Page 215 - Handbook Of Integral Equations

x
1 t
68. y(x)+ f y(t) dt = A cos(ln x).
x x
0
A solution:
AI c AI s
y(x)= cos(ln x)+ sin(ln x),
2
2
I + I s 2 I + I 2 s
c
c
1 1

I c =1 + f(t) cos(ln t) dt, I s = f(t) sin(ln t) dt.
0 0
x 1 t
69. y(x)+ f y(t) dt = A sin(ln x).
0 x x
A solution:
AI s AI c
y(x)= – cos(ln x)+ sin(ln x),
2
2
I + I 2 I + I 2
c s c s
1 1
I c =1 + f(t) cos(ln t) dt, I s = f(t) sin(ln t) dt.
0 0
x
1 t
β
β
70. y(x)+ f y(t) dt = Ax cos(ln x)+ Bx sin(ln x).
x x
0
A solution:
β β
y(x)= px cos(ln x)+ qx sin(ln x),
where
AI c – BI s AI s + BI c
p = , q = ,
2
2
I + I s 2 I + I 2 s
c
c
1 1

β
β
I c =1 + f(t)t cos(ln t) dt, I s = f(t)t sin(ln t) dt.
0 0
x 1 t
71. y(x)+ f y(t) dt = g(x).
0 x x
1 . For a polynomial right-hand side,
◦
N
n
g(x)= A n x
n=0
a solution bounded at zero is given by
N 1
A n n n

y(x)= x , f n = f(z)z dz.
1+ f n 0
n=0
Here its is assumed that f 0 < ∞ and f n ≠ –1(n =0, 1, 2, ... ).
If for some n the relation f n = –1 holds, then a solution differs from the above case in
one term and has the form
n–1 N 1
A m m A m m A n n n
¯
y(x)= x + x + x ln x, f n = f(z)z ln zdz.
¯
1+ f m 1+ f m f n 0
m=0 m=n+1
For arbitrary g(x) expandable into power series, the formulas of item 1 can be used, in
◦
which one should set N = ∞. In this case, the convergence radius of the obtained solution y(x)
is equal to that of the function g(x).
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 194