Advanced book on Mathematics Olympiad

Real Analysis 595

c·

a

b

·

a

ab− 1

x

ab− 1 =

x

cx

ab,

which is equivalent to

c^2 a

ab

=b.

By eliminatingc, we obtain the family of solutions

fb(x)=

√

b

(

x

√

a

)ab

,b> 0.

All such functions satisfy the given condition.
(66th W.L. Putnam Mathematical Competition, 2005, proposed by T. Andreescu)
559.Let us look at the solution to the differential equation
∂y
∂x

=f (x, y),

passing through some point(x 0 ,y 0 ). The condition from the statement implies that along

this solution, df (x,y)dx =0, and so along the solution the functionf is constant. This

means that the solution to the differential equation with the given initial condition is a

line(y−y 0 )=f(x 0 ,y 0 )(x−x 0 ). If for some(x 1 ,y 1 ),f(x 1 ,y 1 )  =f(x 0 ,y 0 ), then the

lines(y−y 0 )=f(x 0 ,y 0 )(x−x 0 )and(y−y 1 )=f(x 1 ,y 1 )(x−x 1 )intersect somewhere,

providing two solutions passing through the same point, which is impossible. This shows

thatfis constant, as desired.

(Soviet Union University Student Mathematical Olympiad, 1976)

560.The equation can be rewritten as

(xy)′′+(xy)= 0.

Solving, we findxy=C 1 sinx+C 2 cosx, and hence

y=C 1

sinx

x

+C 2

cosx

x

,

on intervals that do not contain 0.

561.The functionf′(x)f′′(x)is the derivative of^12 (f′(x))^2. The equation is therefore

equivalent to

(f′(x))^2 =constant.

And becausef′(x)is continuous,f′(x)itself must be constant, which means thatf(x)
is linear. Clearly, all linear functions are solutions.