If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential…
If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5, x \geq 1$, then at $x=2$, the value of cosy is:
$\frac{2 \mathrm{e}^2-\mathrm{e}}{64}$
$\frac{2 \mathrm{e}^2+\mathrm{e}}{64}$
$\frac{2 \mathrm{e}^2-\mathrm{e}}{128}$
$\frac{2 \mathrm{e}^2+\mathrm{e}}{128}$
Solution
$\begin{aligned} & \frac{d y}{d x}=\frac{7 \cot y}{x}-\frac{e^x \operatorname{cosec} y}{x^5} \\ & \frac{d y}{d x}=\frac{7 \cot y}{\sin y \cdot x}-\frac{e^x}{\sin y x^5} \\ & \sin y \frac{d y}{d x}-\cos y \cdot \frac{7}{x}=\frac{-e^x}{x^5} \\ & \text { let }-\cos y=t \\ & \sin y \frac{d y}{d x}=\frac{d t}{d x}\end{aligned}$ $\begin{aligned} & \frac{d t}{d x}+\frac{7 t}{x}=\frac{-e^x}{x^5} \\ & \text { I.F. }=x^7 \\ & \text { t. } x^7=-\int x^2 e^x d x \\ & \operatorname{cosy} x^7=x^2 e^x-2 \int x e^x d x \\ & \operatorname{cosy} x^7=x^2 e^x-2 x e^x+2 e^x+c \\ & x=1, y=\frac{\pi}{2}, c=-e \\ & \operatorname{cosy}=\frac{2 e^2-e}{128} \\ & \text { option }(3)\end{aligned}$