594 Real Analysis
Let us focus on the functionf(t)=tlnt−t. Its derivative isf′(t)=lnt, which is
negative ift<1 and positive ift>1. The minimum offis att=1, and is equal to
−1. An easy application of L’Hôpital’s rule shows that limt→ 0 f(t)=0. It follows that
the equationf(t)=cfails to have a unique solution precisely whenc∈( 0 , 1 )∪( 1 ,e),
in which case it has exactly two solutions.
If we solve algebraically the equationylny−y=xlnx−xon( 1 ,e), we obtain two
possible continuous solutions, one that is greater than 1 and one that is less than 1. The
continuity ofyaterules out the second, so on the interval[ 1 ,∞),y(x)=x.On( 0 , 1 )
again we could have two solutions,y 1 (x)=x, and some other functiony 2 that is greater
than 1 on this interval. Let us show thaty 2 cannot be extended to a solution having
continuous derivative atx =1. On( 1 ,∞),y 2 (x)=x, hence limx→ 1 +y′ 2 (x)=1.
On( 0 , 1 ), as seen above,y 2 ′lny 2 =lnx,soy 2 ′ =lnx/lny 2 <0, sincex<1, and
y 2 (x) >1. Hence limx→ 1 −y 2 ′(x)≤0, contradicting the continuity ofy 2 ′atx =1.
Hence the only solution to the problem isy(x)=xfor allx∈( 0 ,∞).
(R. Gelca)
558.Define
g(x)=f(x)f′(a
x)
,x∈( 0 ,∞).We want to show thatgis a constant function.
Substitutingx→axin the given condition yields
f(a
x)
f′(x)=
a
x,
for allx>0. We have
g′(x)=f′(x)f(a
x)
+f(x)f′(a
x)(
−
a
x^2)
=f′(x)f(a
x)
−
a
x^2f(a
x)
f(x)=
a
x−
a
x= 0 ,
sogis identically equal to some positive constantb. Using the original equation we can
write
b=g(x)=f(x)f(a
x)
=f(x)·a
x·
1
f′(x),
which gives
f′(x)
f(x)=
a
bx.
Integrating both sides, we obtain lnf(x)=ablnx+lnc, wherec>0. It follows that
f(x)=cx
ab
, for allx>0. Substituting back into the original equation yields