Value of Parameter for Solution Curve of Differential Equation

To find the solution curve $y = y(x)$ of the given differential equation, we start by separating the variables:

$(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$

Rearranging the terms: $(1 + \sin x) \frac{dy}{dx} = -(y + 1) \cos x$

Separating the variables $x$ and $y$: $\frac{1}{y + 1} \, dy = -\frac{\cos x}{1 + \sin x} \, dx$

Integrating both sides with respect to their respective variables: $\int \frac{1}{y + 1} \, dy = -\int \frac{\cos x}{1 + \sin x} \, dx$

$\ln |y + 1| = -\ln |1 + \sin x| + C_1$

Combining the logarithmic terms gives: $\ln |(y + 1)(1 + \sin x)| = C_1$

Exponentiating both sides: $(y + 1)(1 + \sin x) = C$

We are given the initial condition $y(0) = 0$. Substituting $x = 0$ and $y = 0$ into the equation: $(0 + 1)(1 + \sin 0) = C \implies C = 1 \cdot (1 + 0) = 1$

Thus, the particular solution to the differential equation is: $(y + 1)(1 + \sin x) = 1$

Now, we are given that the curve passes through the point $\left( \alpha, -\frac{1}{2} \right)$. Substituting $x = \alpha$ and $y = -\frac{1}{2}$: $\left( -\frac{1}{2} + 1 \right) (1 + \sin \alpha) = 1$

$\frac{1}{2} (1 + \sin \alpha) = 1$

$1 + \sin \alpha = 2$

$\sin \alpha = 1$

Since $\sin \alpha = 1$, a possible value for $\alpha$ is: $\alpha = \frac{\pi}{2}$

This matches Option D.