QJEEM24Ma2ef1: JEE Main 2024 Mathematics - Class 12 JEEM24Ma2ef1 Solution: Let be the solution curve of the differential equation . Then…

$\begin{aligned}& \sec y \frac{d y}{d x}+2 x \sin y=x^3 \cos y \\& \Rightarrow \sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3\end{aligned}$

Let $z=\tan y$

$\begin{aligned}& \frac{d z}{d x}=\sec ^2 y \frac{d y}{d x} \\& \Rightarrow \frac{d z}{d x}+2 x z=x^3\end{aligned}$

$\begin{aligned}& \text { I.F. }=e^{x^2} \\& \Rightarrow z . e^{x^2}=\int e^{x^2} \cdot x^3 d x+c\end{aligned}$

$\Rightarrow \tan y \cdot e^{x^2}=\frac{1}{2}\left(x^2 e^{x^2}-e^{x^2}\right)+c$

$\Rightarrow \tan (0) \cdot e=\frac{1}{2}(1 \cdot e-e)+c$

$\Rightarrow c=0$

$\Rightarrow \tan y=\frac{x^2-1}{2}$

$f(x)=\tan ^{-1}\left(\frac{x^2-1}{2}\right) \Rightarrow f(\sqrt{3})=\frac{\pi}{4}$