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Question Number 168905 by safojontoshtemirov last updated on 21/Apr/22

Answered by mindispower last updated on 22/Apr/22

$$\iff xy=z;z^{\prime }+2\frac{z^2}{x}ln\left(x\right)=0$$$$-\frac{z^{\prime }}{z^2}=\frac{2ln\left(x\right)}{x}\Rightarrow \frac{1}{z}={ln}^2\left(x\right)+c\Rightarrow z=\frac{1}{{ln}^2\left(x\right)+\mathrm{c}}$$$$\mathrm{y}=\frac{1}{\mathrm{x}({\ln }^2\left(\mathrm{x}\right)+\mathrm{c})}$$

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