Page 46 - Handbook Of Integral Equations

x

14. [cosh(λx) cosh(µt) + cosh(βx) cosh(γt)]y(t) dt = f(x).
a
This is a special case of equation 1.9.15 with g 1 (x) = cosh(λx), h 1 (t) = cosh(µt), g 2 (x)=
cosh(βx), and h 2 (t) = cosh(γt).

x

3
15. cosh [λ(x – t)]y(t) dt = f(x).
a
3
Using the formula cosh β = 1 cosh 3β + 3 cosh β, we arrive at an equation of the form 1.3.8:
4 4
x

1 cosh[3λ(x – t)] + 3 cosh[λ(x – t)] y(t) dt = f(x).

4 4
a
x

3 3

16. cosh (λx) – cosh (λt) y(t) dt = f(x), f(a)= f (a)=0.
x
a

1 d f (x)
x
Solution: y(x)= .
2
3λ dx sinh(λx) cosh (λx)
x
3 3
17. A cosh (λx)+ B cosh (λt) y(t) dt = f(x).
a
3
For B =–A, see equation 1.3.16. This is a special case of equation 1.9.4 with g(x)=cosh (λx).
Solution:
1 d – 3A x – 3B

y(x)= cosh(λx) A+B cosh(λt) A+B f (t) dt .
t
A + B dx a
x
2 2
18. A cosh (λx) cosh(µt)+ B cosh(βx) cosh (γt) y(t) dt = f(x).
a
2
This is a special case of equation 1.9.15 with g 1 (x)= A cosh (λx), h 1 (t) = cosh(µt), g 2 (x)=
2
B cosh(βx), and h 2 (t) = cosh (γt).
x

4
19. cosh [λ(x – t)]y(t) dt = f(x).
a
Let us transform the kernel of the integral equation using the formula
4
3
cosh β = 1 cosh 4β + 1 cosh 2β + , where β = λ(x – t),
8 2 8
and differentiate the resulting equation with respect to x. Then we obtain an equation of the
form 2.3.18:
x

1
y(x)+ λ sinh[4λ(x – t)] + sinh[2λ(x – t)] y(t) dt = f (x).

2 x
a
x
n
20. [cosh(λx) – cosh(λt)] y(t) dt = f(x), n =1, 2, ...
a

The right-hand side of the equation is assumed to satisfy the conditions f(a)= f (a)= ··· =
x
f (n) (a)=0.
x
n+1
sinh(λx) 1 d
Solution: y(x)= f(x).
n
λ n! sinh(λx) dx
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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