Putnam 2009, Problem A-2

Functions $f, g, h$ are differentiable on some open interval around $0$ and satisfy the equations and initial conditions
\[
f' = 2f^2 gh + \frac{1}{gh}, \quad f(0) = 1,
\]
\[
g' = fg^2 h + \frac{4}{fh}, \quad g(0) = 1,
\]
\[
h' = 3fgh^2 + \frac{1}{fg}, \quad h(0) = 1.
\]
Find an explicit formula for $f(x)$, valid in some open interval around $0$.

Solution:
Multiplying the first differential equation by \( gh \), the second by \( fh \), and the third by \( fg \), and summing gives
\[
(fgh)' = 6(fgh)^2 + 6.
\]
Write \( k(x) = f(x)g(x)h(x) \); then \( k' = 6k^2 + 6 \) and \( k(0) = 1 \). One solution for this differential equation with this initial condition is \( k(x) = \tan(6x + \pi/4) \); by standard uniqueness, this must necessarily hold for \( x \) in some open interval around 0. Now the first given equation becomes
\[
\frac{f'}{f} = 2k(x) + \frac{1}{k(x)} = 2\tan(6x + \pi/4) + \cot(6x + \pi/4);
\]
integrating both sides gives
\[
\ln(f(x)) = \frac{-2\ln\cos(6x + \pi/4) + \ln\sin(6x + \pi/4)}{6} + c,
\]
whence \( f(x) = e^c \left( \frac{\sin(6x + \pi/4)}{\cos^2(6x + \pi/4)} \right)^{1/6} \). Substituting \( f(0) = 1 \) gives \( e^c = 2^{-1/12} \) and thus \( f(x) = 2^{-1/12} \left( \frac{\sin(6x + \pi/4)}{\cos^2(6x + \pi/4)} \right)^{1/6} \).

**Remark.** The answer can be put in alternate forms using trigonometric identities. One particularly simple one is
\[
f(x) = (\sec 12x)^{1/12}(\sec 12x + \tan 12x)^{1/4}.
\]