Question

Let \(y=y(x)\) be the solution curve of the differential equation} \[ (1+\sin x)\frac{dy}{dx}+(y+1)\cos x=0,\qquad y(0)=0. \] If the curve passes through the point \( \left(\alpha,-\frac12\right) \), then a value of \( \alpha \) is:

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{2} \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a variable separable differential equation. We rearrange terms to separate \(y\) and \(x\) and then integrate to find the general solution.
Step 2: Key Formula or Approach:
1. Separate variables: \(\frac{1}{y+1} dy = -\frac{\cos x}{1 + \sin x} dx\).
2. Integrate using \(\int \frac{f'(x)}{f(x)} dx = \ln|f(x)|\).
Step 3: Detailed Explanation:
The DE is: \((1 + \sin x) \frac{dy}{dx} = -(y+1) \cos x\).
Separating variables:
\[ \frac{dy}{y+1} = -\frac{\cos x}{1 + \sin x} dx \]
Integrating:
\[ \ln |y+1| = -\ln |1 + \sin x| + \ln C \]
\[ \ln |y+1| + \ln |1 + \sin x| = \ln C \]
\[ (y+1)(1 + \sin x) = C \]
Given \(y(0) = 0\):
\[ (0+1)(1 + \sin 0) = C \implies 1(1) = C \implies C = 1 \].
Equation of curve: \((y+1)(1 + \sin x) = 1\).
Substituting the point \((\alpha, -1/2)\):
\[ (-1/2 + 1)(1 + \sin \alpha) = 1 \]
\[ \frac{1}{2}(1 + \sin \alpha) = 1 \implies 1 + \sin \alpha = 2 \implies \sin \alpha = 1 \].
So, \(\alpha = \pi/2\).
Step 4: Final Answer:
A value of \(\alpha\) is \(\pi/2\).