Page 408 - Handbook Of Integral Equations

b
35. y(x)+ f(t)y(t)y(x + λt) dt =0, λ >0.
a
1 . Solutions:
◦
1 I 2 – I 1 x
y 1 (x)= – exp(–Cx), y 2 (x)= exp(–Cx),
2
I 0 I – I 0 I 2
1
b
m

I m = t exp –C(λ +1)t f(t) dt, m =0, 1, 2,
a
where C is an arbitrary constant.
k
2 . There are more complicated solutions of the form y(x)=exp(–Cx) n A k x , where C
◦
k=0
is an arbitrary constant and the coefficients A k can be found from the corresponding system
of algebraic equations.
◦
3 . The equation also has the trivial solution y(x) ≡ 0.
βx
◦
4 . The substitution y(x)= e w(x) leads to a similar equation:
b

w(x)+ g(t)w(t)w(x + λt) dt =0, g(t)= e β(λ+1)t f(t).
a
b
36. y(x)+ f(t)y(x + λt)y(t) dt = Ae –µx , λ >0.
a
◦
1 . Solutions:
y 1 (x)= k 1 e –µx , y 2 (x)= k 2 e –µx ,
where k 1 and k 2 are the roots of the quadratic equation
b

2
Ik + k – A =0, I = e –µ(λ+1)t f(t) dt.
a
m
2 . There are more complicated solutions of the form y(x)= e –µx n B m x , where the B m
◦
m=0
can be found from the corresponding system of algebraic equations.
βx
3 . The substitution y(x)= e w(x) leads to an equation of the same form,
◦
b

w(x)+ g(t)w(t)w(x – t) dt = Ae (λ–β)x , g(t)= f(t)e β(λ+1)t .
a
b
37. y(x)+ f(t)y(t)y(x – t) dt =0.
a
◦
1 . Solutions:
1 I 2 – I 1 x b m
y 1 (x)= – exp(Cx), y 2 (x)= exp(Cx), I m = t f(t) dt,
2
I 0 I – I 0 I 2 a
1
where C is an arbitrary constant and m = 0,1,2.
k
2 . There are more complicated solutions of the form y(x)=exp(Cx) n A k x , where C is
◦
k=0
an arbitrary constant and the coefficients A k can be found from the corresponding system of
algebraic equations.
◦
3 . The equation also has the trivial solution y(x) ≡ 0.
◦
4 . The substitution y(x)=exp(Cx)w(x) leads to an equation of the same form:
b
w(x)+ f(t)w(t)w(x – t) dt =0.
a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 388