Let the solution curve y=y(x) of the differential equation [(1+e^2 x)(d y~d x+y)=1] pass t…

Question

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 :

Accepted Answer

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

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

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

Solutions · 1