Page 65 - Handbook Of Integral Equations

x

4. cos(λx + βt)y(t) dt = f(x).
a
Differentiating the equation with respect to x twice yields
x

cos[(λ + β)x]y(x) – λ sin(λx + βt)y(t) dt = f (x), (1)

x
a
x

2

cos[(λ + β)x]y(x) – λ sin[(λ + β)x]y(x) – λ cos(λx + βt)y(t) dt = f (x). (2)
x xx
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 tan[(λ + β)x]w = f (x)+ λ f(x), w = cos[(λ + β)x]y(x). (3)
xx
x
Setting x = a in (1) yields the initial condition w(a)= f (a). On solving equation (3) under this

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

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

5. cos(λx) – cos(λt) y(t) dt = f(x).
a
This is a special case of equation 1.9.2 with g(x) = cos(λx).
1 d f (x)

x
Solution: y(x)= – .
λ dx sin(λx)
x

6. A cos(λx)+ B cos(λt) y(t) dt = f(x).
a
This is a special case of equation 1.9.4 with g(x) = cos(λx). For B = –A, see equation 1.5.5.
Solution with B ≠ –A:
A x B
sign cos(λx) d – –
y(x)= cos(λx) A+B cos(λt) A+B f (t) dt .

t
A + B dx a
x

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

8. A 1 cos[λ 1 (x – t)] + A 2 cos[λ 2 (x – t)] y(t) dt = f(x).
a
The equation is equivalent to the equation
x

B 1 sin[λ 1 (x – t)] + B 2 sin[λ 2 (x – t)] y(t) dt = F(x),
a
x
A 1 A 2
B 1 = , B 2 = , F(x)= f(t) dt.
λ 1 λ 2 a
which has the form 1.5.41. (Differentiation of this equation yields the original integral
equation.)
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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