Page 67 - Handbook Of Integral Equations

x

3
15. cos [λ(x – t)]y(t) dt = f(x).
a
3
Using the formula cos β = 1 cos 3β + 3 cos β, we arrive at an equation of the form 1.5.8:
4 4
x
1 3

4 cos[3λ(x – t)] + 4 cos[λ(x – t)] y(t) dt = f(x).
a
x

3 3

16. cos (λx) – cos (λt) y(t) dt = f(x), f(a)= f (a)=0.
x
a
1 d f (x)

x
Solution: y(x)= – .
2
3λ dx sin(λx) cos (λx)
x

3 3
17. A cos (λx)+ B cos (λt) y(t) dt = f(x).
a
3
For B = –A, see equation 1.3.16. This is a special case of equation 1.9.4 with g(x) = cos (λx).
Solution:
1 d – 3A x – 3B

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

2 2
18. cos (λx) cos(µt) + cos(βx) cos (γt) y(t) dt = f(x).
a
2
This is a special case of equation 1.9.15 with g 1 (x)=cos (λx), h 1 (t)=cos(µt), g 2 (x)=cos(βx),
2
and h 2 (t) = cos (γt).
x

4
19. cos [λ(x – t)]y(t) dt = f(x).
a
4
Let us transform the kernel of the integral equation using the trigonometric formula cos β =
1 1 3
8
8 cos 4β + 2 cos 2β + , where β = λ(x – t), and differentiate the resulting equation with
respect to x. Then we arrive at an equation of the form 2.5.18:
x

1
y(x) – λ sin[4λ(x – t)] + sin[2λ(x – t)] y(t) dt = f (x).

2 x
a
x
n
20. cos(λx) – cos(λt) y(t) dt = f(x), n =1, 2, ...
a

The right-hand side of the equation is assumed to satisfy the conditions f(a)= f (a)= ··· =
x
f x (n) (a)=0.
(–1) n 1 d n+1
Solution: y(x)= sin(λx) f(x).
n
λ n! sin(λx) dx
x √
21. cos t – cos xy(t) dt = f(x).
a
This is a special case of equation 1.9.38 with g(x)=1 – cos x.
Solution: x
2 1 d
2 sin tf(t) dt
y(x)= sin x √ .
π sin x dx cos t – cos x
a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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