<s>[INST] Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $? I was playing(/fiddling) around with some maths and I saw this pattern(
where $ A_n $ is a constant.):
$ A_1 x $ is a soultion of:
$$ \frac{y}{x} - \frac{dy}{dx} = 0 $$
$ A_1 x + A_2 x^2 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} =0 $$
$ A_1 x + A_2 x^2 + A_3 x^3 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} - \frac{x^2}{3!} \frac{d^3y}{dx^3} =0 $$
It continues so on. Can someone prove the solution of 
$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $ is:
$$ \sum_{k=0}^{n} (-1)^k \frac{x^{k-1}}{k!} \frac{d^k y}{d x^k} = 0 $$  
 [/INST] Let $z=\frac yx$.  Then we have
$$\dfrac{d^nz}{dx^n}=\dfrac{d^n(yx^{-1})}{dx^n}=\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}\dfrac{d^{n-k}(x^{-1})}{dx^{n-k}}=$$
$$\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}(-1)^{n-k}(n-k)!x^{-1-n+k}=$$
$$\sum_{k=0}^n(-1)^{n-k}\dfrac{n!}{k!}x^{-1-n+k}\dfrac{d^ky}{dx^k}$$
So we have
$$(\pm1)n!x^{n+1}\dfrac{d^nz}{dx^n}=\sum_{k=0}^n(-1)^k\dfrac{x^k}{k!}\dfrac{d^ky}{dx^k}$$
Setting this to $0$, since $x=0$ won't work, we have $\dfrac{d^nz}{dx^n}=0$.  Integrate $n$ times, then $y=zx$.
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