Question Number 68923 by ramirez105 last updated on 16/Sep/19
Commented by kaivan.ahmadi last updated on 16/Sep/19
$$\frac{\partial{u}}{\partial{x}}=\frac{\mathrm{1}}{{y}}\Rightarrow{u}\left({x},{y}\right)=\int\frac{\mathrm{1}}{{y}}{dx}=\frac{{x}}{{y}}+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}=−\frac{{x}}{{y}^{\mathrm{2}} }=−\frac{{x}}{{y}^{\mathrm{2}} }+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=\mathrm{0}\Rightarrow{h}={c}' \\ $$$$ \\ $$$$\Rightarrow{u}\left({x},{y}\right)={c}\Rightarrow\frac{{x}}{{y}}={c}\Rightarrow{x}−{cy}=\mathrm{0} \\ $$
Commented by ramirez105 last updated on 17/Sep/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$