Page 74 - Handbook Of Integral Equations

x

λ
58. (sin x – sin t) y(t) dt = f(x), 0 < λ <1.
a
Solution:
1 d cos tf(t) dt sin(πλ)

2 x
y(x)= k cos x , k = .
cos x dx a (sin x – sin t) λ πλ
x

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

1 d f (x)
x
Solution: y(x)= .
µ dx cos x sin µ–1 x
x

µ µ
60. A| sin(λx)| + B| sin(λt)| y(t) dt = f(x).
a
µ
This is a special case of equation 1.9.4 with g(x)= | sin(λx)| .
Solution:
1 d – Aµ x – Bµ

y(x)= sin(λx) A+B sin(λt) A+B f (t) dt .
t
A + B dx a
x y(t) dt
61. = f(x), 0 < µ <1.
a [sin(λx) – sin(λt)] µ
This is a special case of equation 1.9.42 with g(x) = sin(λx) and h(x) ≡ 1.
Solution:
λ sin(πµ) d x cos(λt)f(t) dt
y(x)= .
π dx a [sin(λx) – sin(λt)] 1–µ
x

62. (x – t) sin[λ(x – t)]y(t) dt = f(x), f(a)= f (a)= f (a)=0.
x xx
a
Double differentiation yields
x x

2λ cos[λ(x – t)]y(t) dt – λ 2 (x – t) sin[λ(x – t)]y(t) dt = f (x).

xx
a a
Eliminating the second integral on the left-hand side of this equation with the aid of the
original equation, we arrive at an equation of the form 1.5.1:
x
1 2
cos[λ(x – t)]y(t) dt = f (x)+ λ f(x) .

xx
a 2λ
Solution:
1 1 3 x
y(x)= f xxx (x)+ λf (x)+ λ f(t) dt.

x
2λ 2 a
x

β γ
63. Ax + B sin (λt)+ C]y(t) dt = f(x).
a
γ
β
This is a special case of equation 1.9.6 with g(x)= Ax and h(t)= B sin (λt)+ C.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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