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February 25th, 2019, 09:38 AM   #1
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Non-order equation

$\displaystyle y^{(1)} y^{(2)}\cdot .... \cdot y^{(n)} =e^{nx} \; \;$ , $\displaystyle n\in \mathbb{N}$ .

Another way to write it better : $\displaystyle \prod_{i=1}^{n} \frac{d^i y}{dx^i } =e^{nx}$ .
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February 26th, 2019, 05:25 AM   #2
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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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