Question:medium

If \( y = y(x) \) and \[ \left( \frac{2+\sin x}{y+1} \right)\frac{dy}{dx} = -\cos x, \] with the initial condition \( y(0) = 1 \), then find \[ y\left(\frac{\pi}{2}\right) = \, ? \] 

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For equations of the form: \[ f(x)\frac{dy}{dx}=g(y)h(x) \] try separating variables first. It usually makes the integration much easier.
Updated On: May 14, 2026
  • \(\frac{1}{3}\)
  • \(\frac{2}{3}\)
  • \(-\frac{1}{3}\)
  • \(1\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
The given problem is a first-order differential equation. We observe that the variables \(x\) and \(y\) can be separated easily to find the general solution.
Step 2: Key Formula or Approach:
We use the method of separation of variables:
\[ \int f(y) dy = \int g(x) dx \] Integration of \(\frac{1}{y+1}\) is \(\ln(y+1)\) and the substitution method is used for the \(x\) integral.
Step 3: Detailed Explanation:
Given equation: \(\left( \frac{2+\sin x}{y+1} \right) \frac{dy}{dx} = -\cos x\)
Separating the variables:
\[ \frac{dy}{y+1} = \frac{-\cos x}{2+\sin x} dx \] Integrating both sides:
\[ \int \frac{1}{y+1} dy = -\int \frac{\cos x}{2+\sin x} dx \] Let \(2+\sin x = t \implies \cos x dx = dt\).
\[ \ln(y+1) = -\ln(2+\sin x) + \ln C \] \[ \ln(y+1) + \ln(2+\sin x) = \ln C \implies (y+1)(2+\sin x) = C \] Using \(y(0) = 1\):
\[ (1+1)(2+\sin 0) = C \implies 2 \times 2 = C \implies C = 4 \] So, the equation is \((y+1)(2+\sin x) = 4\).
Now, at \(x = \frac{\pi}{2}\):
\[ (y+1)(2 + \sin \frac{\pi}{2}) = 4 \implies (y+1)(2+1) = 4 \] \[ 3(y+1) = 4 \implies y+1 = \frac{4}{3} \implies y = \frac{1}{3} \] Step 4: Final Answer:
The value of \(y(\pi/2)\) is \(\frac{1}{3}\).
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