If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-\mathrm{e}^x \operatorname{cosec} y\right) \frac{\mathrm{d} x}{\mathrm{~d} y}=x^5, x \geq 1$, then at $x=2$, the value of $\cos y$ is :

<p>$$\begin{aligned} & \left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5 \\ & x^5 \frac{d y}{d x}-7 x^4 \cot y=-e^x \operatorname{cosec} y \\ & \frac{d y}{d x}-\frac{7}{x} \cot y=-\frac{e^x}{x^5} \operatorname{cosec} y \\ & \sin y \frac{d y}{d x}-\frac{7}{x} \cos y=-\frac{e^x}{x^5} \\ & \text { Let }-\cos y=t \\ & \sin y \frac{d y}{d x}=\frac{d t}{d x} \\ & \therefore \frac{d t}{d x}+\frac{7}{x} t=-\frac{e^x}{x^5} \end{aligned}$$</p> <p>$$\begin{aligned} & \therefore \text { I.F. }=e^{\int \frac{7}{x} d x}=x^7 \\ & t \cdot x^7=\int \frac{-e^x}{x^5} \cdot x^7 d x \\ & -\cos y \cdot x^7=-\int e^x x^2 d x \\ & \cos y x^7=e^x\left(x^2-2 x+2\right)+c \\ & \because \quad x=1 \text { then } y=\frac{\pi}{2} \Rightarrow c=-e \\ & \therefore \quad \cos y \cdot x^7=e^x\left(x^2-2 x+2\right)-e \\ & \text { When } x=2 \text { then } \cos y=\frac{2 e^2-e}{128} \end{aligned}$$</p>