If y(x) satisfies the differential equation

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

subject to the condition y(π/2) = π/2, then y(π/6) is

The correct option/answer is $\frac{\pi }{3}$ (which corresponds to π/3).

Step-by-Step Explanation:

We are given the following first-order ordinary differential equation:
$$(\sin x)\frac{dy}{dx}+y\cos x=1$$
Observe the left-hand side of this differential equation. By applying the product rule of differentiation, $\frac{d}{dx}(u·v)=u\frac{dv}{dx}+v\frac{du}{dx}$, we can see that the left-hand side is the exact derivative of the product $y\sin x$:
$$\frac{d}{dx}(y\sin x)=(\sin x)\frac{dy}{dx}+y\cos x$$

Therefore, we can rewrite the differential equation as:
$$\frac{d}{dx}(y\sin x)=1$$

To solve for $y$, we integrate both sides with respect to $x$:
$$\int \frac{d}{dx}(y\sin x)dx=\int 1dx$$
This simplifies to:
$$y\sin x=x+C$$
where $C$ is the constant of integration.

Next, we determine the constant $C$ using the given boundary condition $y(\pi /2)=\pi /2$. This means that when $x=\frac{\pi }{2}$, we have $y=\frac{\pi }{2}$:
$$\left(\frac{\pi }{2}\right)\sin \left(\frac{\pi }{2}\right)=\frac{\pi }{2}+C$$
Since $\sin (\pi /2)=1$, this equation becomes:
$$\frac{\pi }{2}=\frac{\pi }{2}+C$$
Subtracting $\frac{\pi }{2}$ from both sides gives:
$$C=0$$

Substituting $C=0$ back into the general solution yields the particular solution:
$$y\sin x=x$$
Which can be solved explicitly for $y$ as:
$$y(x)=\frac{x}{\sin x}$$

Finally, we calculate $y(\pi /6)$ by substituting $x=\frac{\pi }{6}$:
$$y\left(\frac{\pi }{6}\right)=\frac{\pi /6}{\sin (\pi /6)}$$
Knowing that $\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$, we compute:
$$y\left(\frac{\pi }{6}\right)=\frac{\pi /6}{1/2}=\frac{\pi }{6}\times 2=\frac{\pi }{3}$$