Page 70 - Handbook Of Integral Equations

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 cot[(λ + β)x]w = f (x)+ λ f(x), w = sin[(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) under this
x
condition, after some transformation we obtain the solution of the original integral equation
in the form
1 λ cos[(λ + β)x]

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

37. sin(λx) – sin(λt) y(t) dt = f(x).
a
This is a special case of equation 1.9.2 with g(x) = sin(λx).

1 d f (x)
x
Solution: y(x)= .
λ dx cos(λx)
x

38. A sin(λx)+ B sin(λt) y(t) dt = f(x).
a
This is a special case of equation 1.9.4 with g(x) = sin(λx). For B = –A, see equation 1.5.37.
Solution with B ≠ –A:
sign sin(λx) d – A x – B

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

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

40. µ sin[λ(x – t)] – λ sin[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
3
λµ – λ µ
x

41. A 1 sin[λ 1 (x – t)] + A 2 sin[λ 2 (x – t)] y(t) dt = f(x), f(a)= f (a)=0.
x
a
This equation can be solved in the same manner as equation 1.3.41, i.e., by reducing it to a
second-order linear ordinary differential equation with constant coefficients.
Let
A 1 λ 2 + A 2 λ 1
∆ = –λ 1 λ 2 .
A 1 λ 1 + A 2 λ 2
◦
1 . Solution for ∆ >0:
x
(A 1 λ 1 + A 2 λ 2 )y(x)= f (x)+ Bf(x)+ C sinh[k(x – t)]f(t) dt,

xx
a
√
2
2 2
2
2
2
2
k = ∆, B = ∆ + λ + λ , C = √ 1 ∆ +(λ + λ )∆ + λ λ .
1 2 1 2 1 2
∆
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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