Question Number 184085 by Shrinava last updated on 02/Jan/23
$$\left(\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}^{\mathrm{2}} \right)\mathrm{y}^{'} \:+\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{yx}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$
Answered by mr W last updated on 02/Jan/23
$$\frac{{y}^{\mathrm{2}} {dy}}{{y}−\mathrm{1}}=\frac{{x}^{\mathrm{2}} {dx}}{{x}+\mathrm{1}} \\ $$$$\int\frac{{y}^{\mathrm{2}} {dy}}{{y}−\mathrm{1}}=\int\frac{{x}^{\mathrm{2}} {dx}}{{x}+\mathrm{1}} \\ $$$$\int\frac{\left({y}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right){dy}}{{y}−\mathrm{1}}=\int\frac{\left({x}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right){dx}}{{x}+\mathrm{1}} \\ $$$$\int\left({y}+\mathrm{1}+\frac{\mathrm{1}}{{y}−\mathrm{1}}\right){dy}=\int\left({x}−\mathrm{1}+\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx} \\ $$$$\Rightarrow\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{y}+\mathrm{ln}\:\left({y}−\mathrm{1}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}+\mathrm{ln}\:\left({x}+\mathrm{1}\right)+{C} \\ $$
Commented by Shrinava last updated on 02/Jan/23
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{professor},\:\mathrm{cool} \\ $$