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February 25th, 2019, 09:38 AM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 486 Thanks: 75  Nonorder equation
$\displaystyle y^{(1)} y^{(2)}\cdot .... \cdot y^{(n)} =e^{nx} \; \;$ , $\displaystyle n\in \mathbb{N}$ . Another way to write it better : $\displaystyle \prod_{i=1}^{n} \frac{d^i y}{dx^i } =e^{nx}$ . 
February 26th, 2019, 05:25 AM  #2 
Senior Member Joined: Dec 2015 From: iPhone Posts: 486 Thanks: 75 
Let $\displaystyle b_{n} =e^{nx}\; \; $ then $\displaystyle y^{(n+1)} =\frac{b_{n+1} }{b_n }=e^x$ . The equation $\displaystyle y^{(n+1)} =e^x \;$ has solution $\displaystyle y=c+e^x $. 

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