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(D) Given the differential equation: $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$. Dividing by $x \cos x$, we get: $\frac{dy}{dx} + (	an x +

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$x y \sec x - 	an x = C$

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(D) Given the differential equation: $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$. Dividing by $x \cos x$, we get: $\frac{dy}{dx} + (	an x + \frac{1}{x}) y = \frac{\sec x}{x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = 	an x + \frac{1}{x}$ and $Q = \frac{\sec x}{x}$. The Integrating Factor $(IF)$ is given by: $IF = e^{\int P dx} = e^{\int (	an x + \frac{1}{x}) dx} = e^{\ln(\sec x) + \ln(x)} = e^{\ln(x \sec x)} = x \sec x$. The solution is given by: $y \cdot (IF) = \int Q \cdot (IF) dx + C$. Substituting the values: $y(x \sec x) = \int \frac{\sec x}{x} \cdot (x \sec x) dx + C$. $x y \sec x = \int \sec^2 x dx + C$. $x y \sec x = 	an x + C$. Thus, $x y \sec x - 	an x = C$.

The solution of the differential equation $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$ is

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