Question Number 130893 by EDWIN88 last updated on 30/Jan/21
$${Find}\:{a}\:{particular}\:{solution} \\ $$$${of}\:{the}\:{equation}\:{f}\:'\left({x}\right)\:=\:{f}\left({x}\right)\: \\ $$$${such}\:{that}\:{f}\left({x}\right)=\mathrm{2}\:{for}\:{x}=\mathrm{2}\: \\ $$
Answered by mr W last updated on 30/Jan/21
$${y}'={y} \\ $$$$\frac{{dy}}{{y}}={dx} \\ $$$$\mathrm{ln}\:{y}={x}+{C} \\ $$$$\Rightarrow{y}={ce}^{{x}} \\ $$$$\Rightarrow{f}\left(\mathrm{2}\right)={ce}^{\mathrm{2}} =\mathrm{2}\:\Rightarrow{c}=\mathrm{2}{e}^{−\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)={y}=\mathrm{2}{e}^{{x}−\mathrm{2}} \\ $$
Answered by benjo_mathlover last updated on 30/Jan/21
$$\:\mathrm{y}'−\mathrm{y}\:=\:\mathrm{0}\:\Rightarrow\lambda−\mathrm{1}=\mathrm{0}\:;\:\lambda=\mathrm{1}\: \\ $$$$\:\mathrm{y}\left(\mathrm{x}\right)\:=\:\mathrm{C}\:\mathrm{e}^{\mathrm{x}} \:\Rightarrow\mathrm{y}\left(\mathrm{2}\right)=\:\mathrm{Ce}^{\mathrm{2}} \:=\:\mathrm{2}\:;\:\mathrm{C}\:=\:\mathrm{2e}^{−\mathrm{2}} \: \\ $$$$\therefore\:\mathrm{y}\left(\mathrm{x}\right)\:=\:\mathrm{2e}^{−\mathrm{2}} .\mathrm{e}^{\mathrm{x}} \:=\:\mathrm{2e}^{\mathrm{x}−\mathrm{2}} \\ $$