Page 277 - Handbook Of Integral Equations

b
2
3
24. y(x) – λ (Bxt + Ct )y(t) dt = f(x).
a
2
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
2
25. y(x) – λ (Ax t + Bxt )y(t) dt = f(x).
a
2
This is a special case of equation 4.9.17 with g(x)= x 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 4.9.17.

b

3
2
26. y(x) – λ (Ax + Bxt )y(t) dt = f(x).
a
2
3
This is a special case of equation 4.9.18 with g 1 (x)= x , h 1 (t)= A, g 2 (x)= x, and h 2 (t)= Bt .
Solution:
3
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 4.9.18.
b
2
2
3
27. y(x) – λ (Ax + Bx t + Cx + D)y(t) dt = f(x).
a
2
3
2
This is a special case of equation 4.9.18 with g 1 (x)= Ax + Cx + D, h 1 (t)=1, g 2 (x)= x ,
and h 2 (t)= Bt.
Solution:
2
3
2
y(x)= f(x)+ λ[A 1 (Ax + Cx + D)+ A 2 x ],
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.18.
b

2
3
2
28. y(x) – λ (Axt + Bt + Ct + D)y(t) dt = f(x).
a
2
This is a special case of equation 4.9.18 with g 1 (x)= x, h 1 (t)= At , g 2 (x)=1, and h 2 (t)=
3
2
Bt + Ct + D.
Solution:
y(x)= f(x)+ λ(A 1 x + A 2 ),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.18.
b
3
29. 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 =3.
b

3
30. 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 =3.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 256