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

The given differential equation is $$\left(7x^{4}\cot y-e^{x}\csc y\right)\dfrac{dx}{dy}=x^{5},\;x\ge 1$$ and the curve passes through $$\left(1,\dfrac{\pi}{2}\right)$$.

First convert the equation to the more familiar $$\dfrac{dy}{dx}$$ form.
Taking reciprocal,

$$\dfrac{dy}{dx}=\dfrac{7x^{4}\cot y-e^{x}\csc y}{x^{5}}= \dfrac{7\cot y}{x}-\dfrac{e^{x}\csc y}{x^{5}}\;-(1)$$

Since the final answer is asked in terms of $$\cos y$$, set $$u=\cos y$$.
Then $$\dfrac{du}{dx}=-\sin y\,\dfrac{dy}{dx}\;-(2)$$

Substituting $$(1)$$ into $$(2)$$ and using $$\cot y=\dfrac{\cos y}{\sin y}= \dfrac{u}{\sin y}$$ and $$\csc y=\dfrac{1}{\sin y}$$:

$$\dfrac{du}{dx}= -\sin y\left(\dfrac{7u}{x}-\dfrac{e^{x}}{x^{5}\sin y}\right)= -\dfrac{7u}{x}+\dfrac{e^{x}}{x^{5}}\;-(3)$$

Equation $$(3)$$ is a linear first-order ODE in $$u(x)$$:

$$\dfrac{du}{dx}+\dfrac{7}{x}u=\dfrac{e^{x}}{x^{5}}\;-(4)$$

For a linear ODE $$\dfrac{du}{dx}+P(x)u=Q(x)$$, the integrating factor is $$\exp\!\bigl(\int P(x)\,dx\bigr)$$.
Here $$P(x)=\dfrac{7}{x}\;\Rightarrow\;\text{I.F.}=e^{\int 7/x\,dx}=e^{7\ln x}=x^{7}.$$

Multiplying $$(4)$$ by $$x^{7}$$ gives

$$x^{7}\dfrac{du}{dx}+7x^{6}u=x^{2}e^{x}\;-(5)$$

The left-hand side of $$(5)$$ is the derivative of $$x^{7}u$$:

$$\dfrac{d}{dx}\left(x^{7}u\right)=x^{2}e^{x}\;-(6)$$

Integrate $$(6)$$ with respect to $$x$$:

$$x^{7}u=\int x^{2}e^{x}\,dx+C\;-(7)$$

Evaluate the integral on the right by repeated integration by parts:

$$\int x^{2}e^{x}\,dx=e^{x}(x^{2}-2x+2)+C_{1}.$$

Absorbing constants, write $$(7)$$ as

$$x^{7}u=e^{x}(x^{2}-2x+2)+C\;-(8)$$

Use the initial condition. At $$x=1$$, $$y=\dfrac{\pi}{2}\Rightarrow u=\cos\!\left(\dfrac{\pi}{2}\right)=0$$.
Substituting $$x=1,\,u=0$$ in $$(8)$$:

$$0=e(1^{2}-2\cdot1+2)+C=e(1)+C\;\Rightarrow\;C=-e.$$

Thus $$(8)$$ becomes

$$x^{7}u=e^{x}(x^{2}-2x+2)-e\;-(9)$$

Finally, at $$x=2$$:

$$u(2)=\cos y\bigl|_{x=2}=\dfrac{e^{2}(2^{2}-2\cdot2+2)-e}{2^{7}}=\dfrac{e^{2}(4-4+2)-e}{128}=\dfrac{2e^{2}-e}{128}.$$

Therefore, $$\cos y=\dfrac{2e^{2}-e}{128}$$ at $$x=2$$.

The correct option is Option C.