Page 95 - Handbook Of Integral Equations

x

60. e µ(x–t) sin[λ(x – t)]y(t) dt = f(x), f(a)= f (a)=0.
x
a
2
2

Solution: y(x)= 1 f (x) – 2µf (x)+(λ + µ )f(x) .
λ xx x
x

61. e µ(x–t) A 1 sin[λ 1 (x – t)] + A 2 sin[λ 2 (x – t)] y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.5.41:
x

–µx
A 1 sin[λ 1 (x – t)] + A 2 sin[λ 2 (x – t)] w(t) dt = e f(x).
a
x
2
62. e µ(x–t) sin [λ(x – t)]y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.5.43:
x

2
sin [λ(x – t)]w(t) dt = e –µx f(x).
a
x
3
63. e µ(x–t) sin [λ(x – t)]y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.5.49:
x
3
sin [λ(x – t)]w(t) dt = e –µx f(x).
a
x

n
64. e µ(x–t) sin [λ(x – t)]y(t) dt = f(x), n =2, 3, ...
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.5.54:
x

n
sin [λ(x – t)]w(t) dt = e –µx f(x).
a
x √

65. e µ(x–t) sin k x – t y(t) dt = f(x).
a
Solution: √
2 µx d 2 x e –µt cosh k x – t
y(x)= e √ f(t) dt.
πk dx 2 a x – t
x
√
66. e µ(x–t) sin x – sin ty(t) dt = f(x).
a
Solution: x
2 µx 1 d
2 e –µt cos tf(t) dt
y(x)= e cos x √ .
π cos x dx sin x – sin t
a
x e µ(x–t) y(t) dt
67. √ = f(x).
a sin x – sin t
Solution: x
1 µx d e –µt cos tf(t) dt
y(x)= e √ .
π dx sin x – sin t
a

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 73