Question Number 168905 by safojontoshtemirov last updated on 21/Apr/22
Answered by mindispower last updated on 22/Apr/22
$$\Leftrightarrow{xy}={z};{z}'+\mathrm{2}\frac{{z}^{\mathrm{2}} }{{x}}{ln}\left({x}\right)=\mathrm{0} \\ $$$$−\frac{{z}'}{{z}^{\mathrm{2}} }=\frac{\mathrm{2}{ln}\left({x}\right)}{{x}}\Rightarrow\frac{\mathrm{1}}{{z}}={ln}^{\mathrm{2}} \left({x}\right)+{c}\Rightarrow{z}=\frac{\mathrm{1}}{{ln}^{\mathrm{2}} \left({x}\right)+\mathrm{c}} \\ $$$$\mathrm{y}=\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{c}\right)} \\ $$