Page 260 - Handbook Of Integral Equations
P. 260
x
∞
β
25. g y(t) dt = f(x), β >0, λ >0.
0 t λ
By setting
x = e z/β , t = e τ/λ , y(t)= e –τ/λ w(τ), g(ξ)= G(ln ξ), f(ξ)= 1 F(β ln ξ),
λ
we arrive at an integral equation with difference kernel of the form 3.8.15:
∞
G(z – τ)w(τ) dτ = F(z).
–∞
b
3.8-5. Equations of the Form K(x, t)y(···) dt = F (x)
a
b
26. f(t)y(xt) dt = Ax + B.
a
Solution:
A B b b
y(x)= x + , I 0 = f(t) dt, I 1 = tf(t) dt.
I 1 I 0 a a
b
β
27. f(t)y(xt) dt = Ax .
a
Solution:
A β b β
y(x)= x , B = f(t)t dt.
B a
b
28. f(t)y(xt) dt = A ln x + B.
a
Solution:
y(x)= p ln x + q,
where
A B AI l b b
p = , q = – 2 , I 0 = f(t) dt, I l = f(t)ln tdt.
I 0 I 0 I a a
0
b
β
29. f(t)y(xt) dt = Ax ln x.
a
Solution:
β
β
y(x)= px ln x + qx ,
where
A AI 2 b β b β
p = , q = – 2 , I 1 = f(t)t dt, I 2 = f(t)t ln tdt.
I 1 I a a
1
b
30. f(t)y(xt) dt = A cos(ln x).
a
Solution:
AI c AI s
y(x)= cos(ln x)+ sin(ln x),
2
2
I + I 2 I + I 2
c s c s
b b
I c = f(t) cos(ln t) dt, I s = f(t) sin(ln t) dt.
a a
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Page 239