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x ϕ (x)

17. y(x)+ x + ϕ(t)g (x) – ϕ (x)g(t) h(t) y(t) dt = f(x).

x
x
a ϕ(t)
◦
1 . This equation is equivalent to the equation
x
x
ϕ(x)
+ ϕ(t)g(x) – ϕ(x)g(t) h(t) y(t) dt = F(x), F(x)= f(x) dx, (1)
ϕ(t)
a a
obtained by differentiating the original equation with respect to x. Equation (1) is a special
case of equation 1.9.15 with
1
g 1 (x)= g(x), h 1 (t)= ϕ(t)h(t), g 2 (x)= ϕ(x), h 2 (t)= – g(t)h(t).
ϕ(t)
2 . Solution:
◦
x
2
1 d F(t) ϕ (t)h(t)
y(x)= Ξ(x) dt ,
ϕ(x)h(x) dx a ϕ(t) t Ξ(t)
x
x
g(t)
2
F(x)= f(x) dx, Ξ(x)=exp – ϕ (t)h(t) dt .
a a ϕ(t) t
x ϕ (t)

18. y(x) – t + ϕ(x)g (t) – ϕ (t)g(x) h(x) y(t) dt = f(x).

t
t
a ϕ(x)
1 . Let f(a) = 0. The change
◦
x
y(x)= w(t) dt (1)
a
followed by the integration by parts leads to the equation
x
ϕ(t)

+ ϕ(x)g(t) – ϕ(t)g(x) h(x) w(t) dt = f(x), (2)
ϕ(x)
a
which is a special case of equation 1.9.15 with
1
g 1 (x)= – g(x)h(x), h 1 (t)= ϕ(t), g 2 (x)= ϕ(x)h(x), h 2 (t)= g(t).
ϕ(x)
The solution of equation (2) is given by

x
1 d 2 f(t) dt
y(x)= ϕ (x)h(x)Φ(x) ,
ϕ(x) dx a ϕ(t)h(t) t Φ(t)
x
g(t)
2
Φ(x)=exp ϕ (t)h(t) dt .
a ϕ(t) t
2 . Let f(a) ≠ 0. The substitution y(x)= ¯y(x)+ f(a) leads to the integral equation ¯y(x) with
◦
¯
¯
◦
the right-hand side f(x) satisfying the condition f(a) = 0. Thus we obtain case 1 .
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 176