Page 390 - Handbook Of Integral Equations

x

25. y(x)+ sin[λ(x – t)]f t, y(t) dt = g(x).
a
Differentiating the equation with respect to x twice yields
x

y (x)+ λ cos[λ(x – t)]f t, y(t) dt = g (x), (1)
x x
a
x

2
y (x)+ λf x, y(x) – λ sin[λ(x – t)]f t, y(t) dt = g (x). (2)

xx xx
a
Eliminating the integral term from (2) with the aid of the original equation, we arrive at
the second-order nonlinear ordinary differential equation
2
2
y + λf(x, y)+ λ y – λ g(x) – g (x) = 0. (3)

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

y(a)= g(a), y (a)= g (a). (4)
x
x
Equation (3) under conditions (4) defines the solution of the original integral equation. For
the exact solutions of the second-order differential equation (3) 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).
5.8-4. Other Equations
1 x t
26. y(x)+ f , y(t) dt = A.
x 0 x
A solution: y(x)= λ, where λ is a root of the algebraic (or transcendental) equation
1

λ + F(λ) – A =0, F(λ)= f(z, λ) dz.
0
x t y(t)
27. y(x)+ f , dt = Ax.
0 x t
A solution: y(x)= λx, where λ is a root of the algebraic (or transcendental) equation

1

λ + F(λ) – A =0, F(λ)= f(z, λ) dz.
0
∞

–λx
28. y(x)+ f t – x, y(t – x) y(t) dt = ae .
x
Solutions: y(x)= b k e –λx , where b k are roots of the algebraic (or transcendental) equation
∞

b + bI(b)= a, I(b)= f(z, be –λz )e –λz dz.
0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 370