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Question Number 137558 by liberty last updated on 04/Apr/21
$$\left({x}+{y}\right){dx}\:+\:\left({x}+{y}^{\mathrm{2}} \right){dy}\:=\:\mathrm{0}\: \\ $$
Answered by Ñï= last updated on 04/Apr/21
$$\left({x}+{y}\right){dx}+\left({x}+{y}^{\mathrm{2}} \right){dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{d}\left({x}^{\mathrm{2}} \right)+{ydx}+{xdy}+\frac{\mathrm{1}}{\mathrm{3}}{d}\left({y}^{\mathrm{3}} \right) \\ $$$$={d}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{y}^{\mathrm{3}} \right)+{d}\left({xy}\right) \\ $$$$={d}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{y}^{\mathrm{3}} +{xy}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{y}^{\mathrm{3}} +{xy}={C} \\ $$