Question Number 293 by 123456 last updated on 25/Jan/15
$${u}:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R} \\ $$$$\frac{\partial{u}}{\partial{x}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}{xu}=\mathrm{0} \\ $$
Answered by prakash jain last updated on 19/Dec/14
$$\frac{{du}}{{u}}=\frac{\mathrm{2}{xdx}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={t}\Rightarrow\mathrm{2}{xdx}={dt} \\ $$$$\mathrm{ln}\:{u}=\mathrm{ln}\:{t}+{C}=\mathrm{ln}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+{C} \\ $$$${u}=\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\centerdot{C}_{\mathrm{1}} \\ $$