Page 68 - Handbook Of Integral Equations

x
y(t) dt
22. √ = f(x).
a cos t – cos x
Solution: x
1 d sin tf(t) dt
y(x)= √ .
π dx a cos t – cos x
x

λ
23. (cos t – cos x) y(t) dt = f(x), 0 < λ <1.
a
Solution:
x

1 d sin tf(t) dt sin(πλ)

2
y(x)= k sin x , k = .
sin x dx (cos t – cos x) λ πλ
a
x

µ
µ
24. (cos x – cos t)y(t) dt = f(x).
a
µ
This is a special case of equation 1.9.2 with g(x) = cos x.
1 d f (x)

x
Solution: y(x)= – .
µ dx sin x cos µ–1 x
x

µ µ
25. A cos x + B cos t y(t) dt = f(x).
a
µ
For B = –A, see equation 1.5.24. This is a special case of equation 1.9.4 with g(x) = cos x.
Solution:
1 d – Aµ x – Bµ

y(x)= cos x A+B cos t A+B f (t) dt .
t
A + B dx a
x
y(t) dt

26. = f(x), 0 < λ <1.
a (cos t – cos x) λ
Solution: x
sin(πλ) d sin tf(t) dt
y(x)= .
π dx a (cos t – cos x) 1–λ
x

27. (x – t) cos[λ(x – t)]y(t) dt = f(x), f(a)= f (a)=0.
x
a
Differentiating the equation twice yields
x x

y(x) – 2λ sin[λ(x – t)]y(t) dt – λ 2 (x – t) cos[λ(x – t)]y(t) dt = f (x).

xx
a a
Eliminating the third term on the left-hand side with the aid of the original equation, we arrive
at an equation of the form 2.5.16:
x

2

y(x) – 2λ sin[λ(x – t)]y(t) dt = f (x)+ λ f(x).
xx
a
√
cos λ x – t
x
28. √ y(t) dt = f(x).
a x – t
Solution: √
x
1 d cosh λ x – t

y(x)= √ f(t) dt.
π dx x – t
a
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Page 46