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Question Number 93818 by i jagooll last updated on 15/May/20

$$(\mathrm{y}-{\mathrm{xy}}^2)\mathrm{dx}+(\mathrm{x}+{\mathrm{x}}^2{\mathrm{y}}^2)\mathrm{dy}=0$$

Answered by i jagooll last updated on 15/May/20

$$\mathrm{ydx}+\mathrm{xdy}={\mathrm{xy}}^2\mathrm{dx}-{\mathrm{x}}^2{\mathrm{y}}^2\mathrm{dy}$$$$\mathrm{d}\left(\mathrm{xy}\right)={\mathrm{x}}^2{\mathrm{y}}^2(\frac{\mathrm{dx}}{\mathrm{x}}-\mathrm{dy})$$$$\frac{\mathrm{d}\left(\mathrm{xy}\right)}{{\left(\mathrm{xy}\right)}^2}=\frac{\mathrm{dx}}{\mathrm{x}}-\mathrm{dy}$$$$\int \frac{\mathrm{d}\left(\mathrm{xy}\right)}{{\left(\mathrm{xy}\right)}^2}=\int \frac{\mathrm{dx}}{\mathrm{x}}-\int \mathrm{dy}$$$$-\frac{1}{\mathrm{xy}}=\ln \left(\mathrm{x}\right)-\mathrm{y}+\mathrm{c}$$$$\frac{1}{\mathrm{xy}}=\mathrm{y}+\ln \left(\mathrm{x}\right)+\mathrm{C}\Rightarrow \mathrm{xy}=\frac{1}{\mathrm{y}+\ln \left(\mathrm{x}\right)+\mathrm{C}}$$

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