Question:
The solution of the differential equation \( \frac{dy}{dx} - \sin x \sin y = 0 \) is
The solution of the differential equation \( \frac{dy}{dx} - \sin x \sin y = 0 \) is
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For separable differential equations, separate the variables, integrate both sides, and use standard integrals to solve.
Updated On: Apr 22, 2026
- \( e^{\cos x} \cdot \tan \frac{y}{2} = C \)
- \( e^{\cos x} \cdot \tan y = C \)
- \( \cos x \cdot \tan y = C \)
- \( \cos x \cdot \sin y = C \)
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The Correct Option is A
Solution and Explanation
Step 1: Separate variables.
We are given the differential equation: \[ \frac{dy}{dx} - \sin x \sin y = 0 \] Rearrange the equation to separate the variables: \[ \frac{dy}{dx} = \sin x \sin y \] This can be written as: \[ \frac{dy}{\sin y} = \sin x \, dx \]
Step 2: Integrate both sides.
Now, integrate both sides of the equation: \[ \int \frac{dy}{\sin y} = \int \sin x \, dx \]
Step 3: Use standard integrals.
The integral of \( \frac{1}{\sin y} \) is \( \ln | \tan \frac{y}{2} | \), and the integral of \( \sin x \) is \( -\cos x \). Thus, we have: \[ \ln \left| \tan \frac{y}{2} \right| = -\cos x + C \]
Step 4: Simplify the result.
Exponentiating both sides to eliminate the logarithm: \[ \left| \tan \frac{y}{2} \right| = e^{-\cos x + C} \] Since \( e^C \) is just a constant, we can write: \[ \tan \frac{y}{2} = C e^{\cos x} \]
Step 5: Conclusion.
Thus, the solution to the differential equation is: \[ e^{\cos x} \cdot \tan \frac{y}{2} = C \] corresponding to option (A).
We are given the differential equation: \[ \frac{dy}{dx} - \sin x \sin y = 0 \] Rearrange the equation to separate the variables: \[ \frac{dy}{dx} = \sin x \sin y \] This can be written as: \[ \frac{dy}{\sin y} = \sin x \, dx \]
Step 2: Integrate both sides.
Now, integrate both sides of the equation: \[ \int \frac{dy}{\sin y} = \int \sin x \, dx \]
Step 3: Use standard integrals.
The integral of \( \frac{1}{\sin y} \) is \( \ln | \tan \frac{y}{2} | \), and the integral of \( \sin x \) is \( -\cos x \). Thus, we have: \[ \ln \left| \tan \frac{y}{2} \right| = -\cos x + C \]
Step 4: Simplify the result.
Exponentiating both sides to eliminate the logarithm: \[ \left| \tan \frac{y}{2} \right| = e^{-\cos x + C} \] Since \( e^C \) is just a constant, we can write: \[ \tan \frac{y}{2} = C e^{\cos x} \]
Step 5: Conclusion.
Thus, the solution to the differential equation is: \[ e^{\cos x} \cdot \tan \frac{y}{2} = C \] corresponding to option (A).
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