Let $y=y(x)$ be the solution curve of the differential equation $\sec y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y, y(1)=0$. Then $y(\sqrt{3})$ is equal to:

<p>$$\begin{aligned} & \sec y \frac{d y}{d x}+2 x \sin y=x^3 \cos y \\ & \Rightarrow \sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3 \end{aligned}$$</p> <p>Let $z=\tan y$</p> <p>$$\begin{aligned} & \frac{d z}{d x}=\sec ^2 y \frac{d y}{d x} \\ & \Rightarrow \frac{d z}{d x}+2 x z=x^3 \end{aligned}$$</p> <p>$$\begin{aligned} & \text { I.F. }=e^{x^2} \\ & \Rightarrow z . e^{x^2}=\int e^{x^2} \cdot x^3 d x+c \end{aligned}$$</p> <p>$\Rightarrow \tan y \cdot e^{x^2}=\frac{1}{2}\left(x^2 e^{x^2}-e^{x^2}\right)+c$</p> <p>$\Rightarrow \tan (0) \cdot e=\frac{1}{2}(1 \cdot e-e)+c$</p> <p>$\Rightarrow c=0$</p> <p>$\Rightarrow \tan y=\frac{x^2-1}{2}$</p> <p>$$f(x)=\tan ^{-1}\left(\frac{x^2-1}{2}\right) \Rightarrow f(\sqrt{3})=\frac{\pi}{4}$$</p>