QJEEM22Ma1e41: JEE Main 2022 Mathematics - Class 12 JEEM22Ma1e41 Solution: Let the solution curve of the differential equation pass…

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

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

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