Page 233 - Handbook Of Integral Equations

a

λt
10. x – ke – k y(t) dt = f(x).

0
This is a special case of equation 3.8.6 with g(t)= ke λt – k.
b

2
11. exp(λx ) – exp(λt ) y(t) dt = f(x), λ >0.
2
a
2
This is a special case of equation 3.8.3 with g(x) = exp(λx ).
Solution:

1 d 1
2

y(x)= exp(–λx )f (x) .
x
4λ dx x
The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2 ◦
of equation 3.8.3).
1 ∞ t 2
12. √ exp – y(t) dt = f(x).
πx 0 4x
Applying the Laplace transformation to the equation, we obtain
√
˜ y( p ) ∞ –pt
˜
˜
√ = f(p), f(p)= e f(t) dt.
p
0
˜
2
2
Substituting p by p and solving for the transform ˜y,we find that ˜y(p)= pf(p ). The inverse
Laplace transform provides the solution of the original integral equation:
–1
pt
2
˜
–1
y(t)= L {pf(p )}, L {g(p)}≡ 1 c+i∞ e g(p) dp.
2πi
c–i∞
∞

2
13. exp[–g(x)t ]y(t) dt = f(x).
0
Assume that g(0) = ∞, g(∞) = 0, and g <0.

x
1
The substitution z = leads to equation 3.2.12:
4g(x)
1 ∞ t 2
√ exp – y(t) dt = F(z),
πz 0 4z
2 1
where the function F(z) is determined by the relations F = √ f(x) g(x) and z =
π 4g(x)
by means of eliminating x.
3.3. Equations Whose Kernels Contain Hyperbolic
Functions
3.3-1. Kernels Containing Hyperbolic Cosine
b

1. cosh(λx) – cosh(λt) y(t) dt = f(x).

a
This is a special case of equation 3.8.3 with g(x) = cosh(λx).
Solution:

1 d f (x)

x
y(x)= .
2λ dx sinh(λx)
The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2 ◦
of equation 3.8.3).
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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