Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$$\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{6}\:\rightarrow\begin{cases}{\mathrm{y}'=\mathrm{2x}+\mathrm{4}}\\{\mathrm{y}''=\mathrm{2}}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{2}−\left(\mathrm{2x}+\mathrm{4}\right)+\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)\:= \\ $$$$\:\:\:\:\:\:\:\mathrm{2}−\mathrm{2x}−\mathrm{4}+\mathrm{2x}+\mathrm{2}\:=\:\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com