Let's try this exercise

Find all functions $ f : RR rarr RR $ s.t. (so that) $EE F ,F' = f $, and $[xf(x)-2F(x)]*[F(x)-x^2] = 0 $, $AA x in RR.$


I know it's not the right place here, but let's see what happens.

I have found two functions $f:RR->RR$ which
satisfy this property:
$f(x)=(x/2) e^(x^2/4)
and
$f(x)=2x
Notice that they are both onto and one-to-one functions.

Pls explain in detail how you arrived to the results indicated in your post.
Congratulations to Fireball for being the first to reply in english .
Hope many others will follow .

Remark: let $f_1$ be an integrable function which satisfy $xf(x)-2F(x)=0$ and let $f_2$ integrable satisfying the second one $F(x)-x^2=0$. Fixed $A$ and $B$ disjoint subsets of $\RR$ with $A\cup B=\RR$; then the function $f=f_1$ on $A$ and $f=f_2$ on $B$ is a solution of the problem (since other regularity properties are not required), and this show that there are infinite solutions of this equation.

Let $[a,b]$ be a close and limitated subset of $RR$, and $f in ccR(a,b)$,
that is, $f$ belongs to the Riemann-integrables in $[a,b]$ functions subset.
Let $c in [a,b]$. Then, $F(x)=int_c^x f(x) dx " " AAc,x in [a,b]$.
We observe that $(xf(x)-2F(x))*(F(x)-x^2) = 0 $
implies that $xf(x)-2F(x)=0 vv F(x)-x^2=0$.
First, let's consider the second relation, from which
we can immediately say: $F(x)=x^2=>f(x)=2x$.
Now, the first relation is a differential equation:
$xf(x)=2F(x)<=>(2F(x))/(f(x))=x<=>(2F(x))/(F'(x))=x
and, integrating both the members of the equation in $[0,x]$, with $F(0)-=1$:
$2logF(x)=(x^2)/2<=>F(x)=e^((x^2)/4)=>f(x)=(x/2)e^((x^2)/4)$.
This concludes the proof of my results.

Be careful, Fireball, something must be wrong since : $ f= (x/2)e^ (x^2/4) $ doesn't satisfy the initial relation.
Furtherly the exercise requires to find all functions that ....etc.

I have two questions concerning the exercise :

* The solution indicated by Fireball, i.e. $ f(x) = 2x $ is correct but also the more general solution $ f(x) = bx (b in RR)$ is correct as can be easily checked . Where does such solution come from ?
*The two equations :
$x f -2F = 0 $
and
$ F -x^2 = 0 $

provide the same solution .
Is it correct ??

The second equation doesn't provide $f(x)=bx, " " b!=2$ as solution...