Problem:

Given real functions \(f\), \(g\), and \(h\) which satisfy the following:

\(f’=2 f^{2} g h +\displaystyle\frac{1}{g h}\),

\(g’=f g^{2} h +\displaystyle\frac{4}{f h}\),

\(h’=3 f g h^{2} +\displaystyle\frac{1}{f g}\),

with \(f(0)=1\), \(g(0)=1\), and \(h(0)=1\), find the function \(f(x)\).

Solution:

Clearly, one can divide each equation above by the functions \(f\), \(g\), and \(h\), respectively. Adding the equations together produces a single equation for the function \(u=f g h\):

\(\displaystyle\frac{u’}{u}=6 \left (u+\displaystyle\frac{1}{u}\right )\),

from which a solution for \(u\) is produced:

\(6 x+C=\displaystyle\int^{u}\displaystyle\frac{du’}{1+u’^{2}}=\tan^{-1}u\)Because \(u(0)=1\), \(u(x)=\tan \left (6 x+\displaystyle\frac{\pi}{4}\right )\).

This result is plugged back into the first equation, which may be written as follows:

\(\displaystyle\frac{f’}{f}=2 u+\displaystyle\frac{1}{u}=2 \tan \left (6 x+\displaystyle\frac{\pi}{4}\right )+\cot \left (6 x+\displaystyle\frac{\pi}{4}\right )\).

Integrating both sides, we can solve for the unknown function \(f\):

\(f(x)=-\displaystyle\frac{2}{6}\log \left [\cos \left (6 x+\displaystyle\frac{\pi}{4}\right) \right ] + \frac{1}{6}\log \left [\sin\left (6 x+\displaystyle\frac{\pi}{4}\right) \right ] + C’\).

Using \(f(0)=1\), the solution takes the form:

\(f(x)=2^{-\frac{1}{12}} \left [ \displaystyle\frac{\sin\left (6 x+\displaystyle\frac{\pi}{4}\right)}{\cos^{2}\left (6 x+\displaystyle\frac{\pi}{4}\right)} \right ]^{\frac{1}{6}}\)Here is a plot of the solution.