Page 28 - Handbook Of Integral Equations

x

2
3
18. (Ax + Bxt )y(t) dt = f(x).
a
2
3
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t)=1, g 2 (x)= Bx, and h 2 (t)= t .
x
3
2
19. (Ax + Bx t)y(t) dt = f(x).
a
2
3
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t)=1, g 2 (x)= Bx , and
h 2 (t)= t.
x
3
2
20. (Ax t + Bt )y(t) dt = f(x).
a
3
2
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t)= t, g 2 (x)= B, and h 2 (t)= t .
x

3
2
21. (Axt + Bt )y(t) dt = f(x).
a
3
2
This is a special case of equation 1.9.15 with g 1 (x)= Ax, h 1 (t)= t , g 2 (x)= B, and h 2 (t)= t .
x
3 3 2 2
22. A 3 x + B 3 t + A 2 x + B 2 t + A 1 x + B 1 t + C y(t) dt = f(x).
a
2
3
This is a special case of equation 1.9.6 with g(x)= A 3 x + A 2 x + A 1 x + C and h(t)=
3
2
B 3 t + B 2 t + B 1 t.
1.1-4. Kernels Containing Higher-Order Polynomials in x and t
x
n
23. (x – t) y(t) dt = f(x), n =1, 2, ...
a

It is assumed that the right-hand of the equation satisfies the conditions f(a)= f (a)= ··· =
x
f x (n) (a)=0.
1 (n+1)
Solution: y(x)= f x (x).
n!
m
Example. For f(x)= Ax , where m is a positive integer, m > n, the solution has the form
Am!
y(x)= x m–n–1 .
n!(m – n – 1)!
x
n
n

24. (x – t )y(t) dt = f(x), f(a)= f (a)=0, n =1, 2, ...
x
a
1 d f (x)

x
Solution: y(x)= .
n dx x n–1
x
n n+1 n n+1
25. t x – x t y(t) dt = f(x), n =2, 3, ...
a
n
This is a special case of equation 1.9.11 with g(x)= x n+1 and h(x)= x .
1 d 2 f(x)
Solution: y(x)= .
n
x dx 2 x n
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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