Page 122 - Handbook Of Integral Equations

∞

31. K(x – t)y(t) dt = A sinh(λx).
x
Solution:
A λx A –λx 1 A A
1 A A
y(x)= e – e = – cosh(λx)+ + sinh(λx),
2B + 2B – 2 B + B – 2 B + B –
∞ ∞

B + = K(–z)e λz dz, B – = K(–z)e –λz dz.
0 0
∞

32. K(x – t)y(t) dt = A cos(λx).
x
Solution:
A
y(x)= B c cos(λx)+ B s sin(λx) ,
2
B + B 2 s
c
∞ ∞

B c = K(–z) cos(λz) dz, B s = K(–z) sin(λz) dz.
0 0
∞

33. K(x – t)y(t) dt = A sin(λx).
x
Solution:
A
y(x)= B c sin(λx) – B s cos(λx) ,
B + B 2
2
c s
∞ ∞

B c = K(–z) cos(λz) dz, B s = K(–z) sin(λz) dz.
0 0
∞

34. K(x – t)y(t) dt = Ae µx cos(λx).
x
Solution:
A µx
y(x)= e B c cos(λx)+ B s sin(λx) ,
2
B + B s 2
c
∞ ∞

B c = K(–z)e µz cos(λz) dz, B s = K(–z)e µz sin(λz) dz.
0 0
∞

35. K(x – t)y(t) dt = Ae µx sin(λx).
x
Solution:
A µx
y(x)= e B c sin(λx) – B s cos(λx) ,
B + B 2
2
c s

∞ ∞
B c = K(–z)e µz cos(λz) dz, B s = K(–z)e µz sin(λz) dz.
0 0
∞

36. K(x – t)y(t) dt = f(x).
x
k
◦
1 . For a polynomial right-hand side of the equation, f(x)= n A k x , the solution has the
k=0
form
n
k
y(x)= B k x ,
k=0
where the constants B k are found by the method of undetermined coefficients. The solution
◦
can also be obtained by the formula given in 1.9.27 (item 4 ).
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 100