Question Number 68923 by ramirez105 last updated on 16/Sep/19
![](https://www.tinkutara.com/question/9353.png)
Commented by kaivan.ahmadi last updated on 16/Sep/19
![(∂u/∂x)=(1/y)⇒u(x,y)=∫(1/y)dx=(x/y)+h(y) (∂u/∂y)=−(x/y^2 )=−(x/y^2 )+(dh/dy)⇒(dh/dy)=0⇒h=c′ ⇒u(x,y)=c⇒(x/y)=c⇒x−cy=0](https://www.tinkutara.com/question/Q68927.png)
$$\frac{\partial{u}}{\partial{x}}=\frac{\mathrm{1}}{{y}}\Rightarrow{u}\left({x},{y}\right)=\int\frac{\mathrm{1}}{{y}}{dx}=\frac{{x}}{{y}}+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}=−\frac{{x}}{{y}^{\mathrm{2}} }=−\frac{{x}}{{y}^{\mathrm{2}} }+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=\mathrm{0}\Rightarrow{h}={c}' \\ $$$$ \\ $$$$\Rightarrow{u}\left({x},{y}\right)={c}\Rightarrow\frac{{x}}{{y}}={c}\Rightarrow{x}−{cy}=\mathrm{0} \\ $$
Commented by ramirez105 last updated on 17/Sep/19
![thank you sir!](https://www.tinkutara.com/question/Q68943.png)
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$