Question Number 227221 by Spillover last updated on 06/Jan/26
$${Solve}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\left({x}^{\mathrm{2}} +{xy}\right)\frac{{dy}}{{dx}}={xy}−{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$
Answered by som(math1967) last updated on 07/Jan/26
$$\:\frac{{dy}}{{dx}}=\frac{{xy}−{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{xy}} \\ $$$$\:{put}\:{y}={vx}\Rightarrow\frac{{dy}}{{dx}}={v}+{x}\frac{{dv}}{{dx}} \\ $$$$\:{v}+{x}\frac{{dv}}{{dx}}=\frac{{vx}^{\mathrm{2}} −{v}^{\mathrm{2}} {x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} {v}} \\ $$$$\Rightarrow{v}+{x}\frac{{dv}}{{dx}}=\frac{{vx}^{\mathrm{2}} \left(\mathrm{1}−{v}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{v}\right)} \\ $$$$\Rightarrow{x}\frac{{dv}}{{dx}}=\frac{{v}−{v}^{\mathrm{2}} −{v}−{v}^{\mathrm{2}} }{\left(\mathrm{1}+{v}\right)} \\ $$$$\Rightarrow\int\frac{\left(\mathrm{1}+{v}\right){dv}}{\mathrm{2}{v}^{\mathrm{2}} }=−\int\frac{{dx}}{{x}} \\ $$$$\Rightarrow\int\frac{{dv}}{\mathrm{2}{v}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dv}}{{v}}+\int\frac{{dx}}{{x}}=\mathrm{0} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}{v}}+{ln}\sqrt{{v}}+{lnx}={C} \\ $$$$\:\:{where}\:{v}=\frac{{y}}{{x}} \\ $$
Commented by Spillover last updated on 07/Jan/26
$${thanks} \\ $$