Page 386 - Handbook Of Integral Equations

x

7. y(x)+ g(t)f y(t) dt = A.
a
Solution in an implicit form:
du
y x
+ g(t) dt =0.
f(u)
A a

x f y(t)
8. y(x)+ dt = A.
0 ax + bt
A solution: y(x)= λ, where λ is a root of the algebraic (or transcendental) equation
b

ln 1+ f(λ)+ bλ – Ab =0.
a
x
f y(t)
9. y(x)+ √ dt = A.
2
0 ax + bt 2
A solution: y(x)= λ, where λ is a root of the algebraic (or transcendental) equation
1
dz
kf(λ)+ λ – A =0, k = √ .
0 a + bz 2
x

dt
10. y(x)+ x f y(t) = A.
2
0 x + t 2
A solution: y(x)= λ, where λ is a root of the algebraic (or transcendental) equation
1
λ + πf(λ)= A.
4
x

λt
11. y(x)+ e f y(t) dt = Be λx + C.
a
By differentiation, this integral equation can be reduced to a separable ordinary differential
equation.
Solution in an implicit form:
y du
λ + e λx – e λa =0, y 0 = Be λa + C.
f(u) – Bλ
y 0
x

12. y(x)+ e λ(x–t) f y(t) dt = A.
a
Solution in an implicit form:
y
du

= x – a.
λu – f(u) – λA
A
x

13. y(x)+ e λ(x–t) f y(t) dt = Ae λx + B.
a
Solution in an implicit form:
y
du λa

= x – a, y 0 = Ae + B.
λu – f(u) – λB
y 0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 366