Page 37 - Handbook Of Integral Equations

1.2. Equations Whose Kernels Contain Exponential
Functions

1.2-1. Kernels Containing Exponential Functions

x
1. e λ(x–t) y(t) dt = f(x).
a
Solution: y(x)= f (x) – λf(x).

x
Example. In the special case a = 0 and f(x)= Ax, the solution has the form y(x)= A(1 – λx).
x

2. e λx+βt y(t) dt = f(x).
a

Solution: y(x)= e –(λ+β)x f (x) – λf(x) .
x
Example. In the special case a = 0 and f(x)= A sin(γx), the solution has the form y(x)= Ae –(λ+β)x ×
[γ cos(γx) – λ sin(γx)].
x

λ(x–t)
3. e – 1 y(t) dt = f(x), f(a)= f (a)=0.

x
a
Solution: y(x)= 1 λ xx x
f (x) – f (x).

x
λ(x–t)
4. e + b y(t) dt = f(x).
a
For b = –1, see equation 1.2.3. Differentiating with respect to x yields an equation of the
form 2.2.1:

λ x λ(x–t) f (x)
x
y(x)+ e y(t) dt = .
b +1 a b +1
Solution:

f (x) λ x λb
x

y(x)= – exp (x – t) f (t) dt.
t
b +1 (b +1) 2 b +1
a
x
λx+βt
5. e + b y(t) dt = f(x).
a
λx
For β = –λ, see equation 1.2.4. This is a special case of equation 1.9.15 with g 1 (x)= e ,
βt
h 1 (t)= e , g 2 (x) = 1, and h 2 (t)= b.
x
λx λt
6. e – e y(t) dt = f(x), f(a)= f (a)=0.
x
a
λx
This is a special case of equation 1.9.2 with g(x)= e .

1
–λx

Solution: y(x)= e f (x) – f (x) .
x
xx
λ
x
λx λt
7. e – e + b y(t) dt = f(x).
a
λx
For b = 0, see equation 1.2.6. This is a special case of equation 1.9.3 with g(x)= e .
Solution: x
1 λ λx e λt – e λx
y(x)= f (x) – e exp f (t) dt.

x
t
b b 2 a b
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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