Page 66 - Handbook Of Integral Equations

x

2
9. cos [λ(x – t)]y(t) dt = f(x).
a
Differentiating yields an equation of the form 2.5.16:
x

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

x
a
Solution:
2λ 2 x √

y(x)= f (x)+ sin[k(x – t)]f (t) dt, where k = λ 2.
x t
k a
x

2 2

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

x
Solution: y(x)= – .
λ dx sin(2λx)
x

2 2
11. A cos (λx)+ B cos (λt) y(t) dt = f(x).
a
2
For B = –A, see equation 1.5.10. This is a special case of equation 1.9.4 with g(x) = cos (λx).
Solution:
2A x 2B
1 d – –

y(x)= cos(λx) A+B cos(λt) A+B f (t) dt .
t
A + B dx a
x
2 2
12. A cos (λx)+ B cos (µt)+ C y(t) dt = f(x).
a
2
2
This is a special case of equation 1.9.6 with g(x)= A cos (λx) and h(t)= B cos (µt)+ C.
x

13. cos[λ(x – t)] cos[λ(x + t)]y(t) dt = f(x).
a
Using the trigonometric formula

cos(α – β) cos(α + β)= 1 cos(2α) + cos(2β) , α = λx, β = λt,
2
we reduce the original equation to an equation of the form 1.5.6 with A = B =1:
x

cos(2λx) + cos(2λt) y(t) dt =2f(x).
a
Solution with cos(2λx)>0:
d 1 x f (t) dt

t
y(x)= √ √ .
dx cos(2λx) a cos(2λt)
x

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

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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