QJEEM24Mab4ae: JEE Main 2024 Mathematics - Class 12 JEEM24Mab4ae Solution: If is the solution of the differential equation and , then is…

Differential equation :-

$\begin{aligned}& x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x \\& \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x\end{aligned}$

Divide both sides by $\mathrm{x}^2$

$\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}$

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

$\begin{aligned}& \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \\& \cos \mathrm{t~dt}=\frac{1}{\mathrm{x}} \mathrm{dx}\end{aligned}$

Integrating both sides

$\begin{aligned}& \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} \\& \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c}\end{aligned}$

Using $\mathrm{y}(1)=\frac{\pi}{3}$ , we get $\mathrm{c}=\frac{\sqrt{3}}{2}$

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