Question Number 13976 by tawa tawa last updated on 25/May/17
$$\mathrm{Solve}: \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{y}\:+\:\mathrm{1}\right)\:+\:\mathrm{y}^{\mathrm{2}} \left(\mathrm{x}\:−\:\mathrm{1}\right)\mathrm{y}'\:=\:\mathrm{0} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17
$${y}^{'} =−\frac{{x}^{\mathrm{2}} \left({y}+\mathrm{1}\right)}{{y}^{\mathrm{2}} \left({x}−\mathrm{1}\right)},\:\:{y}={mx} \\ $$$${y}^{'} =−\frac{{mx}+\mathrm{1}}{{m}^{\mathrm{2}} \left({x}−\mathrm{1}\right)}=−\frac{{mx}−{m}+{m}+\mathrm{1}}{{m}^{\mathrm{2}} \left({x}−\mathrm{1}\right)}= \\ $$$$=−\frac{\mathrm{1}}{{m}}+\frac{{m}+\mathrm{1}}{{m}^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{1}−{x}}\Rightarrow{y}+{C}=−\frac{{x}}{{m}}−\frac{{m}+\mathrm{1}}{{m}^{\mathrm{2}} }{ln}\left(\mathrm{1}−{x}\right)\Rightarrow \\ $$$${y}+{C}=−\frac{{x}^{\mathrm{2}} }{{y}}−\frac{\frac{{y}}{{x}}+\mathrm{1}}{\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}{ln}\left(\mathrm{1}−{x}\right)\Rightarrow \\ $$$$−{y}+{B}=\frac{{x}^{\mathrm{2}} }{{y}}+\frac{{xy}+{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }{ln}\left(\mathrm{1}−{x}\right)\:\:\left({B}=−{C}\right) \\ $$$$\Rightarrow−{y}^{\mathrm{3}} +{By}^{\mathrm{2}} ={yx}^{\mathrm{2}} +{x}\left({x}+{y}\right){ln}\left(\mathrm{1}−{x}\right)\:\:. \\ $$$$\left({x}=\mathrm{1}\Rightarrow{y}=−\mathrm{1}\right)\Rightarrow\mathrm{1}+{B}=−\mathrm{1}\Rightarrow{B}=−\mathrm{2} \\ $$$$\Rightarrow{y}\left({y}^{\mathrm{2}} +\mathrm{2}{y}+{x}^{\mathrm{2}} \right)+{x}\left({x}+{y}\right){ln}\left(\mathrm{1}−{x}\right)=\mathrm{0}\:.\blacksquare \\ $$
Commented by tawa tawa last updated on 26/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$