Page 402 - Handbook Of Integral Equations

b
13. f(t)y(x – t)y(t) dt = Ae λx .
a
Solutions:

λx
λx
y 1 (x)= A/I 0 e , y 2 (x)= – A/I 0 e ,
λx
λx
y 3 (x)= q(I 0 x – I 1 )e , y 4 (x)= –q(I 0 x – I 1 )e ,
where

b
A
m
I m = t f(t) dt, q = , m =0, 1, 2.
2 2
a I 0 I – I I 2
1
0
k
The integral equation has more complicated solutions of the form y(x)= e λx n B k x , where
k=0
the constants B k can be found from the corresponding system of algebraic equations.
b

14. f(t)y(t)y(x – t) dt = A sinh λx.
a
A solution:
y(x)= p sinh λx + q cosh 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 the notation
b b

I 0 = f(t) dt, I cs = f(t) cosh(λt) sinh(λt) dt,
a a
b b

2
2
I cc = f(t) cosh (λt) dt, I ss = f(t) sinh (λt) dt
a a
is used. Different solutions of system (2) generate different solutions (1) of the integral
equation.

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 sinh λx ± k cosh λx , y 3,4 (x)= –p sinh λx ± k cosh λx ,

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

15. f(t)y(t)y(x – t) dt = A cosh λx.
a
A solution:
y(x)= p sinh λx + q cosh 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 6.2.14. Different solutions of system (2) generate
different solutions (1) of the integral equation.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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