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(B) Given differential equation is $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$. This can be rewritten as $d((e^y+1) \sin x) = 0$. Integrating both s

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$3$

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(B) Given differential equation is $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$. This can be rewritten as $d((e^y+1) \sin x) = 0$. Integrating both sides,we get $(e^y+1) \sin x = C$. Since the solution passes through the point $(\frac{\pi}{2}, 0)$,we substitute $x = \frac{\pi}{2}$ and $y = 0$: $(e^0+1) \sin(\frac{\pi}{2}) = C \Rightarrow (1+1)(1) = C \Rightarrow C = 2$. Thus,the equation of the curve is $(e^y+1) \sin x = 2$. Now,we need to find the value of $e^{y(\frac{\pi}{6})}$. Substitute $x = \frac{\pi}{6}$ into the equation: $(e^y+1) \sin(\frac{\pi}{6}) = 2$. Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$,we have $(e^y+1) \cdot \frac{1}{2} = 2$. $e^y+1 = 4$. $e^y = 3$. Therefore,the value of $e^{y(\frac{\pi}{6})}$ is $3$.

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 $(\frac{\pi}{2}, 0)$,then the value of $e^{y(\frac{\pi}{6})}$ is equal to ...........

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