<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] Use induction. Suppose the property holds for 
$$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $$
Prove that it works for $ n+1$ terms by writing the expression, use the fact that $\dfrac{\mathrm d^{n+1}y}{\mathrm dx^{n+1}}f = \dfrac{\mathrm d^{n}y}{\mathrm dx^{n}}\left(\dfrac{\mathrm dy}{\mathrm dx}f\right)$ and then show that your new equation is equivalent to one with $n$ terms.
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