Question Number 68825 by peter frank last updated on 15/Sep/19
![](https://www.tinkutara.com/question/9330.png)
Answered by mind is power last updated on 16/Sep/19
![dx(1+e^(x/y) )+e^(x/y) (1−(x/y))dy p=1+e^(x/y) q=e^(x/y) (1−(x/y)) (∂p/∂y)=−x(e^(x/y) /y^(2 ) ) (∂q/∂x)=(e^(x/y) /y)(1−(x/y))−(e^(x/y) /y)=−x(e^(x/y) /y^2 )=(∂p/∂y) so exact equation](https://www.tinkutara.com/question/Q68893.png)
$${dx}\left(\mathrm{1}+{e}^{\frac{{x}}{{y}}} \right)+{e}^{\frac{{x}}{{y}}} \left(\mathrm{1}−\frac{{x}}{{y}}\right){dy} \\ $$$${p}=\mathrm{1}+{e}^{\frac{{x}}{{y}}} \\ $$$${q}={e}^{\frac{{x}}{{y}}} \left(\mathrm{1}−\frac{{x}}{{y}}\right) \\ $$$$\frac{\partial{p}}{\partial{y}}=−{x}\frac{{e}^{\frac{{x}}{{y}}} }{{y}^{\mathrm{2}\:} } \\ $$$$\frac{\partial{q}}{\partial{x}}=\frac{{e}^{\frac{{x}}{{y}}} }{{y}}\left(\mathrm{1}−\frac{{x}}{{y}}\right)−\frac{{e}^{\frac{{x}}{{y}}} }{{y}}=−{x}\frac{{e}^{\frac{{x}}{{y}}} }{{y}^{\mathrm{2}} }=\frac{\partial{p}}{\partial{y}} \\ $$$${so}\:{exact}\:{equation} \\ $$
Commented by peter frank last updated on 22/Sep/19
![thank you](https://www.tinkutara.com/question/Q69266.png)
$${thank}\:{you} \\ $$