Page 367 - Handbook Of Integral Equations

x
t
λ
16. y(x)+ A x –λ–1 f y(t)y(x – t) dt = Bx .
x
0
λ
λ
Solutions: y 1 (x)= β 1 x and y 2 (x)= β 2 x , where β 1,2 are the roots of the quadratic equation
1
λ
2
λ
AIβ + β – B =0, I = f(z)z (1 – z) dz.
0
∞

17. y(x)+ f(t – x)y(t – x)y(t) dt = ae –λx .
x
Solutions: y(x)= b k e –λx , where b k (k = 1, 2) are the roots of the quadratic equation
∞

2
b I + b – a =0, I = f(z)e –2λz dz.
0
To calculate the integral I, it is convenient to use tables of Laplace transforms (with parameter
p =2λ).
5.3. Equations With Power-Law Nonlinearity

5.3-1. Equations Containing Arbitrary Parameters

x
λ k
1. y(x)+ A t y (t) dt = Bx λ+1 + C.
a
By differentiation, this integral equation can be reduced to a separable ordinary differential
equation.
Solution in an implicit form:
y du
(λ +1) + x λ+1 – a λ+1 =0, y 0 = Ba λ+1 + C.
k
Au – B(λ +1)
y 0
x k
y (t)
2. y(x)+ dt = A.
0 ax + bt
A solution: y(x)= λ, where λ is a root of the algebraic (or transcendental) equation
b k

ln 1+ λ + bλ – Ab =0.
a
k
x y (t) dt
3. y(x)+ Ax 2 2 = B.
0 x + t
A solution: y(x)= λ, where λ is a root of the algebraic (or transcendental) equation

1
k
λ + Aπλ = B.
4
k
x y (t) dt
4. y(x)+ √ = A.
2
0 ax + bt 2
A solution: y(x)= λ, where λ is a root of the algebraic (or transcendental) equation
1
dz
k
Iλ + λ – A =0, I = √ .
0 a + bz 2

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 347