While solving functional equations I came across questions where the functional equation is not symmetric in $x,y $

Here's an example $f (x+y)=f (x)f (y)+2x^2y+0.5xy^2$ for all $x,y$ belonging to the set of real numbers and $f (x) $ is differentiable everywhere. If $f'(0)=1$, then Find $f'(2)-f (2)$

Here's my attempt to the solution- Partial differentiating the equation w.r.t x $f'(x+y)=f'(x)f (y)+4xy+0.5y^2$

Now put $x=0$ $f'(y)=f'(0)f (y)+0.5y^2$

Since $f'(0)=1$

$f'(y)=f (y) + 0.5y^2$ Putting$ y=2$

$f'(2)-f (2)=2$

Now if I had solved the equation by differentiating w.r.t. y, the answer comes out to be 8. $f'(x+y)=f (x)f'(y)+2x^2+xy $

Put $y=0$

$f'(x)=f (x)+ 2x^2$

Put $x=2$

$f'(2)-f (2)=8$

How is this true?

Also, consider another unsymmetrical functional equation $f(x+y)=f (x)+f (y)+2x $ Now if I interchange $x$ and $y$ $f (x+y)=f (x)+f (y)+2y $

This implies $x=y $ What is its meaning in context of the problem? I think it means that this equation is not a two variable functional equation but rather a single variable functional equation. Am I right?

  • $\begingroup$ Does the functional equation make sense? For any $x,y$ we have $f(x+y)=f(y+x)\implies f(x)f(y)+2x^2y+.5xy^2=f(x)f(y)+2y^2x+.5x^2y\implies 2x^2y+.5xy^2=2y^2x+.5x^2y\;\forall\;x,y$. $\endgroup$ – lulu May 31 '18 at 13:52
  • $\begingroup$ @lulu Yes it doesn't make any sense. Can we conclude that functional equations which are not symmetric in x,y don't make any sense? Or they do make sense but for some values of x and y. Which one is true? $\endgroup$ – user185887 May 31 '18 at 13:55
  • $\begingroup$ When the equation is written down its domain should be stated explicitly. Trying to guess what the author had in mind seems pointless. $\endgroup$ – lulu May 31 '18 at 14:07

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