Solving the differential equation

Since we know that $F_n(x)$ is a polynomial of degree $n$, we can put

$$F_n(x) := \sum_{r=0}^n a_r x^r = a_n x^n + \sum_{r=0}^{n-1} a_r x^r$$ (4)

and it follows that

$$F_n'(x) = \sum_{r=0}^n r a_r x^{r-1} = \sum_{r=1}^n r a_r x^{r-1} = \sum_{r=0}^{n-1} (r+1) a_{r+1} x^r$$

Substituting these into (3) gives

$$a_n x^n + \sum_{r=0}^{n-1} a_r x^r + \sum_{r=0}^{n-1} (r+1) a_{r+1} x^r \equiv x^n$$

$$a_n x^n + \sum_{r=0}^{n-1} (a_r + (r+1) a_{r+1}) x^r \equiv x^n$$

Comparing coefficients of $x^m$ on both sides, we find $a_n = 1$ and

$$a_r + (r+1) a_{r+1} = 0$$ (5)

(for $r = 0, \ldots, n-1$).

Using (5), we obtain

$$a_n = 1$$

$$a_{n-1} = -n a_n = -n$$

$$a_{n-2} = -(n-1) a_{n-1} = (n-1) n$$

$$a_{n-3} = -(n-2) a_{n-2} = -(n-2) (n-1) n$$

There is a clear pattern emerging and we can make the conjecture

$$a_{n-k} = (-1)^k \frac{n!}{(n-k)!}$$ (6)

To prove (6), we note that

$$a_{n-k-1} = -(n - k) a_{n-k}$$

$$= -(n - k) (-1)^k \frac{n!}{(n-k)!}$$

$$= (-1)^{k+1} \frac{n!}{(n-k-1)!}$$

$$= a_{n-k-1}$$

It follows that (6) is correct, by induction.

To find $a_r$ we simply substitute $r = n - k$ into (6) and immediately obtain

$$a_r = (-1)^{n-r} \frac{n!}{r!}$$

Substituting into (4), we obtain the¹ polynomial solution to (3),

$$F_n(x) = \sum_{r=0}^n (-1)^{n-r} \frac{n!}{r!} x^r$$