Page 321 - Handbook Of Integral Equations

∞
√
4. y(x)+ λ J ν 2 xt y(t) dt = f(x).
0
1
2
1 2
By setting x = z , t = τ , y(x)= Y (z), and f(x)= F(z), we arrive at an equation of the
2 2
form 4.8.3:

∞
Y (z)+ λ τJ ν (zτ)Y (τ) dτ = F(z).
0
b

5. y(x) – λ [A + B(x – t)J ν (βt)]y(t) dt = f(x).
a
This is a special case of equation 4.9.8 with h(t)= J ν (βt).
b
6. y(x) – λ [A + B(x – t)J ν (βx)]y(t) dt = f(x).
a
This is a special case of equation 4.9.10 with h(x)= J ν (βx).
b
7. y(x) – λ [AJ µ (αx)+ BJ ν (βt)]y(t) dt = f(x).
a
This is a special case of equation 4.9.5 with g(x)= AJ µ (αx) and h(t)= BJ ν (βt).

b
8. y(x) – λ [AJ µ (x)J ν (t)+ BJ ν (x)J µ (t)]y(t) dt = f(x).
a
This is a special case of equation 4.9.17 with g(x)= J µ (x) and h(t)= J ν (t).
b

9. y(x) – λ Y ν (βx)y(t) dt = f(x).
a
This is a special case of equation 4.9.1 with g(x)= Y ν (βx) and h(t)=1.
b

10. y(x) – λ Y ν (βt)y(t) dt = f(x).
a
This is a special case of equation 4.9.1 with g(x) = 1 and h(t)= Y ν (βt).
b
11. y(x) – λ [A + B(x – t)Y ν (βt)]y(t) dt = f(x).
a
This is a special case of equation 4.9.8 with h(t)= Y ν (βt).
b

12. y(x) – λ [A + B(x – t)Y ν (βx)]y(t) dt = f(x).
a
This is a special case of equation 4.9.10 with h(x)= Y ν (βx).
b

13. y(x) – λ [AY µ (αx)+ BY ν (βt)]y(t) dt = f(x).
a
This is a special case of equation 4.9.5 with g(x)= AY µ (αx) and h(t)= BY ν (βt).
b
14. y(x) – λ [AY µ (x)Y µ (t)+ BY ν (x)Y ν (t)]y(t) dt = f(x).
a
This is a special case of equation 4.9.14 with g(x)= Y µ (x) and h(t)= Y ν (t).

b
15. y(x) – λ [AY µ (x)Y ν (t)+ BY ν (x)Y µ (t)]y(t) dt = f(x).
a
This is a special case of equation 4.9.17 with g(x)= Y µ (x) and h(t)= Y ν (t).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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