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Question Number 122732 by fajri last updated on 19/Nov/20
$${find}\:{critical}\:{point}\:{from}\: \\ $$$$\:{Differential}\:{Equation}\:{System} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}\:=\:{x}\:−\:{y}\: \\ $$$$\frac{{dy}}{{dx}}\:=\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:−\:\mathrm{2}\: \\ $$
Answered by bemath last updated on 19/Nov/20
$$\left(\mathrm{1}\right)\:{let}\:{y}\:=\:{x}−{u}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:\mathrm{1}−\frac{{du}}{{dx}} \\ $$$$\Rightarrow\:\mathrm{1}−\frac{{du}}{{dx}}\:=\:{x}−\left({x}−{u}\right) \\ $$$$\Rightarrow\:\mathrm{1}−{u}\:=\:\frac{{du}}{{dx}}\:;\:\frac{{du}}{\mathrm{1}−{u}}\:=\:{dx} \\ $$$$\:\Rightarrow\mathrm{ln}\:\mid\mathrm{1}−{u}\mid\:=\:{x}\:+\:{c}\: \\ $$$$\Rightarrow\:\mathrm{1}−{u}\:=\:\pm{Ke}^{{x}} \:;\:\mathrm{1}−\left({x}−{y}\right)=\pm{Ke}^{{x}} \\ $$$$\Rightarrow\:{y}\:=\:\pm{Ke}^{{x}} +{x}−\mathrm{1} \\ $$$$\Rightarrow{y}'\:=\:\pm{Ke}^{{x}} +\mathrm{1}\:=\:\mathrm{0}\:;\: \\ $$$$\Rightarrow\:{e}^{{x}} \:=\:\frac{\mathrm{1}}{{K}}\:,\:{where}\:\frac{\mathrm{1}}{{K}}>\mathrm{0} \\ $$$$\Rightarrow\:{x}\:=\:−\mathrm{ln}\:{K}\:{and}\:{y}\:=\:−{Ke}^{−\mathrm{ln}\:{K}} −\mathrm{ln}\:{K}+\mathrm{1}\: \\ $$$${y}\:=\:−{K}\left(\frac{\mathrm{1}}{{K}}\right)−\mathrm{ln}\:{K}\:+\mathrm{1}\:=\:−\mathrm{ln}\:{K}\: \\ $$$${Hence}\:{critical}\:{point}\:{of}\:{y}={f}\left({x}\right)=−{Ke}^{{x}} +{x}−\mathrm{1} \\ $$$${is}\:\left(\:−\mathrm{ln}\:{K},\:−\mathrm{ln}\:{K}\:\right). \\ $$$$\: \\ $$
Commented by fajri last updated on 19/Nov/20
$$\left.{thank}\:{Sir},\:{I}\:{Like}\:{it}\::\right) \\ $$