ABSTRACT
Differentiating (1) with respect to x yields∫ x 0
y(t) dt – ∫ 1 x
y(t) dt = f ′x(x). (2)
Differentiating (2) yields the solution
y(x) = 12 f ′′xx(x). (3)
2◦. Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = 0 and x = 1 in (1), we obtain two corollaries
ty(t) dt = f (0)
and ∫ 1 0 (1 – t)y(t) dt = f (1), which can be rewritten in the form
ty(t) dt = f (0), ∫ 1 0
y(t) dt = f (0) + f (1). (4)
Substitute y(x) of (3) into (4). Integration by parts yields f ′x(1) =f (1)+f (0) and f ′x(1)–f ′x(0) = 2f (1) + 2f (0). Hence, we obtain the desired constraints for f (x):
f ′x(1) = f (0) + f (1), f ′x(0) + f ′x(1) = 0. (5) Conditions (5) make it possible to find the admissible general form of the right-hand side
of the integral equation: f (x) = F (x) + Ax + B,
A = – 12 [ F ′x(1) + F ′x(0)
] , B = 12
[ F ′x(1) – F (1) – F (0)
] ,
whereF (x) is an arbitrary bounded twice differentiable functionwith bounded first derivative.
