If a curve y=y(x) passes through the point $\left(1,\dfrac{\pi}{2}\right)$ and satisfies the differential equation $(7x^{4}\cot y-e^{x}\csc y),\dfrac{dx}{dy}=x^{5},\ x\ge1$, then at $x=2$, the value of $\cos y$ is:

Given: $(7x^{4}\cot y - e^{x}\csc y)\frac{dx}{dy}=x^{5}$

⇒ $\frac{dy}{dx}=\frac{7x^{4}\cot y - e^{x}\csc y}{x^{5}}$

Multiply by $\sin y$:
$\sin y\frac{dy}{dx}=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$

Now note:
$\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}$

So,
$-\frac{d}{dx}(\cos y)=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$

⇒ $\frac{d}{dx}(\cos y)+\frac{7}{x}\cos y=\frac{e^{x}}{x^{5}}$

This is linear in $\cos y$

IF $=e^{\int \frac{7}{x}dx}=x^{7}$

$\frac{d}{dx}(x^{7}\cos y)=x^{2}e^{x}$

Integrate:
$x^{7}\cos y=\int x^{2}e^{x}dx$

$\int x^{2}e^{x}dx=e^{x}(x^{2}-2x+2)$

So,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)+C$

Use $(1,\frac{\pi}{2})$ ⇒ $\cos y=0$
$0=e(1-2+2)+C=e+C$ ⇒ $C=-e$

Thus,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)-e$

At $x=2$:
$2^{7}\cos y=e^{2}(4-4+2)-e=2e^{2}-e$

$\cos y=\frac{2e^{2}-e}{128}=\frac{e(2e-1)}{128}$

$\boxed{\frac{e(2e-1)}{128}}$