Page 27 - Handbook Of Integral Equations

x

2 2
9. Ax + Bt + Cx + Dt + E y(t) dt = f(x).
a
2
2
This is a special case of equation 1.9.6 with g(x)= Ax + Cx and h(t)= Bt + Dt + E.
x
2
10. Axt + Bt + Cx + Dt + E y(t) dt = f(x).
a
This is a special case of equation 1.9.15 with g 1 (x)= x, h 1 (t)= At + C, g 2 (x) = 1, and
2
h 2 (t)= Bt + Dt + E.
x
2
11. Ax + Bxt + Cx + Dt + E y(t) dt = f(x).
a
2
This is a special case of equation 1.9.15 with g 1 (x)= Bx+D, h 1 (t)= t, g 2 (x)= Ax +Cx+E,
and h 2 (t)=1.
1.1-3. Kernels Cubic in the Arguments x and t
x
3
12. (x – t) y(t) dt = f(x), f(a)= f (a)= f (a)= f (a)=0.

x xx xxx
a
1
Solution: y(x)= f (x).
6 xxxx
x

3 3

13. (x – t )y(t) dt = f(x), f(a)= f (a)=0.
x
a
3
This is a special case of equation 1.9.2 with g(x)= x .
1

Solution: y(x)= xf xxx (x) – 2f (x) .

x
3x 3
x
3 3
14. Ax + Bt y(t) dt = f(x).
a
3
For B = –A, see equation 1.1.13. This is a special case of equation 1.9.4 with g(x)= x .
1 d – 3A x – 3B

Solution with 0 ≤ a ≤ x: y(x)= x A+B t A+B f (t) dt .
t
A + B dx a
x

3 3
15. Ax + Bt + C y(t) dt = f(x).
a
3
This is a special case of equation 1.9.5 with g(x)= x .
x

2
2

16. (x t – xt )y(t) dt = f(x), f(a)= f (a)=0.
x
a
2
This is a special case of equation 1.9.11 with g(x)= x and h(x)= x.
1 d 2 1
Solution: y(x)= f(x) .
x dx 2 x
x

2
2
17. (Ax t + Bxt )y(t) dt = f(x).
a
2
For B = –A, see equation 1.1.16. This is a special case of equation 1.9.12 with g(x)= x and
h(x)= x.
Solution:
1 d – A x – B d 1
y(x)= x A+B t A+B f(t) dt .
(A + B)x dx a dt t
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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