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

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Aicharya · Accepted Answer

$\begin{aligned} & \left(\mathrm{e}^{\mathrm{y}}+1\right) \cos \mathrm{x} \mathrm{dx}+\mathrm{e}^{\mathrm{y}} \sin \mathrm{x} d \mathrm{y}=0 \\ & \Rightarrow \mathrm{d}\left(\left(\mathrm{e}^{\mathrm{y}}+1\right) \sin \mathrm{x}\right)=0 \\ & \left(\mathrm{e}^{\mathrm{y}}+1\right) \sin \mathrm{x}=\mathrm{C} \end{aligned}$
It passes through $\left(\frac{\pi}{2}, 0\right)$ $\Rightarrow \mathrm{c}=2$ Now, $x=\frac{\pi}{6}$ $\Rightarrow \mathrm{e}^{\mathrm{y}}=3$

If the solution $y(x)$ of the given differential equation $\left(\mathrm{e}^y+1\right) \cos x \mathrm{~d}…

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