Step 1: Understanding the Concept:
The given problem is a first-order ordinary differential equation (ODE).
In the realm of calculus, specifically differential equations, the first step is identifying the type of equation.
This equation is categorized as "Variable Separable."
The hallmark of a separable equation is that the terms involving the dependent variable \( y \) and the independent variable \( x \) can be moved to opposite sides of the equality sign.
Once the variables are separated, the problem reduces to finding antiderivatives for both sides.
Furthermore, we are provided with an initial condition \( y(0) = 2 \).
Initial conditions are vital because differential equations usually yield a general solution containing a constant \( C \).
This condition allows us to determine a unique, particular solution that describes a specific curve in the family of possible curves.
After finding this particular solution, we can evaluate the function at any point, in this case, \( x = \frac{\pi}{2} \).
Step 2: Key Formula or Approach:
For a separable equation of the form \( f(y) dy = g(x) dx \), the approach is:
1. Separate the variables so that \( y \) terms are with \( dy \) and \( x \) terms are with \( dx \).
2. Integrate both sides: \( \int f(y) dy = \int g(x) dx + C \).
3. Apply the initial condition \( y(x_0) = y_0 \) to find the constant \( C \).
4. Substitute the target value of \( x \) to find the corresponding \( y \).
Step 3: Detailed Explanation:
Starting with the given equation:
\[ \left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = -\cos x \]
To separate the variables, we multiply both sides by \( dx \) and rearrange the terms:
\[ \frac{1}{1 + y} dy = \frac{-\cos x}{2 + \sin x} dx \]
Now that the variables are separated, we integrate both sides:
\[ \int \frac{1}{1 + y} dy = -\int \frac{\cos x}{2 + \sin x} dx \]
The integral on the left is a standard logarithmic form:
\[ \int \frac{1}{1 + y} dy = \ln|1 + y| \]
For the right side, we observe that the numerator \( \cos x \) is the derivative of the denominator part \( \sin x \).
Let \( u = 2 + \sin x \). Then \( du = \cos x dx \).
Substituting these into the right-hand integral:
\[ -\int \frac{1}{u} du = -\ln|u| = -\ln|2 + \sin x| \]
Combining the results and adding the constant of integration \( C \):
\[ \ln|1 + y| = -\ln|2 + \sin x| + C \]
Rearrange to combine logarithmic terms:
\[ \ln|1 + y| + \ln|2 + \sin x| = C \]
Using the property \( \ln A + \ln B = \ln(AB) \):
\[ \ln|(1 + y)(2 + \sin x)| = C \]
Exponentiate both sides to eliminate the natural log:
\[ (1 + y)(2 + \sin x) = e^C = K \]
Where \( K \) is a positive constant.
Apply the initial condition \( y(0) = 2 \):
When \( x = 0 \), \( y = 2 \).
\[ (1 + 2)(2 + \sin 0) = K \]
\[ 3 \times (2 + 0) = K \implies K = 6 \]
The specific solution is:
\[ (1 + y)(2 + \sin x) = 6 \]
To find \( y\left(\frac{\pi}{2}\right) \), substitute \( x = \frac{\pi}{2} \) into the equation:
\[ (1 + y)\left(2 + \sin \frac{\pi}{2}\right) = 6 \]
Knowing \( \sin \frac{\pi}{2} = 1 \):
\[ (1 + y)(2 + 1) = 6 \]
\[ 3(1 + y) = 6 \implies 1 + y = 2 \implies y = 1 \]
Step 4: Final Answer:
The value of \( y\left(\frac{\pi}{2}\right) \) is determined to be 1.
This matches Option (D).