Page 372 - Handbook Of Integral Equations

x

8. y(x)+ k exp λ(x – t)+ βy(t) dt = A.
a
Solution in an implicit form:
dt
y
= x – a.
λt – ke βt – λA
A
x

9. y(x)+ k exp λ(x – t)+ βy(t) dt = Ae λx + B.
a
Solution in an implicit form:
y
dt λa

= x – a, y 0 = Ae + B.
λt – ke βt – λB
y 0
x

10. y(x)+ k sinh[λ(x – t)] exp[βy(t)] dt = Ae λx + Be –λx + C.
a
βy
This is a special case of equation 5.8.14 with f(y)= ke .
x
11. y(x)+ k sinh[λ(x – t)] exp[βy(t)] dt = A cosh(λx)+ B.
a
βy
This is a special case of equation 5.8.15 with f(y)= ke .
x
12. y(x)+ k sinh[λ(x – t)] exp[βy(t)] dt = A sinh(λx)+ B.
a
βy
This is a special case of equation 5.8.16 with f(y)= ke .
x

13. y(x)+ k sin[λ(x – t)] exp[βy(t)] dt = A sin(λx)+ B cos(λx)+ C.
a
βy
This is a special case of equation 5.8.17 with f(y)= ke .
5.4-2. Equations Containing Arbitrary Functions
x

14. y(x)+ f(t) exp[λy(t)] dt = A.
a
Solution:
1 x –Aλ
y(x)= – ln λ f(t) dt + e .
λ a

x
15. y(x)+ g(t) exp[λy(t)] dt = f(x).
a
◦
1 . By differentiation, this integral equation can be reduced to the first-order ordinary differ-
ential equation
y + g(x)e λy = f (x) (1)

x
x
under the initial condition y(a)= f(a). The substitution w = e –λy reduces (1) to the linear
equation

w + λf (x)w – λg(x)=0, w(a)=exp –λf(a) .
x x
2 . Solution:
◦
x
1
y(x)= f(x) – ln 1+ λ g(t)exp λf(t) dt .
λ a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 352