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 $y\left(\frac{\pi}{2}\right)$ is equal to

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

Use substitution when denominator has form \(a + \sin x\).

Answer 1

The Correct Option is D

Concept:
This is a separable differential equation. Step 1: Rearrange equation. \[ \frac{2+\sin x}{1+y} \frac{dy}{dx} = -\cos x \] \[ \frac{dy}{1+y} = -\frac{\cos x}{2+\sin x} dx \]
Step 2: Integrate both sides. LHS: \[ \int \frac{dy}{1+y} = \ln(1+y) \] RHS: Let \(u = 2+\sin x \Rightarrow du = \cos x dx\) \[ \int \frac{-\cos x}{2+\sin x} dx = -\int \frac{du}{u} = -\ln(2+\sin x) \]
Step 3: Combine results. \[ \ln(1+y) = -\ln(2+\sin x) + C \] \[ \ln[(1+y)(2+\sin x)] = C \]
Step 4: Apply initial condition \(y(0)=2\). \[ (1+2)(2+\sin 0) = 3 \times 2 = 6 \] So, \[ (1+y)(2+\sin x) = 6 \]
Step 5: Find \(y\left(\frac{\pi}{2}\right)\). \[ (1+y)(2+\sin \frac{\pi}{2}) = (1+y)(3) = 6 \] \[ 1+y = 2 \Rightarrow y = 1 \]
Step 6: Conclusion. \[ y\left(\frac{\pi}{2}\right) = 1 \]