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$$(f'(x+1)+f'(x-1))f(x)-(f(x+1)+f(x-1))f'(x)=0$$

I don't have any ideas about the solution of this problem. How can I solve this differential equation?

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  • $\begingroup$ Yes, it is correct. $\endgroup$
    – ChipotleHS
    Dec 24, 2013 at 13:49
  • $\begingroup$ Comment with not interest I bet : f(x) = Cte is a solution. More seriously, I do not think that f(x) can be a polynomial. $\endgroup$ Dec 24, 2013 at 13:50
  • $\begingroup$ If this sign is a plus, what is the solution for f(x)? $\endgroup$
    – ChipotleHS
    Dec 24, 2013 at 13:56
  • $\begingroup$ In that case, you can write it the derivative of $(f(x+1) + f(x-1))f(x) = c$. $\endgroup$
    – Amzoti
    Dec 24, 2013 at 13:59
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    $\begingroup$ $f(x)=\alpha x$, $\alpha\in\Bbb{R}$ is also a solution to the original problem.... $\endgroup$ Dec 24, 2013 at 14:14

2 Answers 2

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Notice that $f(x)=0$ is a solution. Now assume $f(x)\neq0$ and we can divide both sides by $f^2(x)$ to get: $(\frac{f(x+1)}{f(x)})'+(\frac{f(x-1)}{f(x)})'=0$

$c f(x)=f(x+1)+f(x-1)$ And now we use characteristic polynomial to solve this recurrence:

$x^2-cx+1=0$ Hence if $c\neq2$

$f(x)=a (\frac{c+\sqrt{c^2-4}}{2})^x+b (\frac{c-\sqrt{c^2-4}}{2})^x$

And if $c=2$

$f(x)=ax+b$

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    $\begingroup$ You might want to use \neq in stead of <>. $\endgroup$ Dec 24, 2013 at 15:46
  • $\begingroup$ We have $f(x)$ a function over the real numbers. You need to introduce an arbitrary function $g(x)$ defined on $[0,2)$. $\endgroup$
    – abnry
    Dec 24, 2013 at 19:35
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You told that there is symmetry with the minus sigh, which is correct.

Notice that $f'(x-1)+f'(x+1)=[f(x-1)+f(x+1)]'$, thus $\frac{f(x-1)+f(x+1)}{f(x)}=c$ yield $f(x)[f'(x-1)+f'(x+1)]-f'(x)[f(x-1)+f(x+1)]=0$.

Is it helpful?

Edit: the above is true if $f(x) \neq 0$. We need to check separately if $f(x)=0$ is a solution and it is.

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