Page 51 - Handbook Of Integral Equations

x

42. A sinh[λ(x – t)] + B sinh[µ(x – t)] + C sinh[β(x – t)] y(t) dt = f(x).
a
It assumed that f(a)= f (a) = 0. Differentiating the integral equation twice yields

x
x

2 2
(Aλ + Bµ + Cβ)y(x)+ Aλ sinh[λ(x – t)] + Bµ sinh[µ(x – t)] y(t) dt
a
x

+ Cβ 2 sinh[β(x – t)]y(t) dt = f (x).
xx
a
Eliminating the last integral with the aid of the original equation, we arrive at an equation of
the form 2.3.18:
(Aλ + Bµ + Cβ)y(x)
x
2 2 2 2 2
+ A(λ – β ) sinh[λ(x – t)] + B(µ – β ) sinh[µ(x – t)] y(t) dt = f (x) – β f(x).
xx
a
In the special case Aλ + Bµ + Cβ = 0, this is an equation of the form 1.3.41.
x

2
43. sinh [λ(x – t)]y(t) dt = f(x), f(a)= f (a)= f (a)=0.
x xx
a
Differentiating yields an equation of the form 1.3.34:
x 1

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

Solution: y(x)= λ f (x) – 2f (x).
2 xxx x
x

2 2
44. sinh (λx) – sinh (λt) y(t) dt = f(x), f(a)= f (a)=0.

x
a
1 d f (x)

x
Solution: y(x)= .
λ dx sinh(2λx)
x
2 2
45. A sinh (λx)+ B sinh (λt) y(t) dt = f(x).
a
2
For B = –A, see equation 1.3.44. This is a special case of equation 1.9.4 with g(x) = sinh (λx).
Solution:
1 d – 2A x – 2B

y(x)= sinh(λx) A+B sinh(λt) A+B f (t) dt .
t
A + B dx
a
x
2 2
46. A sinh (λx)+ B sinh (µt) y(t) dt = f(x).
a
2
2
This is a special case of equation 1.9.6 with g(x)= A sinh (λx) and h(t)= B sinh (µt).
x

47. sinh[λ(x – t)] sinh[λ(x + t)]y(t) dt = f(x).
a
Using the formula
1
sinh(α – β) sinh(α + β)= [cosh(2α) – cosh(2β)], α = λx, β = λt,
2
we reduce the original equation to an equation of the form 1.3.5:
x
[cosh(2λx) – cosh(2λt)]y(t) dt =2f(x).
a
1 d f (x)

x
Solution: y(x)= .
λ dx sinh(2λx)
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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