Page 404 - Handbook Of Integral Equations

b
2
6.2-2. Equations of the Form y(x)+ K(x, t)y (t) dt = F (x)
a
b

2
19. y(x)+ f(x)y (t) dt =0.
a
b –1
2
Solutions: y 1 (x)=0 and y 2 (x)= λf(x), where λ = – f (t) dt .
a
b
2
20. y(x)+ f(x)g(t)y (t) dt =0.
a
This is a special case of equation 6.8.29.
b –1

2
Solutions: y 1 (x)=0 and y 2 (x)= λf(x), where λ = – f (t)g(t) dt .
a
b

2
21. y(x)+ A y (t) dt = f(x).
a
This is a special case of equation 6.8.27.
A solution: y(x)= f(x)+ λ, where λ is determined by the quadratic equation

b b
2
2
A(b – a)λ +(1+2AI 1 )λ + AI 2 = 0, where I 1 = f(t) dt, I 2 = f (t) dt.
a a
b

2
22. y(x)+ g(t)y (t) dt = f(x).
a
This is a special case of equation 6.8.29.
A solution: y(x)= f(x)+ λ, where λ is determined by the quadratic equation

b
2
m
I 0 λ +(1+2I 1 )λ + I 2 = 0, where I m = f (t)g(t) dt, m = 0,1,2.
a
b

2
23. y(x)+ g(x)y (t) dt = f(x).
a
Solution: y(x)= λg(x)+ f(x), where λ is determined by the quadratic equation

2
I gg λ +(1+2I fg )λ + I ff =0,
b b b
2
2
I gg = g (t) dt, I fg = f(t)g(t) dt, I ff = f (t) dt.
a a a
b

2
24. y(x)+ g 1 (x)h 1 (t)+ g 2 (x)h 2 (t) y (t) dt = f(x).
a
A solution: y(x)= λ 1 g 1 (x)+ λ 2 g 2 (x)+ f(x), where the constants λ 1 and λ 2 can be found
from a system of two second-order algebraic equations (this system can be obtained from the
more general system presented in 6.8.42).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 384