Page 273 - Handbook Of Integral Equations

3 . The solution with λ = λ 2 ≠ λ 1 and f 1 = f 2 = 0 is given by the formulas of item 2 in
◦
◦
which one must replace λ 1 and y 1 (x)by λ 2 and y 2 (x), respectively.
6
◦
4 . Solution with λ = λ 1,2 = λ ∗ and f 1 = f 2 = 0, where λ ∗ = is the double
3
3
(A + B)(b – a )
characteristic value:
y(x)= f(x)+ C 1 y ∗ (x),
where C 1 is an arbitrary constant and y ∗ (x) is an eigenfunction of the equation corresponding
to λ ∗ :
3
3
(A – B)(b – a )
2
y ∗ (x)= x – .
6A(b – a)

b
2
10. y(x) – λ (xt – t )y(t) dt = f(x).
a
This is a special case of equation 4.9.8 with A =0, B = 1, and h(t)= t.
Solution:
y(x)= f(x)+ λ(A 1 + A 2 x),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.8.

b
2
11. y(x) – λ (x – xt)y(t) dt = f(x).
a
This is a special case of equation 4.9.10 with A =0, B = 1, and h(x)= x.
Solution:
2
y(x)= f(x)+ λ(E 1 x + E 2 x),
where E 1 and E 2 are the constants determined by the formulas presented in 4.9.10.

b
2
12. y(x) – λ (Bxt + Ct )y(t) dt = f(x).
a
This is a special case of equation 4.9.9 with A = 0 and h(t)= t.
Solution:
y(x)= f(x)+ λ(A 1 + A 2 x),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.9.

b
2
13. y(x) – λ (Bx + Cxt)y(t) dt = f(x).
a
This is a special case of equation 4.9.11 with A = 0 and h(x)= x.
Solution:
2
y(x)= f(x)+ λ(A 1 x + A 2 x),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.11.

b
2
14. y(x) – λ (Axt + Bx + Cx + D)y(t) dt = f(x).
a
2
This is a special case of equation 4.9.18 with g 1 (x)= Bx + Cx + D, h 1 (t)=1, g 2 (x)= x,
and h 2 (t)= At.
Solution:
2
y(x)= f(x)+ λ[A 1 (Bx + Cx + D)+ A 2 x],
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.18.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 252