µ+q)z  dz.
                                        –∞                        –∞
                       ∞

               22.       g(xt)y(t) dt = f(x).
                      0
                     By setting
                                  z
                                                    τ
                                         –τ
                             x = e ,  t = e ,  y(t)= e w(τ),  g(ξ)= G(ln ξ),  f(ξ)= F(ln ξ),
                     we arrive at an integral equation with difference kernel of the form 3.8.15:
                                                ∞

                                                  G(z – τ)w(τ) dτ = F(z).
                                               –∞

                       ∞    x
               23.       g     y(t) dt = f(x).
                      0     t
                     By setting
                                         τ
                                                   –τ
                                  z
                             x = e ,  t = e ,  y(t)= e w(τ),  g(ξ)= G(ln ξ),  f(ξ)= F(ln ξ),
                     we arrive at an integral equation with difference kernel of the form 3.8.15:
                                                ∞

                                                  G(z – τ)w(τ) dτ = F(z).
                                               –∞

                       ∞
                             β λ
               24.       g x t  y(t) dt = f(x),   β >0,  λ >0.
                      0
                     By setting
                         x = e z/β ,  t = e –τ/λ ,  y(t)= e τ/λ w(τ),  g(ξ)= G(ln ξ),  f(ξ)=  1  F(β ln ξ),
                                                                                 λ
                     we arrive at an integral equation with difference kernel of the form 3.8.15:
                                                ∞

                                                  G(z – τ)w(τ) dτ = F(z).
                                               –∞



                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
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