# Find the function

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.

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