Let the solution curve y = y x of the differential equation 1 + e 2 x d y d x + y = 1 pass through the point…

Given,

$\left(1+e^{2x}\right)\left(\frac{dy}{dx}+y\right)=1$

Now on rearranging we get,

$\Rightarrow \frac{dy}{dx}+y=\frac{1}{1+e^{2x}}$

We can see it is a linear differential equation,

So integrating factor is $e^{\int 1·dx}=e^x$

So solution will be

$y·IF=\int \frac{1}{1+e^{2x}}\times IF dx$

$\Rightarrow y{e}^x=\int \frac{1}{1+e^{2x}}\times e^x dx$

 $\Rightarrow y·e^x={\tan }^{-1}\left(e^x\right)+c$

Now as curve is passing through $\left(0,\frac{\pi }{2}\right)$ so 

$\Rightarrow c=\frac{\pi }{4}$

Now calculating the limit $\underset{x\to \infty }{\lim }\left(y·e^x\right)$ we get,

$\Rightarrow \underset{x\to \infty }{\lim }\left(y·e^x\right)=\underset{x\to \infty }{\lim }\left({\tan }^{-1}\left(e^x\right)+\frac{\pi }{4}\right)=\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{4}=\frac{3\pi }{4}$