Page 189 - Handbook Of Integral Equations

x

2. y(x) – A J ν (λx)y(t) dt = f(x).
a
This is a special case of equation 2.9.2 with g(x)= AJ ν (λx) and h(t)=1.

x

3. y(x) – A J ν (λt)y(t) dt = f(x).
a
This is a special case of equation 2.9.2 with g(x)= A and h(t)= J ν (λt).

x J ν (λx)
4. y(x) – A y(t) dt = f(x).
a J ν (λt)
Solution:
x
J ν (λx)
y(x)= f(x)+ A e A(x–t) f(t) dt.
a J ν (λt)
x J ν (λt)
5. y(x) – A y(t) dt = f(x).
a J ν (λx)
Solution:
x J ν (λt)
y(x)= f(x)+ A e A(x–t) f(t) dt.
a J ν (λx)
∞

6. y(x)+ A J ν [λ(t – x)]y(t) dt = f(x).
x
This is a special case of equation 2.9.62 with K(x)= AJ ν (–λx).

x

7. y(x) – AJ ν (kx)+ B – AB(x – t)J ν (kx) y(t) dt = f(x).
a
This is a special case of equation 2.9.7 with λ = B and g(x)= AJ ν (kx).
x

8. y(x)+ AJ ν (kt)+ B + AB(x – t)J ν (kt) y(t) dt = f(x).
a
This is a special case of equation 2.9.8 with λ = B and g(t)= AJ ν (kt).

x

9. y(x) – λ e µ(x–t) J 0 (x – t)y(t) dt = f(x).
0
Solution:
x
y(x)= f(x)+ R(x – t)f(t) dt,
0
where
2
√ λ √
2
2
R(x)= e µx λ cos 1 – λ x + √ sin 1 – λ x +
1 – λ 2
λ x √ J 1 (t)
2
√ sin 1 – λ (x – t) dt .
1 – λ 2 0 t
x
10. y(x) – A Y ν (λx)y(t) dt = f(x).
a
This is a special case of equation 2.9.2 with g(x)= AY ν (λx) and h(t)=1.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 168