Question Number 68860 by ramirez105 last updated on 16/Sep/19
Commented by kaivan.ahmadi last updated on 16/Sep/19
$${set}\:{M}={x}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:\:{and}\:\:\:{N}=−{y}+\frac{{xy}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\frac{\partial{M}}{\partial{y}}=\frac{{y}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}=\frac{\partial{N}}{\partial{x}} \\ $$$$\begin{cases}{\frac{\partial{u}}{\partial{x}}={x}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}\\{\frac{\partial{u}}{\partial{y}}=−{y}+\frac{{xy}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}}\end{cases}\Rightarrow \\ $$$${u}\left({x},{y}\right)=\int\left({x}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\right){dx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}+{h}\left({y}\right) \\ $$$$\Rightarrow\frac{\partial{u}}{\partial{y}}=−{y}+\frac{{xy}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}=\frac{{xy}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=−{y}\Rightarrow{h}=−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{c}' \\ $$$$\Rightarrow{u}\left({x},{y}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}={c} \\ $$
Commented by ramirez105 last updated on 17/Sep/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$