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

We need to solve $$(e^y + 1)\cos x\, dx + e^y \sin x\, dy = 0$$ with the condition that it passes through $$\left(\frac{\pi}{2}, 0\right)$$, and find $$e^{y(\pi/6)}$$.

Consider the function $$F(x, y) = (e^y + 1)\sin x$$ whose total differential is

$$dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy = (e^y + 1)\cos x\, dx + e^y \sin x\, dy$$,

exactly matching the given equation. Hence

$$d\left[(e^y + 1)\sin x\right] = 0\,. $$

Integration yields

$$(e^y + 1)\sin x = C\,. $$

Applying the condition at $$\left(\frac{\pi}{2}, 0\right)$$ gives

$$(e^0 + 1)\sin\frac{\pi}{2} = C \implies (1 + 1)\cdot 1 = C \implies C = 2\,. $$

At $$x = \frac{\pi}{6}$$, the equation becomes

$$(e^y + 1)\sin\frac{\pi}{6} = 2 \implies (e^y + 1)\cdot\frac{1}{2} = 2 \implies e^y + 1 = 4 \implies e^y = 3\,. $$

Therefore, $$e^{y(\pi/6)} = 3\,. $$