Question Number 133508 by Ahmed1hamouda last updated on 22/Feb/21
Commented by Ahmed1hamouda last updated on 22/Feb/21
$$\boldsymbol{\mathrm{S}}\mathrm{olve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations} \\ $$
Answered by Olaf last updated on 22/Feb/21
![ydx−2xdy = (y−2)e^y dy y(dx/dy)−2x = (y−2)e^y (1) Let x(y) = u(y)e^y (1) : y[(du/dy)e^y +ue^y ]−2ue^y = (y−2)e^y y[(du/dy)+u]−2u = y−2 y(du/dy)+u(y−2) = y−2 y(du/dy) = (y−2)(1−u) −(du/(1−u)) = −((y−2)/y) ln∣1−u∣ = 2ln∣y∣−y+C_1 ln∣1−u∣ = ln(C_2 y^2 e^(−y) ) 1−u = C_2 y^2 e^(−y) u = 1−C_2 y^2 e^(−y) x = ue^y = e^y −C_2 y^2](https://www.tinkutara.com/question/Q133513.png)
$${ydx}−\mathrm{2}{xdy}\:=\:\left({y}−\mathrm{2}\right){e}^{{y}} {dy} \\ $$$${y}\frac{{dx}}{{dy}}−\mathrm{2}{x}\:=\:\left({y}−\mathrm{2}\right){e}^{{y}} \:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{x}\left({y}\right)\:=\:{u}\left({y}\right){e}^{{y}} \\ $$$$\left(\mathrm{1}\right)\::\:{y}\left[\frac{{du}}{{dy}}{e}^{{y}} +{ue}^{{y}} \right]−\mathrm{2}{ue}^{{y}} \:=\:\left({y}−\mathrm{2}\right){e}^{{y}} \\ $$$${y}\left[\frac{{du}}{{dy}}+{u}\right]−\mathrm{2}{u}\:=\:{y}−\mathrm{2} \\ $$$${y}\frac{{du}}{{dy}}+{u}\left({y}−\mathrm{2}\right)\:=\:{y}−\mathrm{2} \\ $$$${y}\frac{{du}}{{dy}}\:=\:\left({y}−\mathrm{2}\right)\left(\mathrm{1}−{u}\right) \\ $$$$−\frac{{du}}{\mathrm{1}−{u}}\:=\:−\frac{{y}−\mathrm{2}}{{y}} \\ $$$$\mathrm{ln}\mid\mathrm{1}−{u}\mid\:=\:\mathrm{2ln}\mid{y}\mid−{y}+\mathrm{C}_{\mathrm{1}} \\ $$$$\mathrm{ln}\mid\mathrm{1}−{u}\mid\:=\:\mathrm{ln}\left(\mathrm{C}_{\mathrm{2}} {y}^{\mathrm{2}} {e}^{−{y}} \right) \\ $$$$\mathrm{1}−{u}\:=\:\mathrm{C}_{\mathrm{2}} {y}^{\mathrm{2}} {e}^{−{y}} \\ $$$${u}\:=\:\mathrm{1}−\mathrm{C}_{\mathrm{2}} {y}^{\mathrm{2}} {e}^{−{y}} \\ $$$${x}\:=\:{ue}^{{y}} \:=\:{e}^{{y}} −\mathrm{C}_{\mathrm{2}} {y}^{\mathrm{2}} \\ $$