QJEEM24Maa7c4: JEE Main 2024 Mathematics - Class 12 JEEM24Maa7c4 Solution: If the solution of the given differential equation passes…

Given the differential equation

$\left(e^y + 1\right) \cos x \, dx + e^y \sin x \, dy = 0,$

we aim to find the value of $e^{y\left(\frac{\pi}{6}\right)}$ given that the solution $y(x)$ passes through the point $\left(\frac{\pi}{2}, 0\right)$ .

First, we recognize that the differential equation can be rearranged as:

$d\left(e^y \sin x\right) + \cos x \, dx = 0.$

Integrating this expression, we obtain:

$e^y \sin x + \sin x = C,$

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 $C$ :

$e^0 \sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = C \Rightarrow 1 + 1 = C \Rightarrow C = 2.$

Thus, the equation simplifies to:

$e^y \sin x + \sin x = 2.$

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:

$\left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \sin\left(\frac{\pi}{6}\right) = 2.$

Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ , the equation becomes:

$\left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \frac{1}{2} = 2 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} + 1 = 4 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} = 3.$

Therefore, the value of $e^{y\left(\frac{\pi}{6}\right)}$ is $3$ .