Page 42 - Handbook Of Integral Equations

x e λ(x–t)
41. √ y(t) dt = f(x).
a x – t
Solution: x
1 λx d e –λt f(t) dt
y(x)= e √ .
π dx x – t
a
x

λ µ(x–t)
42. (x – t) e y(t) dt = f(x), 0 < λ <1.
a
Solution: x
d 2 e –µt f(t) dt sin(πλ)
µx
y(x)= ke , k = .
dx 2 (x – t) λ πλ
a
x λ(x–t)
e
43. y(t) dt = f(x), 0 < µ <1.
a (x – t) µ
Solution: x
sin(πµ) λx d e –λt f(t)
y(x)= e dt.
π dx (x – t) 1–µ
a
x

√ √
λ µ(x–t)
44. x – t e y(t) dt = f(x), 0 < λ <1.
a
The substitution u(x)= e –µx y(x) leads to an equation of the form 1.1.48:
x

√ √
λ –µx
x – t u(t) dt = e f(x).
a
x e µ(x–t) y(t) dt
45. √ λ = f(x), 0 < λ <1.
t
a x – √
The substitution u(x)= e –µx y(x) leads to an equation of the form 1.1.49:

x
u(t) dt –µx

√ √ = e f(x).
a ( x – t) λ
x e λ(x–t)
46. √ y(t) dt = f(x).
2
a x – t 2
2 λx d x te –λt
Solution: y = e √ f(t) dt.
π dx a x – t 2
2
x

2
2
47. exp[λ(x – t )]y(t) dt = f(x).
a
Solution: y(x)= f (x) – 2λxf(x).

x
x
2
2
48. [exp(λx ) – exp(λt )]y(t) dt = f(x).
a
2
This is a special case of equation 1.9.2 with g(x) = exp(λx ).
1 d f (x)

x
Solution: y(x)= .
2
2λ dx x exp(λx )
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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