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(B) The given differential equation is $(1+e^{2x})(\frac{dy}{dx}+y)=1$,which can be rewritten as $\frac{dy}{dx}+y=\frac{1}{1+e^{2x}}$.This is a linear

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$\frac{3\pi}{4}$

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(B) The given differential equation is $(1+e^{2x})(\frac{dy}{dx}+y)=1$,which can be rewritten as $\frac{dy}{dx}+y=\frac{1}{1+e^{2x}}$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=1$ and $Q=\frac{1}{1+e^{2x}}$.The integrating factor is $I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x$.The general solution is $y \cdot e^x = \int Q \cdot I.F. dx + c = \int \frac{e^x}{1+e^{2x}} dx + c$.Let $u=e^x$,then $du=e^x dx$. The integral becomes $\int \frac{1}{1+u^2} du = 	an^{-1}(u) = 	an^{-1}(e^x)$.So,$y \cdot e^x = 	an^{-1}(e^x) + c$.Since the curve passes through $(0, \frac{\pi}{2})$,we have $\frac{\pi}{2} \cdot e^0 = 	an^{-1}(e^0) + c$,which gives $\frac{\pi}{2} = 	an^{-1}(1) + c = \frac{\pi}{4} + c$.Thus,$c = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.The solution is $y \cdot e^x = 	an^{-1}(e^x) + \frac{\pi}{4}$.Taking the limit as $x ightarrow \infty$,$\lim_{x ightarrow \infty} (y \cdot e^x) = \lim_{x ightarrow \infty} (	an^{-1}(e^x) + \frac{\pi}{4}) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.

Let the solution curve $y=y(x)$ of the differential equation $(1+e^{2x})(\frac{dy}{dx}+y)=1$ pass through the point $(0, \frac{\pi}{2})$. Then,$\lim_{x \rightarrow \infty} e^{x} y(x)$ is equal to.

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