Page 166 - Handbook Of Integral Equations

x
ln(λx)
4. y(x) – A y(t) dt = f(x).
ln(λt)
a
Solution:
x
ln(λx)
y(x)= f(x)+ A e A(x–t) f(t) dt.
ln(λt)
a
x ln(λt)
5. y(x) – A y(t) dt = f(x).
a ln(λx)
Solution:
x ln(λt)
y(x)= f(x)+ A e A(x–t) f(t) dt.
a ln(λx)
x
k m
6. y(x) – A ln (λx)ln (µt)y(t) dt = f(x).
a
k
m
This is a special case of equation 2.9.2 with g(x)= A ln (λx) and h(t)=ln (µt).
∞

7. y(x)+ a ln(t – x)y(t) dt = f(x).
x
This is a special case of equation 2.9.62 with K(x)= a ln(–x).
For f(x)= m A k exp(–λ k x), where λ k > 0, a solution of the equation has the form
k=1
m
A k
a
y(x)= exp(–λ k x), B k =1 – (ln λ k + C),
B k λ k
k=1
where C = 0.5772 ... is the Euler constant.
∞

2
8. y(x)+ a ln (t – x)y(t) dt = f(x).
x
2
This is a special case of equation 2.9.62 with K(x)= a ln (–x).
For f(x)= m A k exp(–λ k x), where λ k > 0, a solution of the equation has the form
k=1
m
A k a 1 2 2

y(x)= exp(–λ k x), B k =1 + 6 π + (ln λ k + C) ,
B k λ k
k=1
where C = 0.5772 ... is the Euler constant.

2.4-2. Kernels Containing Power-Law and Logarithmic Functions

x
m
k
9. y(x) – A x ln (λt)y(t) dt = f(x).
a
m
k
This is a special case of equation 2.9.2 with g(x)= Ax and h(t)=ln (λt).
x

k
m
10. y(x) – A t ln (λx)y(t) dt = f(x).
a
m k
This is a special case of equation 2.9.2 with g(x)= A ln (λx) and h(t)= t .
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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