Page 115 - Handbook Of Integral Equations

x

5. [Ag(x)+ Bg(t)+ C]y(t) dt = f(x).
a
For B = –A, see equation 1.9.3. Assume that B ≠ –A and (A + B)g(x)+ C >0.
Solution:

d – A x – B

y(x)= (A + B)g(x)+ C A+B (A + B)g(t)+ C A+B f (t) dt .
t
dx a
x

6. [g(x)+ h(t)]y(t) dt = f(x).
a
Solution:
d Φ(x) x f (t) dt x h (t) dt

t
t
y(x)= , Φ(x)=exp .
dx g(x)+ h(x) a Φ(t) a g(t)+ h(t)
x

7. g(x)+(x – t)h(x) y(t) dt = f(x).
a
This is a special case of equation 1.9.15 with g 1 (x)= g(x)+ xh(x), h 1 (t)=1, g 2 (x)= h(x),
and h 2 (t)= –t.
Solution:
x
x
d h(x) f(t) dt h(t)
y(x)= Φ(x) , Φ(t)=exp – dt .
dx g(x) h(t) Φ(t) g(t)
a t a
x

8. g(t)+(x – t)h(t) y(t) dt = f(x).
a
This is a special case of equation 1.9.15 with g 1 (x)= x, h 1 (t)= h(t), g 2 (x) = 1, and h 2 (t)=
g(t) – th(t).
x

λ µ
9. g(x)+(Ax + Bt )h(x) y(t) dt = f(x).
a
λ
This is a special case of equation 1.9.15 with g 1 (x)= g(x)+ Ax h(x), h 1 (t)=1, g 2 (x)= h(x),
µ
and h 2 (t)= Bt .
x
λ µ
10. g(t)+(Ax + Bt )h(t) y(t) dt = f(x).
a
λ
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t)= h(t), g 2 (x) = 1, and
µ
h 2 (t)= g(t)+ Bt h(t).
x

11. [g(x)h(t) – h(x)g(t)]y(t) dt = f(x), f(a)= f (a)=0.

x
a
For g = const or h = const, see equation 1.9.2.
Solution:
1 d (f/h) x
y(x)= , where f = f(x), g = g(x), h = h(x).
h dx (g/h) x
Here Af + Bg + Ch /≡ 0, with A, B, and C being some constants.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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