Page 30 - Handbook Of Integral Equations

x
y(t) dt
27. = f(x), a >0, a + b >0.
ax + bt
0
n
◦
1 . For a polynomial right-hand side, f(x)= N A n x , the solution has the form
n=0
N 1 n
A n n t dt
y(x)= x , B n = .
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 λ+n
A n n t dt
λ
y(x)= x x , B n = .
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 1 n 1 n
A n n A n C n n t dt t ln t
y(x)=ln x x – 2 x , B n = , C n = dt.
B n B n 0 a + bt 0 a + bt
n=0 n=0
4 . For some other special forms of the right-hand side (see items 4 and 5, equation 1.1.26),
◦
the solution may be found by the method of undetermined coefficients.
x
y(t) dt
28. = f(x), a >0, a + b >0.
2
0 ax + bt 2
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 n+1
t n+1 dt
y(x)= x , B n = 2 .
B n 0 a + bt
n=0
2
Example. For a = b = 1 and f(x)= Ax + Bx + C, the solution of the integral equation is:
2A 4B 2C
2
3
y(x)= x + x + x.
1 – ln 2 4 – π ln 2
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+n+1 dt
y(x)= x λ A n x n+1 , B n = 2 .
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 1 1
t n+1 dt t n+1 ln t
A n n+1 A n C n n+1
y(x)=ln x x – 2 x , B n = 2 , C n = 2 dt.
B n B n 0 a + bt 0 a + bt
n=0 n=0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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