Page 134 - Handbook Of Integral Equations

x

2 3
28. y(x)+ A x t – t y(t) dt = f(x).
a
2
2
The transformation z = x , τ = t , y(x)= w(z) leads to an equation of the form 2.1.4:
z

1
w(z)+ A (z – τ)w(τ) dτ = F(z), F(z)= f(x).
2
a 2
x

2 3
29. y(x)+ Ax t + Bt y(t) dt = f(x).
a
2
2
The transformation z = x , τ = t , y(x)= w(z) leads to an equation of the form 2.1.6:
z

1 1
w(z)+ Az + Bτ w(τ) dτ = F(z), F(z)= f(x).
2 2
a 2
x

3 2
30. y(x)+ B 2x – xt y(t) dt = f(x).
a
This is a special case of equation 2.1.55 with λ =2, µ = 2, and B = –2A.
x

3 2
31. y(x) – A 4x – 3x t y(t) dt = f(x).
a
This is a special case of equation 2.1.55 with λ = 3 and µ =1.
x
3 2 2
32. y(x)+ ABx – ABx t – Ax – B y(t) dt = f(x).
a
2
This is a special case of equation 2.9.7 with g(x)= Ax and λ = B.
Solution:
x
y(x)= f(x)+ R(x – t)f(t) dt,
a
x
3
3
3
3
2

R(x, t)=(Ax + B)exp 1 A(x – t ) + B 2 exp 1 A(s – t )+ B(x – s) ds.
3 3
t
x
2 3 2
33. y(x)+ ABxt – ABt + At + B y(t) dt = f(x).
a
2
This is a special case of equation 2.9.8 with g(t)= At and λ = B.
Solution:
x
y(x)= f(x)+ R(x – t)f(t) dt,
a
x
3
3
2
3
3

R(x, t)= –(At + B)exp 1 A(t – x ) + B 2 exp 1 A(s – x )+ B(t – s) ds.
3 3
t
2.1-4. Kernels Containing Higher-Order Polynomials in x and t
x
n
34. y(x)+ A (x – t) y(t) dt = f(x), n =1, 2, ...
a
1 . Differentiating the equation n + 1 times with respect to x yields an (n + 1)st-order linear
◦
ordinary differential equation with constant coefficients for y = y(x):
y x (n+1) + An! y = f x (n+1) (x).

This equation under the initial conditions y(a)= f(a), y (a)= f (a), ... , y x (n) (a)= f x (n) (a)

x
x
determines the solution of the original integral equation.
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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