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b

8. A sin λ|x – t| + B sin µ|x – t| y(t) dt = f(x), –∞ < a < b < ∞.
a
Let us remove the modulus in the integrand and differentiate the equation with respect to x
twice to obtain
b
2 2
2(Aλ + Bµ)y(x) – Aλ sin λ|x – t| + Bµ sin µ|x – t| y(t) dt = f (x). (1)
xx
a

Eliminating the integral term with sin µ|x – t| from (1) with the aid of the original equation,
we find that
b

2
2
2

2(Aλ + Bµ)y(x)+ A(µ – λ ) sin λ|x – t| y(t) dt = f (x)+ µ f(x). (2)
xx
a
For Aλ+Bµ = 0, this is an equation of the form 3.5.7 and for Aλ+Bµ ≠ 0, this is an equation
of the form 4.5.29.
The right-hand side f(x) must satisfy certain relations, which can be obtained by setting
x = a and x = b in the original equation (a similar procedure is used in 3.5.7).
b

9. sin(λx) – sin(λt) y(t) dt = f(x).

a
This is a special case of equation 3.8.3 with g(x) = sin(λx).
Solution:

1 d f (x)

x
y(x)= .
2λ dx cos(λx)
◦
The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2 of
equation 3.8.3).
a

10. sin(βx) – sin(µt) y(t) dt = f(x), β >0, µ >0.

0
This is a special case of equation 3.8.4 with g(x) = sin(βx) and λ = µ/β.
b

3

11. sin λ|x – t| y(t) dt = f(x).
a
3
1
Using the formula sin β = – sin 3β + 3 sin β, we arrive at an equation of the form 3.5.8:
4 4
b

1 3
– A sin 3λ|x – t| + A sin λ|x – t| y(t) dt = f(x).
4 4
a
n
b

12. A k sin λ k |x – t| y(t) dt = f(x), –∞ < a < b < ∞.
a
k=1
◦
1 . Let us remove the modulus in the kth summand of the integrand:
b x b

I k (x)= sin λ k |x – t| y(t) dt = sin[λ k (x – t)]y(t) dt + sin[λ k (t – x)]y(t) dt. (1)
a a x
Differentiating (1) with respect to x yields
x b

I = λ k cos[λ k (x – t)]y(t) dt – λ k cos[λ k (t – x)]y(t) dt,
k
a x (2)
x b
I =2λ k y(x) – λ 2 k sin[λ k (x – t)]y(t) dt – λ 2 k sin[λ k (t – x)]y(t) dt,

k
a x
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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