Page 371 - Handbook Of Integral Equations

x

2
3. y(x)+ k (x – t) exp[λy(t)] dt = Ax + Bx + C.
a
λy
1 . This is a special case of equation 5.8.5 with f(y)= ke . The solution of this integral
◦
equation is determined by the solution of the second-order autonomous ordinary differential
equation
y + ke λy – 2A =0
xx
under the initial conditions
2

y(a)= Aa + Ba + C, y (a)=2Aa + B.
x
2 . Solution in an implicit form:
◦
y
2 –1/2
4Au – 2F(u)+ B – 4AC du = ±(x – a),
y 0
k λu 2
F(u)= e – e λy 0 , y 0 = Aa + Ba + C.
λ
x
λ
4. y(x)+ A t exp[βy(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.
Ae βu – B(λ +1)
y 0
x
exp[λy(t)]

5. y(x)+ dt = A.
0 ax + bt
A solution: y(x)= β, where β is a root of the transcendental equation
b λβ

ln 1+ e + bβ – Ab =0.
a
x exp[λy(t)]
6. y(x)+ √ dt = A.
2
0 ax + bt 2
A solution: y(x)= β, where β is a root of the transcendental equation

1 dz
ke λβ + β – A =0, k = √ .
0 a + bz 2
x

λx
7. y(x)+ A exp λt + βy(t) dt = Be + 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.
Ae βu – Bλ
y 0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 351