JEE Main 2024 — Differential Equations Question with Solution

Given the differential equation

$$\left(e^y+1\right)\cos xdx+e^y\sin xdy=0,$$

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:

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

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 $3$.