If the solution $y(x)$ of the given differential equation $\left(e^y+1\right) \cos x \mathrm{~d} x+\mathrm{e}^y \sin x \mathrm{~d} y=0$ passes through the point $\left(\frac{\pi}{2}, 0\right)$, then the value of $e^{y\left(\frac{\pi}{6}\right)}$ is equal to _________.

<p>Given the differential equation</p> <p>$\left(e^y + 1\right) \cos x \, dx + e^y \sin x \, dy = 0,$</p> <p>we aim to find the value of $ e^{y\left(\frac{\pi}{6}\right)} $ given that the solution $ y(x) $ passes through the point $\left(\frac{\pi}{2}, 0\right)$.</p> <p>First, we recognize that the differential equation can be rearranged as:</p> <p>$d\left(e^y \sin x\right) + \cos x \, dx = 0.$</p> <p>Integrating this expression, we obtain:</p> <p>$e^y \sin x + \sin x = C,$</p> <p>where $ C $ is a constant. Given that the solution passes through the point $\left(\frac{\pi}{2}, 0\right)$, we substitute these values into the equation to find $ C $:</p> <p>$$ e^0 \sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = C \Rightarrow 1 + 1 = C \Rightarrow C = 2.$$</p> <p>Thus, the equation simplifies to:</p> <p>$e^y \sin x + \sin x = 2.$</p> <p>We now need to determine the value of $ e^{y\left(\frac{\pi}{6}\right)} $. Substituting $ x = \frac{\pi}{6} $ into the equation, we get:</p> <p>$$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \sin\left(\frac{\pi}{6}\right) = 2.$$</p> <p>Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, the equation becomes:</p> <p>$$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \frac{1}{2} = 2 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} + 1 = 4 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} = 3.$$</p> <p>Therefore, the value of $ e^{y\left(\frac{\pi}{6}\right)} $ is $ 3 $.</p>