Page 86 - Handbook Of Integral Equations

1.6-4. Kernels Containing Arccotangent

x

32. arccot(λx) – arccot(λt) y(t) dt = f(x).
a
This is a special case of equation 1.9.2 with g(x) = arccot(λx).
1 d 2 2
Solution: y(x)= – (1 + λ x ) f (x) .

x
λ dx
x

33. A arccot(λx)+ B arccot(λt) y(t) dt = f(x).
a
For B =–A, see equation 1.6.32. This is a special case of equation 1.9.4 with g(x)=arccot(λx).
Solution:
1 d – A x – B

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

34. A arccot(λx)+ B arccot(µt)+ C y(t) dt = f(x).
a
This is a special case of equation 1.9.6 with g(x)= A arccot(λx) and h(t)= B arccot(µt)+ C.
x
n
35. arccot(λx) – arccot(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.
Solution:
(–1) n 2 2 d n+1
y(x)= (1 + λ x ) f(x).
λ n!(1 + λ x ) dx
n
2 2
x
36. arccot(λt) – arccot(λx) y(t) dt = f(x).
a
Solution:
2 x

2 1 d ϕ(t)f(t) dt 1

y(x)= ϕ(x) √ , ϕ(x)= .
2 2
π ϕ(x) dx a arccot(λt) – arccot(λx) 1+ λ x
x y(t) dt
37. √ = f(x).
a arccot(λt) – arccot(λx)
Solution:
λ d x ϕ(t)f(t) dt 1
y(x)= √ , ϕ(x)= .
π dx a arccot(λt) – arccot(λx) 1+ λ x
2 2
x
µ
38. arccot(λt) – arccot(λx) y(t) dt = f(x), 0 < µ <1.
a
Solution:
2
1 d ϕ(t)f(t) dt
x
y(x)= kϕ(x) ,
ϕ(x) dx a [arccot(λt) – arccot(λx)] µ
1 sin(πµ)
ϕ(x)= , k = .
2 2
1+ λ x πµ

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 64