Question

The general solution of the differential equation $\frac{dy}{dx} + y \cot x = 2 \cos x$ is:

• A. $y \sin x = -\frac{1}{2} \cos 2x + C$
• B. $y \sin x = \frac{1}{2} \cos 2x + C$
• C. $y \sin x = -\cos 2x + C$
• D. $y \sin x = \sin 2x + C$

Hint:

Always remember that $\int \cot x \, dx = \ln|\sin x|$, making the integrating factor simply $\sin x$ for such standard equations.

Answer 1

The Correct Option is A

Step 1: Concept
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$. The general solution is given by: \[ y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \] where the integrating factor is $\text{I.F.} = e^{\int P(x) \, dx}$.

Step 2: Meaning
Here, $P(x) = \cot x$ and $Q(x) = 2 \cos x$. We compute the Integrating Factor first.

Step 3: Analysis
Find the Integrating Factor: \[ \text{I.F.} = e^{\int \cot x \, dx} = e^{\ln|\sin x|} = \sin x \] Now write the general solution: \[ y \sin x = \int (2 \cos x) \sin x \, dx + C \] \[ y \sin x = \int \sin 2x \, dx + C \] \[ y \sin x = -\frac{1}{2} \cos 2x + C \]

Step 4: Conclusion
The general solution is $y \sin x = -\frac{1}{2} \cos 2x + C$. Final Answer: (A)