592 Real Analysis
If on any of these intervalsf−g=f′+g′, then sincef+g=g′−f′it follows
thatf =g′on that interval, and sog′+g=g′−g′′. This implies thatgsatisfies
the equationg′′+g =0, or thatg(x)=Asinx+Bcosxon that interval. Also,
f(x)=g′(x)=Acosx−Bsinx.
Iff−g=−f′−g′on some interval, then using againf+g=g′−f′, we find
thatg=g′on that interval. Henceg(x)=C 1 ex. From the fact thatf=−f′, we obtain
f(x)=C 2 e−x.
Assuming thatf andgare exponentials on the interval( 0 ,x 0 ), we deduce that
C 1 =g( 0 ) =f( 0 )=C 2 and thatC 1 ex^0 =g(x 0 )=f(x 0 )=C 2 e−x. These two
inequalities cannot hold simultaneously, unlessfandgare identically zero, ruled out
by the hypothesis of the problem. Therefore,f(x)=Acosx−Bsinxandg(x)=
Asinx+Bcosxon( 0 ,x 0 ), and consequentlyx 0 =π.
On the intervals(−∞, 0 ]and[x 0 ,∞)the functionsfandgcannot be periodic, since
then the equationf =gwould have infinitely many solutions. So on these intervals
the functions are exponentials. Imposing differentiability at 0 andπ, we obtainB=A,
C 1 =AonI 1 andC 1 =−Ae−πonI 3 and similarlyC 2 =AonI 1 andC 2 =−Aeπon
I 3. Hence the answer to the problem is
f(x)=⎧
⎪⎨
⎪⎩
Ae−x forx∈(−∞, 0 ],
A(sinx+cosx) forx∈( 0 ,π],
−Ae−x+π forx∈(π,∞),g(x)=⎧
⎪⎨
⎪⎩
Aex forx∈(−∞, 0 ],
A(sinx−cosx) forx∈( 0 ,π],
−Aex−π forx∈(π,∞),whereAis some nonzero constant.
(Romanian Mathematical Olympiad, 1976, proposed by V. Matrosenco)
555.The idea is to integrate the equation using an integrating factor. If instead we had
the first-order differential equation(x^2 +y^2 )dx+xydy=0, then the standard method
findsxas an integrating factor. So if we multiply our equation byfto transform it into
(f^3 +fg^2 )f′+f^2 gg′= 0 ,then the new equation is equivalent to
(
1
4
f^4 +1
2
f^2 g^2)′
= 0.
Therefore,fandgsatisfy
f^4 + 2 f^2 g^2 =C,