Page 316 - Handbook Of Integral Equations

2
b – a 2 ∞ 1 x

9. y(x)+ exp –a ln y(t) dt = f(x).
2a 0 t t
Solution with a >0, b > 0, and x >0:
2
a – b 2 ∞ 1 x
y(x)= f(x)+ exp –b ln f(t) dt.
2b t t
0
•
Reference: F. D. Gakhov and Yu. I. Cherskii (1978).

4.7-3. Kernels Containing Exponential and Trigonometric Functions

b

10. y(x) – λ e µt cos(βx)y(t) dt = f(x).
a
µt
This is a special case of equation 4.9.1 with g(x) = cos(βx) and h(t)= e .
b
11. y(x) – λ e µx cos(βt)y(t) dt = f(x).
a
This is a special case of equation 4.9.1 with g(x)= e µx and h(t) = cos(βt).

∞
12. y(x) – λ e µ(x–t) cos(xt)y(t) dt = f(x).
0
Solution:
∞
f(x) λ µ(x–t)
y(x)= π + π e cos(xt)f(t) dt, λ ≠ ± 2/π.
1 – λ 2 1 – λ 2
2 2 0
b
13. y(x) – λ e µ(x–t) cos[β(x – t)]y(t) dt = f(x).
a
This is a special case of equation 4.9.18 with g 1 (x)= e µx cos(βx), h 1 (t)= e –µt cos(βt),
g 2 (x)= e µx sin(βx), and h 2 (t)= e –µt sin(βt).

b

14. y(x) – λ e µt sin(βx)y(t) dt = f(x).
a
µt
This is a special case of equation 4.9.1 with g(x) = sin(βx) and h(t)= e .
b
15. y(x) – λ e µx sin(βt)y(t) dt = f(x).
a
This is a special case of equation 4.9.1 with g(x)= e µx and h(t) = sin(βt).

∞
16. y(x) – λ e µ(x–t) sin(xt)y(t) dt = f(x).
0
Solution:
∞
f(x) λ µ(x–t)
y(x)= π + π e sin(xt)f(t) dt, λ ≠ ± 2/π.
1 – λ 2 1 – λ 2
2 2 0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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