Page 31 - Handbook Of Integral Equations

x
y(t) dt
29. = f(x), a >0, a + b >0, m =1, 2, ...
ax m + bt m
0
n
1 . For a polynomial right-hand side, f(x)= N A n x , the solution has the form
◦
n=0
N 1
A n m+n–1
t m+n–1 dt
y(x)= x , B n = m .
B n 0 a + bt
n=0
n
◦
2 .For f(x)= x λ N A n x , where λ is an arbitrary number (λ > –1), the solution has the
n=0
form
N 1 t λ+m+n–1 dt
y(x)= x λ A n x m+n–1 , B n = m .
B n 0 a + bt
n=0

3 .For f(x)=ln x A n x n , the solution has the form
N
◦
n=0
N N

A n m+n–1 A n C n m+n–1
y(x)=ln x x – 2 x ,
B n B n
n=0 n=0
1 m+n–1 1 m+n–1
t dt t ln t
B n = m , C n = m dt.
0 a + bt 0 a + bt
1.1-6. Kernels Containing Square Roots

x

√
30. x – ty(t) dt = f(x).
a
Differentiating with respect to x, we arrive at Abel’s equation 1.1.36:
x
y(t) dt

√ =2f (x).
x
a x – t
Solution:
2 d 2 x f(t) dt
y(x)= √ .
π dx 2 x – t
a
x
√ √
31. x – t y(t) dt = f(x).
a
1
This is a special case of equation 1.1.44 with µ = .
2
d √

Solution: y(x)=2 xf (x) .
x
dx
x

√ √
32. A x + B t y(t) dt = f(x).
a
1
This is a special case of equation 1.1.45 with µ = .
2
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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