Question:
The general solution of the differential equation $\frac{dy}{dx} + y \cot x = 2 \cos x$ is:
The general solution of the differential equation $\frac{dy}{dx} + y \cot x = 2 \cos x$ is:
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Always remember that $\int \cot x \, dx = \ln|\sin x|$, making the integrating factor simply $\sin x$ for such standard equations.
Updated On: Jun 3, 2026
- $y \sin x = -\frac{1}{2} \cos 2x + C$
- $y \sin x = \frac{1}{2} \cos 2x + C$
- $y \sin x = -\cos 2x + C$
- $y \sin x = \sin 2x + C$
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The Correct Option is A
Solution and Explanation
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)
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)
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