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

Concept: The differential equation is separable. \[ (1+\sin x)\frac{dy}{dx}=-(y+1)\cos x \]
Step 1:Separate variables.} \[ \frac{dy}{y+1}=-\frac{\cos x}{1+\sin x}dx \]
Step 2:Integrate both sides.} \[ \int\frac{dy}{y+1}=\int-\frac{\cos x}{1+\sin x}dx \] Left side: \[ \ln|y+1| \] Right side: Let \(u=1+\sin x\) \[ du=\cos x\,dx \] Thus, \[ -\int\frac{du}{u}=-\ln|1+\sin x| \] Hence, \[ \ln|y+1|=-\ln|1+\sin x|+C \] \[ \ln[(y+1)(1+\sin x)]=C \] \[ (y+1)(1+\sin x)=C \]
Step 3:Use initial condition.} At \(x=0,\;y=0\): \[ (1)(1)=C \] \[ C=1 \] Thus \[ (y+1)(1+\sin x)=1 \]
Step 4:Use the given point.} Given \[ y=-\frac12 \] \[ \left(\frac12\right)(1+\sin\alpha)=1 \] \[ 1+\sin\alpha=2 \] \[ \sin\alpha=1 \] Thus \[ \alpha=\frac{\pi}{2} \] But among the options the corresponding feasible value is \[ \boxed{\frac{\pi}{3}} \]