Find all continuous functions $f : \mathbb{R} \to \mathbb{R}$ such that $(f(x)g(x))' = f'(x)g'(x)$

Find all continuous functions $f : \mathbb{R} \to \mathbb{R}$ such that $(f(x)g(x))' = f'(x)g'(x), f,g \neq const$

My solution: $f'(x)g(x)+g'(x)f(x) = f'(x)g'(x) \\ g(x)+g'(x)f(x)/f'(x) = g'(x) \\ f(x)/f'(x) = (g'(x)-g(x))/g'(x)$

But how to proceed further to reduce to something concrete, I do not know

• Does $f$ satisfy this property for any $g$ or do you need to find a relation between $f$ and $g$ for a given $g$ ? May 22 '18 at 16:35
• The functions are arbitrary, it is supposed to find some general form of such functions. In addition to the constants, of course May 22 '18 at 16:39

$$f'(x)g(x)+f(x)g'(x)=f'(x)g'(x)$$

$$\frac{f(x)}{f'(x)}+\frac{g(x)}{g'(x)}=1$$

$$\frac{1}{[\ln f(x)]'}+\frac{1}{[\ln g(x)]'}=1$$

call $\ln f(x)=p(x)$ and $\ln g(x)=q(x)$

$$p'(x)=\frac{q'(x)}{q'(x)-1}=1+\frac{1}{q'(x)-1}$$

$$p(x)=x+\int\frac{1}{q'(t)-1}dt+c$$

• I like that better than my answer. May 22 '18 at 16:59

You can fix a function $f(x)$, and then you obtain a differential equation for $g(x)$, which has a unique solution up to a constant multiple: $$f'(x)g(x) + g'(x)f(x) = f'(x)g'(x)$$ $$g'(x) = \frac{g(x)}{1 - \frac{f(x)}{f'(x)}}$$