0
$\begingroup$

$$f(x)+f(x+2y)+3xy=2f(2y-x)+2y^2,~x,y\in\mathbb{R}$$

I'm given a functional equation $f(x,y)$. I've been told by my teachers to solve such equations by first partially differenciating the equation wrt one variable ,keeping the other variable constant, and then plugging $0,-1,$ etc into one of the variables. However, when I do so in the above problem, I obtain different answers each time, depending on which variable I partially differenciated it in the first step, and which variable I chose to plug in some value$(0)$ in the second step.

I want to know why does this happen? Shouldn't I get the same answer each time? Please guide me

Also,there is a neat way to do the problem,to plug $x=0$ in the original equation itself,we get $f(x)=f(0)-x^2/2 $ So we know for certain that it is the correct answer.

Some context:Yes,I understand that this problem don't require any derivatives,but This question appeared in a test which is very time-bound,and we don't get much time to focus on a particular question.It would be very unlikely that I would be able to notice that plugging x=0 just does the trick.So I just want to solve every such functional equation in a general manner,using partial derivatives,so I dont waste time looking for alternatives in the test.So,requesting you all to solve this by partial derivatives only

$\endgroup$
  • $\begingroup$ What is it you are trying to solve for exactly? $\endgroup$ – Aaron Zolotor Oct 21 '18 at 12:57
  • $\begingroup$ I'm trying to express the function in terms of one variable only,eliminating the other.Expressing it in a 'normal' way like f(x)= sinx etc(Sorry,dont know the exact mathematical terms I should use) $\endgroup$ – Saplin Oct 21 '18 at 13:22
  • $\begingroup$ I've reread your problem and your comment several times now and I'm still not exactly sure what you're trying to accomplish. Also, what do you mean you get different answers? The partial derivative will be both be different so I'm not quite sure what you're trying to ask here. $\endgroup$ – Aaron Zolotor Oct 21 '18 at 13:32
  • $\begingroup$ What I mean is,if I partially differenciate the function with respect to x,treating y as a constant, I get. [ f'(x)+f'(x+2y)+3y=2f'(2y-x)(-1)].Now plugging x=0,I get [f'(0)+3f'(2y)=-3y],which is equivalent to [f'(t)=-t/2-f'(0)/3].Which means that[ f(t)=-(t^2)/4-f'(0)t/3] $\endgroup$ – Saplin Oct 21 '18 at 13:47
  • $\begingroup$ And this is different from the answer given in the question text above. Why is that so?Shouldn't the functional equation represent a single function? $\endgroup$ – Saplin Oct 21 '18 at 13:48
1
$\begingroup$

Okay, so we take our partial derivatives

$$\frac{\partial}{\partial x} = f'(x)+f'(x+2y)+3y=-f'(2y-x)$$

Notice that we deleted a function of $y$ ($y^2$ in this case) so it is reasonable to expect it to contribute later in some way. Similarly

$$\frac{\partial}{\partial y} 2f'(x+2y)+3x=4f'(2y-x)+4y$$

Now, we want to solve picking smart values. Let $x=0$ and $y=1/2t$. We have

$$f'(t)=4'(t)+2t \implies 3f'(t)=-2t$$

We integrate with respect to $t$ and find $f(t)=t^2+c(x)$ where $c(x)$ is some function of $x$ (what we maybe deleted when we differentiated with respect to $y$). Similarly, $y=0$ we have $f'(x)+f'(x)=-f'(-x)$. See if you can go from here?

$\endgroup$
  • $\begingroup$ I'm sorry,I don't understand how to proceed from here.We want our final answer to be a function of 't' only,so wouldn't finding c(x) bring some terms of variable 'x' into f(t)? Or do we have to somehow eliminate c(x) by expressing it somehow in terms of 't' too? $\endgroup$ – Saplin Oct 21 '18 at 15:46
  • $\begingroup$ Can you complete the solution please?I've been scratching my head for the past couple of days,can't seem to get it. $\endgroup$ – Saplin Oct 24 '18 at 3:45

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.