Page 185 - Handbook Of Integral Equations

x

25. y(x) – e µ(x–t) A sin(kx)+ B – AB(x – t) sin(kx) y(t) dt = f(x).
a
Solution:
x
y(x)= f(x)+ e µ(x–t) M(x, t)f(t) dt,
a
G(x) B 2 x B(x–s) A
M(x, t)=[A sin(kx)+ B] + e G(s) ds, G(x)=exp – cos(kx) .
G(t) G(t) t k
x

26. y(x)+ e µ(x–t) A sin(kt)+ B + AB(x – t) sin(kt) y(t) dt = f(x).
a
Solution:
x

y(x)= f(x)+ e µ(x–t) M(x, t)f(t) dt,
a
G(t) B 2 x B(t–s) A
M(x, t)= –[A sin(kt)+ B] + e G(s) ds, G(x)=exp – cos(kx) .
G(x) G(x) k
t
x

27. y(x) – A e µt tan(λx)y(t) dt = f(x).
a
µt
This is a special case of equation 2.9.2 with g(x)= A tan(λx) and h(t)= e .
x

28. y(x) – A e µx tan(λt)y(t) dt = f(x).
a
This is a special case of equation 2.9.2 with g(x)= Ae µx and h(t) = tan(λt).
x

29. y(x)+ A e µ(x–t) tan(λx) – tan(λt) y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 2.5.32:
x
–µx
w(x)+ A tan(λx) – tan(λt) w(t) dt = e f(x).
a
x

30. y(x) – e µ(x–t) A tan(kx)+ B – AB(x – t) tan(kx) y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 2.9.7 with λ = B and
g(x)= A tan(kx):

x
–µx
w(x) – A tan(kx)+ B – AB(x – t) tan(kx) w(t) dt = e f(x).
a
x

31. y(x)+ e µ(x–t) A tan(kt)+ B + AB(x – t) tan(kt) y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 2.9.8 with λ = B and
g(t)= A tan(kt):
x
–µx
w(x)+ A tan(kt)+ B + AB(x – t) tan(kt) w(t) dt = e f(x).
a

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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