Page 26 - Handbook Of Integral Equations

x

2. (x – t)y(t) dt = f(x).
a
Solution: y(x)= f (x).

xx
x
3. (Ax + Bt + C)y(t) dt = f(x).
a
This is a special case of equation 1.9.5 with g(x)= x.
1 . Solution with B ≠ –A:
◦
d – A x – B
y(x)= (A + B)x + C A+B (A + B)t + C A+B f (t) dt .

t
dx a
◦
2 . Solution with B = –A:
1 d A
x A

y(x)= exp – x exp t f (t) dt .
t
C dx C C
a
1.1-2. Kernels Quadratic in the Arguments x and t
x
2

4. (x – t) y(t) dt = f(x), f(a)= f (a)= f (a)=0.
x xx
a
1
Solution: y(x)= f (x).
2 xxx
x
2
2
5. (x – t )y(t) dt = f(x), f(a)= f (a)=0.

x
a
2
This is a special case of equation 1.9.2 with g(x)= x .
1

Solution: y(x)= xf (x) – f (x) .
x
xx
2x 2
x

2 2
6. Ax + Bt y(t) dt = f(x).
a
2
For B = –A, see equation 1.1.5. This is a special case of equation 1.9.4 with g(x)= x .
1 d – 2A x – 2B

Solution: y(x)= x A+B t A+B f (t) dt .
t
A + B dx a
x

2 2
7. Ax + Bt + C y(t) dt = f(x).
a
2
This is a special case of equation 1.9.5 with g(x)= x .
Solution:
d – A x – B 2
y(x) = sign ϕ(x) |ϕ(x)| A+B |ϕ(t)| A+B f (t) dt , ϕ(x)=(A + B)x + C.

t
dx a
x
2 2

8. Ax +(B – A)xt – Bt y(t) dt = f(x), f(a)= f (a)=0.
x
a
Differentiating with respect to x yields an equation of the form 1.1.3:
x

[2Ax +(B – A)t]y(t) dt = f (x).
x
a
Solution: x
1 d – 2A A–B

y(x)= x A+B t A+B f (t) dt .
tt
A + B dx a
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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