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

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