Page 41 - Handbook Of Integral Equations

x

31. (x – t)e λ(x–t) y(t) dt = f(x), f(a)= f (a)=0.
x
a
2
Solution: y(x)= f (x) – 2λf (x)+ λ f(x).

x
xx
x

32. (Ax + Bt + C)e λ(x–t) y(t) dt = f(x).
a
The substitution u(x)= e –λx y(x) leads to an equation of the form 1.1.3:
x

(Ax + Bt + C)u(t) dt = e –λx f(x).
a
x

33. (Axe λt + Bte µx )y(t) dt = f(x).
a
µx
λt
This is a special case of equation 1.9.15 with g 1 (x)= Ax, h 1 (t)= e , and g 2 (x)= Be ,
h 2 (t)= t.
x
λ(x–t) µ(x–t)
34. Axe + Bte y(t) dt = f(x).
a
µx
λx
This is a special case of equation 1.9.15 with g 1 (x)= Axe , h 1 (t)= e –λt , g 2 (x)= Be , and
h 2 (t)= te –µt .
x

2 λ(x–t)
35. (x – t) e y(t) dt = f(x), f(a)= f (a)= f (a)=0.

x xx
a
3

Solution: y(x)= 1 f (x) – 3λf (x)+3λ f (x) – λ f(x) .

2
2 xxx xx x
x

n λ(x–t)
36. (x – t) e y(t) dt = f(x), n =1, 2, ...
a
It is assumed that f(a)= f (a)= ··· = f x (n) (a)=0.

x
1 λx d n+1 –λx
Solution: y(x)= e e f(x) .
n! dx n+1
x

β
λt
37. (Ax + Be )y(t) dt = f(x).
a
β
λt
This is a special case of equation 1.9.6 with g(x)= Ax and h(t)= Be .
x

β
38. (Ae λx + Bt )y(t) dt = f(x).
a
β
This is a special case of equation 1.9.6 with g(x)= Ae λx and h(t)= Bt .
x
β λt
γ µx
39. (Ax e + Bt e )y(t) dt = f(x).
a
µx
λt
β
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t)= e , g 2 (x)= Be , and
γ
h 2 (t)= t .
x
√
40. e λ(x–t) x – ty(t) dt = f(x).
a
Solution: x
2 λx d 2 e –λt f(t) dt
y(x)= e √ .
π dx 2 x – t
a
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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