Page 153 - Handbook Of Integral Equations

x

23. y(x)+ A (x – t)e λ(x–t) y(t) dt = f(x).
a
◦
1 . Solution with A >0:
x √
y(x)= f(x) – k e λ(x–t) sin[k(x – t)]f(t) dt, k = A.
a
2 . Solution with A <0:
◦
x √
y(x)= f(x)+ k e λ(x–t) sinh[k(x – t)]f(t) dt, k = –A.
a
x
24. y(x)+ A (x – t)e λx+µt y(t) dt = f(x).
a
The substitution u(x)= e –λx y(x) leads to an equation of the form 2.2.22:
x

u(x)+ A (x – t)e (λ+µ)t u(t) dt = f(x)e –λx .
a
x

25. y(x) – (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 2.1.6:

x
u(x) – A (Ax + Bt + C)u(t) dt = f(x)e –λx .
a
x
2 λ(x–t)
26. y(x)+ A x e y(t) dt = f(x).
a
Solution:
x

2
3
3

y(x)= f(x) – A x exp 1 A(t – x )+ λ(x – t) f(t) dt.
3
a
x

27. y(x)+ A xte λ(x–t) y(t) dt = f(x).
a
Solution:
x
1 3 3
y(x)= f(x) – A xt exp A(t – x )+ λ(x – t) f(t) dt.
3
a
x
2 λ(x–t)
28. y(x)+ A t e y(t) dt = f(x).
a
Solution:
x

3
2
3

y(x)= f(x) – A t exp 1 A(t – x )+ λ(x – t) f(t) dt.
3
a
x

2 λ(x–t)
29. y(x)+ A (x – t) e y(t) dt = f(x).
a
Solution:
x
y(x)= f(x) – R(x – t)f(t) dt,
a
√ √ √ 1/3
2
2

R(x)= ke (λ–2k)x – ke (λ+k)x cos 3 kx – 3 sin 3 kx , k = 1 A .
3 3 4
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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