Page 394 - Handbook Of Integral Equations

∞

15. e –λt y(t)y(xt) dt = A, λ >0, 0 ≤ x < ∞.
0
This is a special case of equation 6.2.2 with f(t)= e –λt , a = 0, and b = ∞.
1 . Solutions:
◦
√ √
y 1 (x)= Aλ, y 2 (x)= – Aλ,

y 3 (x)= 1 Aλ (λx – 2), y 4 (x)= – 1 Aλ (λx – 2).
2 2
2 . The integral equation has some other (more complicated) solutions of the polynomial
◦
k
form y(x)= n B k x , where the constants B k can be found from the corresponding system
k=0
of algebraic equations. See 6.2.2 for some other solutions.
1
16. y(t)y(x + λt) dt = A, 0 ≤ x < ∞.
0
This is a special case of equation 6.2.7 with f(t) ≡ 1, a = 0, and b =1.
Solutions:
√ √
y 1 (x)= A, y 2 (x)= – A,

y 3 (x)= 3A/λ (1 – 2x), y 4 (x)= – 3A/λ (1 – 2x).
∞

17. y(t)y(x + λt) dt = Ae –βx , A, λ, β >0, 0 ≤ x < ∞.
0
This is a special case of equation 6.2.9 with f(t) ≡ 1, a = 0, and b = ∞.
Solutions:

–βx
–βx
y 1 (x)= Aβ(λ +1) e , y 2 (x)= – Aβ(λ +1) e ,
–βx –βx
y 3 (x)= B β(λ +1)x – 1 e , y 4 (x)= –B β(λ +1)x – 1 e ,

where B = Aβ(λ +1)/λ.
1

18. y(t)y(x – t) dt = A, –∞ < x < ∞.
0
This is a special case of equation 6.2.10 with f(t) ≡ 1, a = 0, and b =1.
◦
1 . Solutions with A >0:
√ √
y 1 (x)= A, y 2 (x)= – A,
√ √
2
2
y 3 (x)= 5A(6x – 6x + 1), y 4 (x)= – 5A(6x – 6x + 1).
◦
2 . Solutions with A <0:
√ √
y 1 (x)= –3A (1 – 2x), y 2 (x)= – –3A (1 – 2x).
The integral equation has some other (more complicated) solutions of the polynomial
k
form y(x)= n B k x , where the constants B k can be found from the corresponding system
k=0
of algebraic equations.
∞ x

b
19. e –λt y y(t) dt = Ax , λ >0.
t
0
√
b
Solutions: y(x)= ± Aλ x .
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 374