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I came across this problem in an old olympiad paper (Putnam?)

Find all polynomials $p(x)$ with real coefficients satisfying the differential equation

$7\dfrac{d }{dx } [xp(x)]=3p(x)+4p(x+1)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\infty<x<\infty$

I didn't find any "official" solution on internet and it would be interesting to see your approaches on this one.

My approach was the following , i don't know if i'm right. Consider the leading coefficient $a_n$ on both sides. Then we have $7na_n=7a_n$. Suppose $a_n\neq 0$ then we get $n=1$. I plugged in the general solution $p(x)=a_0+a_1x$ in the differential equation and solved for $a_1$ wich turned out to be $0$ , contradicition. The other option is that $a_n=0$ but then $a_{n-1 }$ would be the leading coefficient of the polynomial wich would be again $0$ (by the same reasoning). Hence by induction the only non zero leading coefficient must be $a_0$.

Could you tell me if this is ok? Thanks in advance.

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  • $\begingroup$ @doraemonpaul According to a certain usage and to this wikipedia entry in particular, a differential equation is not exactly considered as a functional equation. $\endgroup$ – Julien May 24 '13 at 20:59
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By the product rule $$ \frac d{dx}\bigl(xp(x)\bigr) = xp'(x) + p(x) $$ so we get $$ 7xp'(x) = 4\bigl(p(x+1) - p(x)\bigr) $$ Obviously, constant $p$s are a solution, if $p$ were not constant, $n = \deg p \ge 1$, say, we have that the degree of $p(x+1) - p(x)$ is less than $n$. On the other hand, $\deg p' = n-1$, hence the degree of $xp'(x)$ is $n$. But that is not possible, so constant polynomials are the only solutions.

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