Advanced book on Mathematics Olympiad

3.4 Equations with Functions as Unknowns 193

is equal to zero. For functions defined on the entire two-dimensional plane, the field is
conservative if and only if it has a scalar potential. This means that there exists a scalar
functionu(x, y)whose differential is the field, i.e.,

∂u

∂x

=p(x, y) and

∂u

∂y

=q(x, y).

For a conservative field, the scalar potential solves the differential equation, giving the
solution in implicit form asu(x, y)=C, withCa constant. Let us apply this method to
a problem by the first author of the book.

Example.Does there exist a differentiable functionydefined on the entire real axis that
satisfies the differential equation

( 2 x+y−e−x

2

)dx+(x+ 2 y−e−y

2

)dy=0?

Solution.Let us assume that such aydoes exist. Because

∂

∂x

(

x+ 2 y−e−y

2 )

=

∂

∂y

(

2 x+y−e−x

2 )

,

the equation can be integrated. The potential function is

u(x, y)=x^2 +xy+y^2 −

∫x

0

e−s

2

ds−

∫y

0

e−t

2

dt.

The differential equation translates into the algebraic equation

(

x+

1

2

y

) 2

+

3

4

y^2 =

∫x

0

e−s

2

ds+

∫y

0

e−t

2

dt+C

for some real constantC. The right-hand side is bounded from above by

√

8 π+C(note
the Gaussian integrals). This means that both squares on the left must be bounded. In
particular,yis bounded, but thenx+^12 yis unbounded, a contradiction. Hence the answer
to the question is no; a solution can exist only on a bounded interval. 

Sometimes the field is not conservative but becomes conservative after the differential
equation is multiplied by a function. This function is called an integrating factor. There
is a standard method for finding integrating factors, which can be found in any textbook.
In particular, any first-order linear equation

y′+p(x)y=q(x)

can be integrated after it is multiplied by the integrating factor exp(

∫

p(x)dx).

It is now time for problems.