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JEE Main 29-Jan-2024 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 29-Jan-2024 Paper - Shift 24 Marks
MCQ
If $\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2}$ is the solution of the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ and $y(1)=\frac{\pi}{3},$ then $\alpha^2$ is equal to
  • A
    3
  • B
    12
  • C
    4
  • D
    9
✓

Answer

$(A)$ Differential equation : $-$
$x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x$
$\cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x$
Divide both sides by $x ^2$
$\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}$
Let $ \frac{y}{x}=t$
$\cos t\left(\frac{dt}{dx}\right)=\frac{1}{x}$
$\cos t \ dt=\frac{1}{x} dx$
Integrating both sides
$\sin t=\ln |x|+c$
$\sin \frac{y}{x}=\ln |x|+c$
Using $y (1)=\frac{\pi}{3},$
we get $c =\frac{\sqrt{3}}{2}$
So, $\alpha=\sqrt{3} $
$\Rightarrow \alpha^2=3$

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