Question

If \( y = y(x) \) satisfies \[ \left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = -\cos x, \] such that \( y(0) = 2 \), then the value of \( y\left( \frac{\pi}{2} \right) \) is:

• A. \( 3 \)
• B. \( 4 \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

Always check if a differential equation allows variable separation. Use substitution when integrals involve compositions like \( \frac{\cos x}{2 + \sin x} \).

Answer 1

The Correct Option is D

We are given the differential equation: \[ \left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = -\cos x \] Step 1: Separate the variables Multiply both sides by \( 1 + y \): \[ \frac{dy}{dx} = -\cos x \cdot \frac{1 + y}{2 + \sin x} \] Rewriting and separating: \[ \frac{dy}{1 + y} = -\frac{\cos x}{2 + \sin x} dx \] Step 2: Integrate both sides Left-hand side: \[ \int \frac{dy}{1 + y} = \ln|1 + y| \] Right-hand side: Let \( u = 2 + \sin x \Rightarrow \frac{du}{dx} = \cos x \Rightarrow dx = \frac{du}{\cos x} \), but more cleanly: \[ \int \frac{-\cos x}{2 + \sin x} dx \] Let \( u = 2 + \sin x \Rightarrow du = \cos x \, dx \Rightarrow dx = \frac{du}{\cos x} \) So: \[ \int \frac{-\cos x}{u} dx = -\int \frac{1}{u} du = -\ln|u| = -\ln|2 + \sin x| \] Thus: \[ \ln|1 + y| = -\ln|2 + \sin x| + C \Rightarrow \ln|1 + y| + \ln|2 + \sin x| = C \Rightarrow \ln| (1 + y)(2 + \sin x) | = C \] Take exponentials: \[ (1 + y)(2 + \sin x) = A \quad \text{where \( A = e^C \)} \] Step 3: Apply initial condition \( y(0) = 2 \) At \( x = 0 \), \( \sin 0 = 0 \Rightarrow 2 + \sin 0 = 2 \) \[ (1 + 2)(2) = A \Rightarrow A = 6 \] So the general solution is: \[ (1 + y)(2 + \sin x) = 6 \Rightarrow 1 + y = \frac{6}{2 + \sin x} \Rightarrow y = \frac{6}{2 + \sin x} - 1 \] Step 4: Evaluate \( y\left( \frac{\pi}{2} \right) \) \[ \sin\left( \frac{\pi}{2} \right) = 1 \Rightarrow y = \frac{6}{2 + 1} - 1 = \frac{6}{3} - 1 = 2 - 1 = \boxed{1} \]