Page 52 - Handbook Of Integral Equations

x

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

3

49. sinh [λ(x – t)]y(t) dt = f(x), f(a)= f (a)= f (a)= f (a)=0.
x xx xxx
a
3
Using the formula sinh β = 1 sinh 3β – 3 sinh β, we arrive at an equation of the form 1.3.41:
4 4
x
1 3

4 sinh[3λ(x – t)] – 4 sinh[λ(x – t)] y(t) dt = f(x).
a
x

3 3
50. sinh (λx) – sinh (λt) y(t) dt = f(x), f(a)= f (a)=0.

x
a
3
This is a special case of equation 1.9.2 with g(x) = sinh (λx).
x

3 3
51. A sinh (λx)+ B sinh (λt) y(t) dt = f(x).
a
3
This is a special case of equation 1.9.4 with g(x) = sinh (λx).
Solution:
1 d – 3A x – 3B
y(x)= sinh(λx) A+B sinh(λt) A+B f (t) dt .

t
A + B dx a
x

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

4
53. sinh [λ(x – t)]y(t) dt = f(x).
a

It is assumed that f(a)= f (a)= ··· = f (a)=0.
x xxxx
Let us transform the kernel of the integral equation using the formula
4
3
sinh 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 arrive at an equation of
the form 1.3.41:
x

1

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

n
54. sinh [λ(x – t)]y(t) dt = f(x), n =2, 3, ...
a
It is assumed that f(a)= f (a)= ··· = f x (n) (a)=0.

x
Let us differentiate the equation with respect to x twice and transform the kernel of the
2
2
resulting integral equation using the formula cosh β = 1 + sinh β, where β = λ(x – t). Then
we have
x x
n
2
2 2

λ n sinh [λ(x – t)]y(t) dt + λ n(n – 1) sinh n–2 [λ(x – t)]y(t) dt = f (x).
xx
a a
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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