Page 88 - Handbook Of Integral Equations

x
n
5. e µ(x–t) cosh(λx) – cosh(λt) y(t) dt = f(x), n =1, 2, ...
a
Solution:
n+1

1 µx 1 d –µx
y(x)= e sinh(λx) F µ (x), F µ (x)= e f(x).
n
λ n! sinh(λx) dx
x √
6. e µ(x–t) cosh x – cosh ty(t) dt = f(x), f(a)=0.
a
Solution:
2 µx 1 d
2 x e –µt sinh tf(t) dt
y(x)= e sinh x √ .
π sinh x dx a cosh x – cosh t
x µ(x–t)
e y(t) dt
7. √ = f(x).
a cosh x – cosh t
Solution:
1 µx d x e –µt sinh tf(t) dt
y(x)= e √ .
π dx a cosh x – cosh t
x

λ
8. e µ(x–t) (cosh x – cosh t) y(t) dt = f(x), 0 < λ <1.
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.3.23:
x

λ
(cosh x – cosh t) w(t) dt = e –µx f(x).
a
x

µ(x–t) λ
9. Ae + B cosh x y(t) dt = f(x).
a
µx
λ
This is a special case of equation 1.9.15 with g 1 (x)= Ae , h 1 (t)= e –µt , g 2 (x)= B cosh x,
and h 2 (t)=1.
x
µ(x–t) λ
10. Ae + B cosh t y(t) dt = f(x).
a
µx
This is a special case of equation 1.9.15 with g 1 (x)= Ae , h 1 (t)= e –µt , g 2 (x)= B, and
λ
h 2 (t) = cosh t.
x
λ
λ
11. e µ(x–t) (cosh x – cosh t)y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.3.24:
x
λ
λ
(cosh x – cosh t)w(t) dt = e –µx f(x).
a
x

λ
λ

12. e µ(x–t) A cosh x + B cosh t y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.3.25:
x
λ λ
–µx
A cosh x + B cosh t w(t) dt = e f(x).
a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 66