Page 147 - Handbook Of Integral Equations

under the initial conditions
Y (a)=0, Y (a)= f(a). (2)

x
A fundamental system of solutions of the homogeneous equation (1) with f ≡ 0 has the
form
A m λx A m λx

Y 1 (x)= Φ ,1; – e , Y 2 (x)= Ψ ,1; – e , m = A + B,
m λ m λ

where Φ α, β; x and Ψ α, β; x are degenerate hypergeometric functions.
Solving the homogeneous equation (1) under conditions (2) for an arbitrary function

f = f(x) and taking into account the relation y(x)= Y (x), we thus obtain the solution of the
x
integral equation in the form
x

y(x)= f(x) – R(x, t)f(t) dt,
a
Γ(A/m) ∂ 2
m λt
R(x, t)= exp e Y 1 (x)Y 2 (t) – Y 2 (x)Y 1 (t) .
λ ∂x∂t λ
x

λ(x+t) 2λt
7. y(x)+ A e – e y(t) dt = f(x).
a
λx
The transformation z = e , τ = e λt leads to an equation of the form 2.1.4.
1 . Solution with Aλ >0:
◦
x
λt
λt

y(x)= f(x) – λk e sin k(e λx – e ) f(t) dt, k = A/λ.
a
2 . Solution with Aλ <0:
◦
x
λt
λt
y(x)= f(x)+ λk e sinh k(e λx – e ) f(t) dt, k = |A/λ|.

a
x

λx+µt (λ+µ)t
8. y(x)+ A e – e y(t) dt = f(x).
a
µt
µx
The transformation z = e , τ = e , Y (z)= y(x) leads to an equation of the form 2.1.52:
A z k k
Y (z)+ (z – τ )Y (τ) dτ = F(z), F(z)= f(x),
µ b
µa
where k = λ/µ, b = e .
x

λx+µt (λ+µ)x
9. y(x)+ A λe – (λ + µ)e y(t) dt = f(x).
a
This equation can be obtained by differentiating an equation of the form 1.2.22:
x x

λx µt µx
1+ Ae (e – e ) y(t) dt = F(x), F(x)= f(t) dt.
a a
Solution:
x
d λx F(t) dt Aµ (λ+µ)x
y(x)= e Φ(x) , Φ(x)=exp e .
dx a e λt t Φ(t) λ + µ
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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