Question:
The integrating factor of the differential equation \( \sin x\, dy = \frac{1}{2}(\sin2x + 2y\cos x)\,dx \) is
The integrating factor of the differential equation \( \sin x\, dy = \frac{1}{2}(\sin2x + 2y\cos x)\,dx \) is
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I.F. = \( e^{\int P(x)dx} \) for linear equations.
Updated On: Apr 21, 2026
- \( \sec x \)
- \( \sin x \)
- \( \tan x \)
- \( \cos x \)
- \( \csc x \)
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The Correct Option is
Solution and Explanation
Concept:
Convert into linear form.
Step 1: Rewrite.
\[ \frac{dy}{dx} = \frac{\sin2x}{2\sin x} + y\frac{\cos x}{\sin x} \] \[ = \cos x + y\cot x \]
Step 2: Linear form.
\[ \frac{dy}{dx} - y\cot x = \cos x \]
Step 3: Find I.F.
\[ IF = e^{-\int \cot x dx} = e^{-\ln(\sin x)} = \frac{1}{\sin x} \] \[ = \csc x \]
Step 1: Rewrite.
\[ \frac{dy}{dx} = \frac{\sin2x}{2\sin x} + y\frac{\cos x}{\sin x} \] \[ = \cos x + y\cot x \]
Step 2: Linear form.
\[ \frac{dy}{dx} - y\cot x = \cos x \]
Step 3: Find I.F.
\[ IF = e^{-\int \cot x dx} = e^{-\ln(\sin x)} = \frac{1}{\sin x} \] \[ = \csc x \]
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