Question Number 227219 by Spillover last updated on 06/Jan/26
$${Solve}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} +{xy}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}−{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$$$ \\ $$
Answered by peace2 last updated on 08/Jan/26
$$\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\left(\frac{\mathrm{1}+{x}}{{y}−\mathrm{1}}\right) \\ $$$$\Leftrightarrow\left(\frac{{y}−\mathrm{1}}{{y}^{\mathrm{2}} }\right){dy}=\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$\left({ln}\left({y}\right)+\frac{\mathrm{1}}{{y}}\right)=−\frac{\mathrm{1}}{{x}}+{ln}\left({x}\right)+{c} \\ $$$${ln}\left({ye}^{\frac{\mathrm{1}}{{y}}} \right)=−\frac{\mathrm{1}}{{x}}+{ln}\left({x}\right)+{c}….{A} \\ $$$${ye}^{\frac{\mathrm{1}}{{y}}} ={a};\frac{\mathrm{1}}{{y}}={s}\Rightarrow{e}^{{s}} ={as}\Rightarrow{se}^{−{s}} ={a}\Rightarrow−{s}={W}\left(−{a}\right) \\ $$$${s}=−{W}\left(−{a}\right) \\ $$$${y}=\frac{\mathrm{1}}{{W}\left(−{a}\right)}…{A}\Rightarrow{ye}^{\frac{\mathrm{1}}{{y}}} ={kxe}^{−\frac{\mathrm{1}}{{x}}} \\ $$$${y}=−\frac{\mathrm{1}}{{W}\left(−{kxe}^{−\frac{\mathrm{1}}{{x}}} \right)} \\ $$