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# Question-68855

Question Number 68855 by ramirez105 last updated on 16/Sep/19
Commented by kaivan.ahmadi last updated on 16/Sep/19
$${set}\:{M}=\mathrm{6}{x}+{y}^{\mathrm{2}} \:\:{and}\:{N}={y}\left(\mathrm{2}{x}−\mathrm{3}{y}\right) \\$$$$\frac{\partial{M}}{\partial{y}}=\mathrm{2}{y}\:=\frac{\partial{N}}{\partial{x}}\Rightarrow{the}\:{equation}\:{is}\:{exact} \\$$$$\frac{\partial{u}}{\partial{x}}=\mathrm{6}{x}+{y}^{\mathrm{2}} \:\:\:{and}\:\:\frac{\partial{u}}{\partial{y}}={y}\left(\mathrm{2}{x}−\mathrm{3}{y}\right)=\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} \\$$$$\Rightarrow{u}\left({x},{y}\right)=\int\left(\mathrm{6}{x}+{y}^{\mathrm{2}} \right){dx}=\mathrm{3}{x}^{\mathrm{2}} +{xy}^{\mathrm{2}} +{h}\left({y}\right) \\$$$$\Rightarrow\frac{\partial{u}}{\partial{y}}=\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} =\mathrm{2}{xy}+\frac{{d}\left({h}\right)}{{dy}}\Rightarrow\frac{{d}\left({h}\right)}{{dy}}=−\mathrm{3}{y}^{\mathrm{2}} \Rightarrow \\$$$${h}=−{y}^{\mathrm{3}} +{c}' \\$$$${u}\left({x},{y}\right)={c}\Rightarrow\mathrm{3}{x}^{\mathrm{2}} +{xy}^{\mathrm{2}} −{y}^{\mathrm{3}} +{c}'={c}\Rightarrow \\$$$${u}\left({x},{y}\right)=\mathrm{3}{x}^{\mathrm{2}} +{xy}^{\mathrm{2}} −{y}^{\mathrm{3}} ={C} \\$$$$\\$$$$\\$$$$\\$$
Commented by ramirez105 last updated on 17/Sep/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\$$$$\\$$