"Let the solution curve $y=y(x)$ of the differential equation $$\left(1+\mathrm{e}^{2 x}\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}+y\right)=1$$ pass through the point $\left(0, \frac{\pi}{2}\right)$. Then, $\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)$ is equal to :"

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Accepted Answer

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D.E. $(1 + {e^{2x}})\left( {{{dy} \over {dx}} + y} \right) = 1$

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

I.F. $= {e^{\int {1\,.\,dx} }} = {e^x}$

$\therefore$ Solution

${e^x}y(x) = \int {{{{e^x}} \over {1 + {e^{2x}}}}dx}$

$\Rightarrow {e^x}y(x) = {\tan ^{ - 1}}({e^x}) + C$

$\because$ It passes through $$\left( {0,{\pi \over 2}} \right),\,C = {\pi \over 2} - {\pi \over 4} = {\pi \over 4}$$

$\therefore$ $$\mathop {\lim }\limits_{x \to \infty } {e^x}y(x) = \mathop {\lim }\limits_{x \to \infty } {\tan ^{ - 1}}({e^x}) + {\pi \over 4}$$

$= {{3\pi } \over 4}$

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Let the solution curve $y=y(x)$ of the differential equation $$\left(1+\mathrm{e}^{2 x}\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}+y\right)=1$$ pass through the point $\left(0, \frac{\pi}{2}\right)$. Then, $\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)$ is equal to :

Solution

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Let the solution curve $y=y(x)$ of the differential equation $$\left(1+\mathrm{e}^{2 x}\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}+y\right)=1$$ pass through the point $\left(0, \frac{\pi}{2}\right)$. Then, $\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)$ is equal to :

Solution

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