Page 136 - Handbook Of Integral Equations

4 .For f(x) = cos(βx) m a k exp(–µ k x), a solution of the equation has the form
◦
k=1
m m

y(x) = cos(βx) B k exp(–µ k x) + sin(βx) C k exp(–µ k x), (4)
k=1 k=1
where the constants B k and C k are found by the method of undetermined coefficients. The
solution can also be constructed using the formulas given in 2.9.60.

◦
5 .For f(x) = sin(βx) m a k exp(–µ k x), a solution of the equation has the form
k=1
m m

y(x) = cos(βx) B k exp(–µ k x) + sin(βx) C k exp(–µ k x), (5)
k=1 k=1

where the constants B k and C k are found by the method of undetermined coefficients. The
solution can also be constructed using the formulas given in 2.9.61.

◦
◦
◦
6 . To obtain the general solution in item 2 –5 , the solution (1) of the homogeneous equation
must be added to each right-hand side of (2)–(5).
x

n
36. y(x)+ A (x – t)t y(t) dt = f(x), n =1, 2, ...
a
This is a special case of equation 2.1.49 with λ = n.
x

n
n
37. y(x)+ A (x – t )y(t) dt = f(x), n =1, 2, ...
a
This is a special case of equation 2.1.52 with λ = n.
x
n+1 n n
38. y(x)+ ABx – ABx t – Ax – B y(t) dt = f(x), n =1, 2, ...
a
n
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
n
R(x, t)=(Ax +B)exp A x n+1 –t n+1 +B 2 exp A s n+1 –t n+1 +B(x–s) ds.
n +1 t n +1
x

n n+1 n
39. y(x)+ ABxt – ABt + At + B y(t) dt = f(x), n =1, 2, ...
a
n
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

n
R(x, t)= –(At +B)exp A t n+1 –x n+1 +B 2 exp A s n+1 –x n+1 +B(t–s) ds.
n +1 t n +1

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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