Advanced book on Mathematics Olympiad

598 Real Analysis

This splits into

d^2 y

dx^2

=0 and

(

dy

dx

) 3

= 1.

The first of these has the solutionsy=ax+b, witha =0, becauseyhas to be one-to-
one, while the second reduces toy′=1, whose family of solutionsy=x+cis included
in the first. Hence the answer to the problem consists of the nonconstant linear functions.
(M. Ghermanescu, ̆ Ecua ̧tii Diferen ̧tiale(Differential Equations), Editura Didactic ̆a
̧si Pedagogica, Bucharest, 1963) ̆

566.First solution: Multiplying the equation bye−xy′and integrating from 0 tox,we
obtain

y^2 (x)−y^2 ( 0 )+ 2

∫x

0

e−ty′y′′dt= 0.

The integral in this expression is positive. To prove this we need the following lemma.

Lemma.Letf :[ 0 ,a]→Rbe a continuous function andφ:[ 0 ,a]→Ra positive,
continuously differentiable, decreasing function withφ( 0 )= 1. Then there existsc∈
[ 0 ,a]such that

∫a

0

φ(t)f(t)dt=

∫c

0

f(t)dt.

Proof.LetF(x)=

∫x
0 f(t)dt,x∈[^0 ,a], and letαbe the negative of the derivative of
φ, which is a positive function. Integrating by parts, we obtain

∫a

0

φ(t)f(t)dt=φ(a)F(a)+

∫a

0

α(t)F(t)dt=F(a)−

∫a

0

(F (a)−F (t))α(t)dt.

We are to show that there exists a pointcsuch that

F(a)−F(c)=

∫a

0

(F (a)−F (t))α(t)dt.

If

∫a
0 α(t)dtwere equal to 1, this would be true by the mean value theorem applied to the
functionF(a)−F(t)and the probability measureα(t)dt. But in general, this integral
is equal to some subunitary numberθ, so we can findc′such that the integral is equal
toθ(F(a)−F(c′)). But this number is betweenF(a)−F(a)andF(a)−F(c′),soby
the intermediate value property, there is acsuch thatθ(F(a)−F(c′))=F(a)−F(c).
This proves the lemma.