Question Number 68946 by ramirez105 last updated on 17/Sep/19
![](https://www.tinkutara.com/question/9359.png)
Commented by kaivan.ahmadi last updated on 17/Sep/19
![(∂u/∂x)=(2xy−tany)⇒u(x,y)=x^2 y−xtany+h(y) (∂u/∂y)=x^2 −xsec^2 y=x^2 −xsec^2 y+(dh/dy)⇒(dh/dy)=0⇒h=c′ u(x,y)=c⇒x^2 y−xtany=c](https://www.tinkutara.com/question/Q68963.png)
$$\frac{\partial{u}}{\partial{x}}=\left(\mathrm{2}{xy}−{tany}\right)\Rightarrow{u}\left({x},{y}\right)={x}^{\mathrm{2}} {y}−{xtany}+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}={x}^{\mathrm{2}} −{xsec}^{\mathrm{2}} {y}={x}^{\mathrm{2}} −{xsec}^{\mathrm{2}} {y}+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=\mathrm{0}\Rightarrow{h}={c}' \\ $$$${u}\left({x},{y}\right)={c}\Rightarrow{x}^{\mathrm{2}} {y}−{xtany}={c} \\ $$