Page 228 - Handbook Of Integral Equations

∞

40. t z–1 y(t) dt = f(z).
0
The left-hand side of this equation is the Mellin transform of y(t)(z is treated as a complex
variable).
Solution:
1 c+i∞ –z 2
y(t)= t f(z) dz, i = –1.
2πi
c–i∞
For specific f(z), one can use tables of Mellin and Laplace integral transforms to calculate
the integral.
•
References: H. Bateman and A. Erd´ elyi (vol. 2, 1954), V. A. Ditkin and A. P. Prudnikov (1965).

3.1-6. Equation Containing the Unknown Function of a Complicated Argument

1

41. y(xt) dt = f(x).
0
Solution:

y(x)= xf (x)+ f(x).
x

The function f(x) is assumed to satisfy the condition xf(x) =0.
x=0
1
λ
42. t y(xt) dt = f(x).
0
x

λ
The substitution ξ = xt leads to equation ξ y(ξ) dξ = x λ+1 f(x). Differentiating with
0
respect to x yields the solution
y(x)= xf (x)+(λ +1)f(x).

x
λ+1
The function f(x) is assumed to satisfy the condition x f(x) =0.
x=0
1

k m
43. Ax + Bt )y(xt) dt = f(x).
0
The substitution ξ = xt leads to an equation of the form 1.1.50:
x

k+m m m+1
Ax + Bξ y(ξ) dξ = x f(x).
0
1
y(xt) dt
44. √ = f(x).
0 1 – t
The substitution ξ = xt leads to Abel’s equation 1.1.36:
x
y(ξ) dξ √

√ = xf(x).
x – ξ
0
y(xt) dt
1
45. λ = f(x), 0 < λ <1.
0 (1 – t)
The substitution ξ = xt leads to the generalized Abel equation 1.1.46:
x
y(ξ) dξ 1–λ

= x f(x).
(x – ξ) λ
0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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