Question Number 128179 by bounhome last updated on 05/Jan/21
$${show}\:{me}\:{that}\:{y}={x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{6}\:{is}\:{the}\:{solution}\:{of} \\ $$$${y}''−{y}'+\mathrm{2}\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$
Answered by bemath last updated on 05/Jan/21
$$\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{6}\:\rightarrow\begin{cases}{\mathrm{y}'=\mathrm{2x}+\mathrm{4}}\\{\mathrm{y}''=\mathrm{2}}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{2}−\left(\mathrm{2x}+\mathrm{4}\right)+\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)\:= \\ $$$$\:\:\:\:\:\:\:\mathrm{2}−\mathrm{2x}−\mathrm{4}+\mathrm{2x}+\mathrm{2}\:=\:\mathrm{0} \\ $$