4
$\begingroup$

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

$\endgroup$
2
  • 2
    $\begingroup$ Does $f$ satisfy this property for any $g$ or do you need to find a relation between $f$ and $g$ for a given $g$ ? $\endgroup$
    – Delta-u
    May 22 '18 at 16:35
  • $\begingroup$ The functions are arbitrary, it is supposed to find some general form of such functions. In addition to the constants, of course $\endgroup$ May 22 '18 at 16:39
3
$\begingroup$

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

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

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

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

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

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

$\endgroup$
1
  • 1
    $\begingroup$ I like that better than my answer. $\endgroup$ May 22 '18 at 16:59
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.