The particular solution of the differential equation $\frac{d y}{d x}=e^{2 y} \cos x$, when $y\left(\frac{\pi}{6}\right)=0$ is

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$\begin{aligned} & \frac{d y}{d x}=e^{2 y} \cos x \Rightarrow \int e^{2 y} d y=\int \cos d x \ & \Rightarrow \frac{e^{2 y}}{-2}=\sin x+c\end{aligned}$ Putting $\mathrm{x}=\frac{\pi}{6}$ and $\mathrm{y}=0$ we get $\mathrm{c}=-1$ $\begin{aligned} & \Rightarrow \frac{e^{-2 y}}{-2}=\sin x-1 \ & \Rightarrow e^{-2 y}=-2 \sin x+2 \ & \Rightarrow 2 \sin x+e^{-2 y}-2=0\end{aligned}$

The particular solution of the differential equation $\frac{d y}{d x}=e^{2 y} \cos x$, when…

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