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∞
19. K(ax + t)y(t) dt = Ae λx .
–∞
Solution:
A λ ∞ λ
y(x)= exp – x , B = K(z)exp – z dz.
B a a
–∞
∞
20. K(ax + t)y(t) dt = f(x).
–∞
The transformation τ = –t, z = ax, y(t)= Y (τ) leads to an equation of the form 3.8.15:
∞
K(z – τ)Y (τ) dt = f(z/a).
–∞
∞
µt
βt
21. [e K(ax + t)+ e M(ax – t)]y(t) dt = Ae λx .
–∞
Solution:
I k (q)e px – I m (p)e qx λ λ
y(x)= A , p = – – β, q = – µ,
I k (p)I k (q) – I m (p)I m (q) a a
where
∞ ∞
I k (q)= K(z)e (β+q)z dz, I m (q)= M(z)e –(µ+q)z dz.
–∞ –∞
∞
22. g(xt)y(t) dt = f(x).
0
By setting
z
τ
–τ
x = e , t = e , y(t)= e w(τ), g(ξ)= G(ln ξ), f(ξ)= F(ln ξ),
we arrive at an integral equation with difference kernel of the form 3.8.15:
∞
G(z – τ)w(τ) dτ = F(z).
–∞
∞ x
23. g y(t) dt = f(x).
0 t
By setting
τ
–τ
z
x = e , t = e , y(t)= e w(τ), g(ξ)= G(ln ξ), f(ξ)= F(ln ξ),
we arrive at an integral equation with difference kernel of the form 3.8.15:
∞
G(z – τ)w(τ) dτ = F(z).
–∞
∞
β λ
24. g x t y(t) dt = f(x), β >0, λ >0.
0
By setting
x = e z/β , t = e –τ/λ , y(t)= e τ/λ w(τ), g(ξ)= G(ln ξ), f(ξ)= 1 F(β ln ξ),
λ
we arrive at an integral equation with difference kernel of the form 3.8.15:
∞
G(z – τ)w(τ) dτ = F(z).
–∞
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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