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JEE Main 31-Jan-2024 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 31-Jan-2024 Paper - Shift 14 Marks
MCQ
The solution curve of the differential equation$y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0$ passing through the point $(e, 1)$ is
  • A
    $\left|\log _e \frac{y}{x}\right|=x$
  • B
    $\left|\log _e \frac{y}{x}\right|=y^2$
  • C
    $\left|\log _{ e } \frac{ x }{ y }\right|= y$
  • D
    $2\left|\log _e \frac{x}{y}\right|=y+1$
✓

Answer

$\frac{d x}{d y}=\frac{x}{y}\left(\ln \left(\frac{x}{y}\right)+1\right)$
Let $\frac{x}{y}=t \Rightarrow x=t y$
$\frac{d x}{d y}=t+y \frac{d t}{d y}$
$t+y \frac{d t}{d y}=t(\ln (t)+1)$
$y \frac{d t}{d y}=t \ln (t) \Rightarrow \frac{d t}{t \ln (t)}=\frac{d y}{y}$
$\Rightarrow \int \frac{d t}{t \cdot \ln (t)}=\int \frac{d y}{y} \frac{1}{y}$
$\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \frac{1}{t} ln t=p$
$\Rightarrow \ln p=\ln y+c$
$\ln (\ln t)=\ln y+c$
$\ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c$
$\text { at } x=e, y=1$
$\ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0$
$\ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y$
$\left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y}$
$\left|\ln \left(\frac{x}{y}\right)\right|=y$

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