Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12

Question:

If the solution of the differential equation $\frac{dy}{dx} = \frac{1}{x \cos y + \sin 2y}$ is $y = ce^{\sin y} - k(1 + \sin y)$, then the value of $k$ is ______.

Step-by-Step Solution

Key Concept: Interchange variables to treat $x$ as a function of $y$, converting the equation to a linear form in $x$.
Given $\frac{dy}{dx} = \frac{1}{x\cos y + 2\sin y \cos y}$, rewrite as $\frac{dx}{dy} = x\cos y + 2\sin y \cos y = x\cos y + \sin 2y$. This is linear in $x$ with integrating factor $e^{\int \cos y \, dy} = e^{\sin y}$. The solution is $xe^{-\sin y} = 2\int e^{-\sin y} \sin y \cos y \, dy = -2\sin y e^{-\sin y} + 2\int (-e^{-\sin y}) \cos y \, dy$.
Correct Answer: 2

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