QJEEM24Ma1bb4: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1bb4 Solution: The solution curve of the differential equation passing through …

$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}}{\mathrm{y}}\left(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1\right)$

Let $\frac{x}{y}=t \Rightarrow x=t y$

$\begin{aligned}& \frac{d x}{d y}=t+y \frac{d t}{d y} \\& t+y \frac{d t}{d y}=t(\ln (t)+1)\end{aligned}$

$\mathrm{y} \frac{\mathrm{dt}}{\mathrm{dy}}=\mathrm{t} \ln (\mathrm{t}) \Rightarrow \frac{\mathrm{dt}}{\mathrm{t} \ln (\mathrm{t})}=\frac{\mathrm{dy}}{\mathrm{y}}$

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t} \cdot \ln (\mathrm{t})}=\int \frac{\mathrm{dy}}{\mathrm{y}}$

$\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \quad$ let $\ln t=p$

$\frac{1}{\mathrm{t}} \mathrm{dt}=\mathrm{dp}$

$\begin{aligned}& \Rightarrow \ln p=\ln y+c \\& \ln (\ln t)=\ln y+c \\& \ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c \\& \text { at } x=e, y=1 \\& \ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0\end{aligned}$

$\begin{aligned}& \ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y \\& \left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y} \\& \left|\ln \left(\frac{x}{y}\right)\right|=y\end{aligned}$