Page 135 - Handbook Of Integral Equations

2 . Solution:
◦
x

y(x)= f(x)+ R(x – t)f(t) dt,
a
n
1
R(x)= exp(σ k x) σ k cos(β k x) – β k sin(β k x) ,
n +1
k=0
where the coefficients σ k and β k are given by
1
2πk 1
2πk
σ k = |An!| n+1 cos , β k = |An!| n+1 sin for A <0,
n +1 n +1
1
2πk + π 1
2πk + π
σ k = |An!| n+1 cos , β k = |An!| n+1 sin for A >0.
n +1 n +1

∞
n
35. y(x)+ A (t – x) y(t) dt = f(x), n =1, 2, ...
x
n
The Picard–Goursat equation. This is a special case of equation 2.9.62 with K(z)= A(–z) .
1 . A solution of the homogeneous equation (f ≡ 0) is
◦
1

y(x)= Ce –λx , λ = –An! n+1 ,
where C is an arbitrary constant and A < 0. This is a unique solution for n = 0,1,2,3.
The general solution of the homogeneous equation for any sign of A has the form
s

y(x)= C k exp(–λ k x). (1)
k=1

Here C k are arbitrary constants and λ k are the roots of the algebraic equation λ n+1 + An!=0
that satisfy the condition Re λ k > 0. The number of terms in (1) is determined by the inequality
n

s ≤ 2 + 1, where [a] stands for the integral part of a number a. For more details about the
4
solution of the homogeneous Picard–Goursat equation, see Subsection 9.11-1 (Example 1).
2 .For f(x)= m a k exp(–β k x), where β k > 0, a solution of the equation has the form
◦
k=1
m
a k β n+1
k
y(x)= exp(–β k x), (2)
β n+1 + An!
k=1 k
where β n+1 + An! ≠ 0. For A > 0, this formula can also be used for arbitrary f(x) expandable
k
into a convergent exponential series (which corresponds to m = ∞).
k
3 .For f(x)= e –βx m a k x , where β > 0, a solution of the equation has the form
◦
k=1
m
k
–βx
y(x)= e B k x , (3)
k=0
where the constants B k are found by the method of undetermined coefficients. The solution
◦
can also be constructed using the formulas given in item 3 , equation 2.9.55.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 114