If $$(2 + \sin x)\frac{dy}{dx} + (y+1)\cos x = 0$$ and $$y(0) = 1$$, then $$y\left(\frac{\pi}{2}\right)$$ is equal to

We have the differential equation

$$ (2+\sin x)\,\frac{dy}{dx} + (y+1)\cos x = 0. $$

First we isolate $$\dfrac{dy}{dx}$$. Dividing both sides by $$(2+\sin x)$$ gives

$$ \frac{dy}{dx} = -\frac{(y+1)\cos x}{\,2+\sin x\,}. $$

Now we separate the variables. Bringing all terms that involve $$y$$ to the left and all terms that involve $$x$$ to the right, we get

$$ \frac{dy}{y+1} = -\frac{\cos x}{\,2+\sin x\,}\,dx. $$

The equation is now in completely separable form, so we integrate both sides:

$$ \int \frac{dy}{y+1} = \int -\frac{\cos x}{\,2+\sin x\,}\,dx. $$

On the left we use the standard result $$\displaystyle\int \frac{du}{u} = \ln|u| + C$$ with $$u = y+1$$. Thus

$$ \int \frac{dy}{y+1} = \ln|y+1| + C_1. $$

On the right we make the substitution $$u = 2+\sin x$$, so $$du = \cos x\,dx$$. Then

$$ \int -\frac{\cos x}{\,2+\sin x\,}\,dx = -\int \frac{du}{u} = -\ln|u| + C_2 = -\ln|2+\sin x| + C_2. $$

Grouping the arbitrary constants into a single constant $$C$$, we can write

$$ \ln|y+1| = -\ln|2+\sin x| + C. $$

Adding $$\ln|2+\sin x|$$ to both sides gives

$$ \ln|y+1| + \ln|2+\sin x| = C. $$

Using the property $$\ln a + \ln b = \ln(ab)$$, this becomes

$$ \ln\!\bigl|(y+1)(2+\sin x)\bigr| = C. $$

Exponentiating both sides removes the logarithm:

$$ (y+1)(2+\sin x) = K, $$

where $$K = e^{\,C}$$ is a non-zero constant.

We now apply the initial condition $$y(0) = 1$$. At $$x = 0$$ we have $$\sin 0 = 0$$, so

$$ (1+1)\,\bigl(2 + 0\bigr) = K \quad\Longrightarrow\quad 2 \times 2 = K \quad\Longrightarrow\quad K = 4. $$

Thus the relation between $$y$$ and $$x$$ is

$$ (y+1)(2+\sin x) = 4. $$

Solving for $$y$$ gives

$$ y + 1 = \frac{4}{\,2+\sin x\,}, \qquad\text{so}\qquad y = \frac{4}{\,2+\sin x\,} - 1. $$

We are asked to find $$y\!\left(\dfrac{\pi}{2}\right)$$. Substituting $$x = \dfrac{\pi}{2}$$, we note that $$\sin\!\left(\dfrac{\pi}{2}\right) = 1$$. Hence

$$ y\!\left(\frac{\pi}{2}\right) = \frac{4}{\,2 + 1\,} - 1 = \frac{4}{3} - 1 = \frac{1}{3}. $$

Hence, the correct answer is Option A.