JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 7)
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 :
$$
\frac{\pi}{4}
$$
$$
\frac{3\pi}{4}
$$
$$
\frac{\pi}{2}
$$
$$
\frac{3\pi}{2}
$$
Explanation
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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