Page 403 - Handbook Of Integral Equations

b
16. f(t)y(t)y(x – t) dt = A sin λx.
a
A solution:
y(x)= p sin λx + q cos λx. (1)

Here p and q are roots of the algebraic system

2
2
2
2
I 0 pq + I cs (p + q )= A, I cc q – I ss p = 0, (2)
where
b b

I 0 = f(t) dt, I cs = f(t) cos(λt) sin(λt) dt,
a a
b b

2 2
I cc = f(t) cos (λt) dt, I ss = f(t) sin (λt) dt.
a a

It follows from the second equation of (2) that q = ± I ss /I cc p. Using this expression to
eliminate q from the first equation of (2), we obtain the following four solutions:

y 1,2 (x)= p sin λx ± k cos λx , y 3,4 (x)= –p sin λx ± k cos λx ,

A
I ss
k = , p = 2 .
I cc (1 + k )I cs ± kI 0
b

17. f(t)y(t)y(x – t) dt = A cos λx.
a
A solution:
y(x)= p sin λx + q cos λx. (1)

Here p and q are roots of the algebraic system
2
2
2
2
I 0 pq + I cs (p + q )=0, I cc q – I ss p = A, (2)
where we use the notation introduced in 6.2.16. Different solutions of system (2) generate
different solutions (1) of the integral equation.
1
18. y(t)y(ξ) dt = A, ξ = f(x)t.
0
◦
1 . Solutions:
√ √
y 1 (t)= A, y 2 (t)= – A,
√ √
y 3 (t)= A (3t – 2), y 4 (t)= – A (3t – 2),
√ √
2
2
y 5 (t)= A (10t – 12t + 3), y 6 (t)= – A (10t – 12t + 3).
2 . The integral equation has some other (more complicated) solutions of the polynomial
◦
k
form y(t)= n B k t , where the constants B k can be found from the corresponding system of
k=0
algebraic equations.
◦
3 . The substitution z = f(x) leads to an equation of the form 6.1.12.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 383