Advanced book on Mathematics Olympiad

Real Analysis 599

Returning to the problem, we see that there existsc∈[ 0 ,x]such that

∫x

0

e−ty′y′′dt=

∫c

0

y′y′′dt=

1

2

[

((y′(c))^2 −(y′( 0 ))^2

]

.

In conclusion,

(y(x))^2 +(y′(c))^2 =(y( 0 ))^2 +(y′( 0 ))^2 , forx> 0 ,

showing thatyis bounded asx→∞.

Second solution: Use an integrating factor as in the previous solution to obtain

y^2 (x)−y^2 ( 0 )+ 2

∫x

0

e−ty′y′′dt= 0.

Then integrate by parts to obtain

y^2 (x)+e−x(y′(x))^2 +

∫x

0

e−t(y′(t))^2 dt=y^2 ( 0 )+(y′( 0 ))^2.

Because every term on the left is nonnegative, it follows immediately that

|y(x)|≤

(

y^2 ( 0 )+(y′( 0 ))^2

) 1 / 2

is bounded, and we are done.

(27th W.L. Putnam Mathematical Competition, 1966)

567.We have

y 1 ′′(t)+y 1 (t)=

∫∞

0

t^2 e−tx

1 +t^2

dt+

∫∞

0

e−tx

1 +t^2

dt=

∫∞

0

e−txdt=

1

x

.

Also, integrating by parts, we obtain

y 2 (x)=

−cost

t+x

∣∣

∣∣

∞

0

−

∫∞

0

cost

(t+x)^2

dt=

1

x

−

sint

(t+x)^2

∣∣

∣∣

∞

0

−

∫∞

0

2 sint

(t+x)^3

dt

=

1

x

−y 2 ′′(x).

Since the functionsy 1 andy 2 satisfy the same inhomogeneous equation, their difference

y 1 −y 2 satisfies the homogeneous equationy′′+y = 0, and hence is of the form

Acosx+Bsinx. On the other hand,

lim

x→∞

(y 1 (x)−y 2 (x))=lim

x→∞

y 1 (x)−lim

x→∞

y 2 (x)= 0 ,