Page 49 - Handbook Of Integral Equations

x

36. sinh(λx + βt)y(t) dt = f(x).
a
For β = –λ, see equation 1.3.34. Assume that β ≠ –λ.
Differentiating the equation with respect to x twice yields

x
sinh[(λ + β)x]y(x)+ λ cosh(λx + βt)y(t) dt = f (x), (1)

x
a
x
2
sinh[(λ + β)x]y(x) + λ cosh[(λ + β)x]y(x)+ λ sinh(λx + βt)y(t) dt = f (x). (2)

xx
x
a
Eliminating the integral term from (2) with the aid of the original equation, we arrive at the
first-order linear ordinary differential equation
2

w + λ coth[(λ + β)x]w = f (x) – λ f(x), w = sinh[(λ + β)x]y(x). (3)

x
xx
Setting x = a in (1) yields the initial condition w(a)= f (a). On solving equation (3) with this

x
condition, after some manipulations we obtain the solution of the original integral equation
in the form
1 λ cosh[(λ + β)x]

y(x)= f (x) – f(x)
x
2
sinh[(λ + β)x] sinh [(λ + β)x]
λβ x k–2 λ
– k+1 f(t) sinh [(λ + β)t] dt, k = .
sinh [(λ + β)x] a λ + β
x

37. [sinh(λx) – sinh(λt)]y(t) dt = f(x), f(a)= f (a)=0.
x
a
This is a special case of equation 1.9.2 with g(x) = sinh(λx).
1 d f (x)

x
Solution: y(x)= .
λ dx cosh(λx)
x

38. [A sinh(λx)+ B sinh(λt)]y(t) dt = f(x).
a
For B = –A, see equation 1.3.37. This is a special case of equation 1.9.4 with g(x) = sinh(λx).
A x B
1 d – –

Solution: y(x)= sinh(λx) A+B sinh(λt) A+B f (t) dt .
t
A + B dx
a
x

39. [A sinh(λx)+ B sinh(µt)]y(t) dt = f(x).
a
This is a special case of equation 1.9.6 with g(x)= A sinh(λx), and h(t)= B sinh(µt).
x

40. µ sinh[λ(x – t)] – λ sinh[µ(x – t)] y(t) dt = f(x).
a
It is assumed that f(a)= f (a)= f (a)= f (a)=0.

x xx xxx
Solution:
2 2
2
2
f xxxx – (λ + µ )f xx + λ µ f

y(x)= , f = f(x).
3
µλ – λµ 3
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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