Page 274 - Handbook Of Integral Equations

b

2
2
15. y(x) – λ (Ax + Bt + Cx + Dt + E)y(t) dt = f(x).
a
2
This is a special case of equation 4.9.18 with g 1 (x)= Ax + Cx, h 1 (t)=1, g 2 (x) = 1, and
2
h 2 (t)= Bt + Dt + E.
Solution:
2
y(x)= f(x)+ λ[A 1 (Ax + Cx)+ A 2 ],
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.18.

b
16. y(x) – λ [Ax + B +(Cx + D)(x – t)]y(t) dt = f(x).
a
2
This is a special case of equation 4.9.18 with g 1 (x)= Cx +(A + D)x + B, h 1 (t)=1,
g 2 (x)= Cx + D, and h 2 (t)= –t.
Solution:
2
y(x)= f(x)+ λ[A 1 (Cx + Ax + Dx + B)+ A 2 (Cx + D)],

where A 1 and A 2 are the constants determined by the formulas presented in 4.9.18.
b

17. y(x) – λ [At + B +(Ct + D)(t – x)]y(t) dt = f(x).
a
2
This is a special case of equation 4.9.18 with g 1 (x)=1, h 1 (t)= Ct +(A+D)t+B, g 2 (x)= x,
and h 2 (t)= –(Ct + D).
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 4.9.18.

b

2
18. y(x) – λ (x – t) y(t) dt = f(x).
a
This is a special case of equation 4.9.19 with g(x)= x, h(t)= –t, and m =2.
b
2
19. y(x) – λ (Ax + Bt) y(t) dt = f(x).
a
This is a special case of equation 4.9.19 with g(x)= Ax, h(t)= Bt, and m =2.

4.1-3. Kernels Cubic in the Arguments x and t

b

3
3
20. y(x) – λ (x + t )y(t) dt = f(x).
a
The characteristic values of the equation:
1 1
λ 1 = , λ 2 = .
1 (b – a )+ 1 (b – a )(b – a) 1 (b – a ) – 1 (b – a )(b – a)
7
4
4
7
4
7
4
7
4 7 4 7
1 . Solution with λ ≠ λ 1,2 :
◦
3
y(x)= f(x)+ λ(A 1 x + A 2 ),
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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