Question Number 50431 by Abdo msup. last updated on 16/Dec/18
$${solve}\:{xy}^{'} \:+{y}\:=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$
Commented by Abdo msup. last updated on 16/Dec/18
$$\left({he}\right)\:\:\:\:\:{xy}^{'} \:+{y}\:=\mathrm{0}\:\Rightarrow{xy}^{'} =−{y}\:\Rightarrow\frac{{y}^{'} }{{y}}=−\frac{\mathrm{1}}{{x}}\:\Rightarrow{ln}\mid{y}\mid=−{ln}\mid{x}\mid\:+{k}\:\Rightarrow \\ $$$${y}\left({x}\right)=\frac{{K}}{\mid{x}\mid}\:\:\:{let}\:{take}\:{x}>\mathrm{0}\Rightarrow{y}=\frac{{K}}{{x}}\:\:{mvc}\:{method}\:{give}\: \\ $$$${y}^{'} =\frac{{K}^{'} }{{x}}\:−\frac{{K}}{{x}^{\mathrm{2}} }\:\:{and}\:\left({e}\right)\:\Leftrightarrow\:\:{K}^{'} \:−\frac{{K}}{{x}}\:+\frac{{K}}{{x}}\:=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:\Rightarrow \\ $$$${K}^{'} \:=\:\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:\Rightarrow\:{K}\left({x}\right)=\:\int\:\:\:\frac{\mathrm{2}{xdx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:\:{changement}\:{x}^{\mathrm{2}} ={sint} \\ $$$${give}\:{K}\left({x}\right)\:=\int\:\:\frac{{cost}\:{dt}}{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}}\:=\int\:{dt}\:+{c}\:={t}\:+\lambda\:\:={arcsin}\left({x}^{\mathrm{2}} \right)\:+\lambda \\ $$$$\Rightarrow\:{y}\left({x}\right)=\:\frac{{arcsin}\left({x}^{\mathrm{2}} \right)\:+\lambda}{{x}}\:=\frac{\lambda}{{x}}\:+\:\frac{{arcsin}\left({x}^{\mathrm{2}} \right)}{{x}}\:. \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Dec/18
$${xdy}+{ydx}=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} }}{dx} \\ $$$${d}\left({xy}\right)=\frac{{d}\left({x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\int{d}\left({xy}\right)=\int\frac{{d}\left({x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$${xy}={sin}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)+{c} \\ $$$$ \\ $$