Question Number 212435 by Ar Brandon last updated on 13/Oct/24
Answered by mr W last updated on 13/Oct/24
$${y}'=−\frac{\mathrm{2}{y}^{\mathrm{2}} +{xy}}{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} } \\ $$$${let}\:{y}={px} \\ $$$${p}+{x}\frac{{dp}}{{dx}}=−\frac{\mathrm{2}{p}^{\mathrm{2}} +{p}}{{p}^{\mathrm{2}} +{p}+\mathrm{1}} \\ $$$${x}\frac{{dp}}{{dx}}=−\frac{{p}\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}{{p}^{\mathrm{2}} +{p}+\mathrm{1}} \\ $$$$\frac{{p}^{\mathrm{2}} +{p}+\mathrm{1}}{{p}\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}{dp}=−{xdx} \\ $$$$\int\left(\frac{\mathrm{1}}{\mathrm{2}{p}}−\frac{\mathrm{1}}{{p}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{2}\left({p}+\mathrm{2}\right)}\right){dp}=−\int{xdx} \\ $$$$\mathrm{ln}\:{p}−\mathrm{2ln}\:\left({p}+\mathrm{1}\right)+\mathrm{3ln}\:\left({p}+\mathrm{2}\right)=−{x}^{\mathrm{2}} −{C}_{\mathrm{1}} \\ $$$$\mathrm{ln}\:\frac{{p}\left({p}+\mathrm{2}\right)^{\mathrm{3}} }{\left({p}+\mathrm{1}\right)^{\mathrm{2}} }=−\left({x}^{\mathrm{2}} +{C}_{\mathrm{1}} \right) \\ $$$$\frac{\frac{{y}}{{x}}\left(\frac{{y}}{{x}}+\mathrm{2}\right)^{\mathrm{3}} }{\left(\frac{{y}}{{x}}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{{C}}{{e}^{{x}^{\mathrm{2}} } } \\ $$$$\Rightarrow\frac{{y}\left({y}+\mathrm{2}{x}\right)^{\mathrm{3}} }{\left({y}+{x}\right)^{\mathrm{2}} }=\frac{{Cx}^{\mathrm{2}} }{{e}^{{x}^{\mathrm{2}} } } \\ $$
Commented by Ar Brandon last updated on 13/Oct/24
Thanks, sir!