Page 366 - Handbook Of Integral Equations

x

2
11. y(x)+ g(x)h(t)y (t) dt = f(x).
a
Differentiating the equation with respect to x yields
x

2
2

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

x x x
a
Eliminating the integral term from (1) with the aid of the original equation, we arrive at a
Riccati ordinary differential equation,
g (x) g (x)

x
x
2

y + g(x)h(x)y – y = f (x) – f(x), (2)
x
x
g(x) g(x)
under the initial condition y(a)= f(a). Equation (2) can be reduced to a second-order linear
ordinary differential equation. For the exact solutions of equation (2) with various specific
functions f, g, and h, see, for example, E. Kamke (1977) and A. D. Polyanin and V. F. Zaitsev
(1995).
x t
2
λ
12. y(x)+ x –λ–1 f y (t) dt = Ax .
0 x
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
1
2
Iβ + β – A =0, I = f(z)z 2λ dz.
0
x

2
13. y(x) – e λt+βx f(x – t)y (t) dt =0.
–∞
This is a special case of equation 5.3.19 with k =2.

∞
2
14. y(x) – e λt+βx f(x – t)y (t) dt =0.
x
A solution:
1 –(λ+β)x ∞ –(λ+2β)z
y(x)= e , A = e f(–z) dz.
A 0
x

5.2-3. Equations of the Form y(x)+ G(···) dt = F (x)
a
x 1 t
15. y(x)+ f y(t)y(x – t) dt = Ae λx .
0 x x
Solutions:
λx
λx
y 1 (x)= B 1 e , y 2 (x)= B 2 e ,
where B 1 and B 2 are the roots of the quadratic equation
1
2
IB + B – A =0, I = f(z) dz.
0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 346