JEE Main 2024 — 29 January, Shift 2 — Question 12

Differential equation :- $\begin{aligned}& x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x &\end{aligned}$

$\cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x$

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

$$\cos t\left(\frac{d t}{d x}\right)=\frac{1}{x}$$ $$\operatorname{cost} \mathrm{dt}=\frac{1}{\mathrm{x}} \mathrm{dx}$$

Integrating both sides $\begin{aligned}& \sin t=\ln |x|+c &\end{aligned}$

$\sin \frac{y}{x}=\ln |x|+c$

$\operatorname{Using} \mathrm{y}(1)=\frac{\pi}{3}$,

we get $\mathrm{c}=\frac{\sqrt{3}}{2}$

So, $\alpha=\sqrt{3} \Rightarrow \alpha^{2}=3$