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

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Differential Equations

Let the solution curve

y = y (x)

of the differential equation

(1 + e^{2 x}) (\frac{d y}{d x} + y) = 1

pass through the point

(0, \frac{π}{2})

. Then,

lim_{x \to \infty} e^{x} y (x)

is equal to :

Answer: (B)

Rewriting the differential equation:

\frac{d y}{d x} + y = \frac{1}{1 + e ^{2 x}}

Integrating factor:

e^{x}

Solution:

y \cdot e^{x} = \int \frac{e ^{x}}{1 + e ^{2 x}} d x = tan^{- 1} (e^{x}) + c

Using initial condition

(0, \frac{π}{2})

\frac{π}{2} \cdot 1 = tan^{- 1} (1) + c = \frac{π}{4} + c

c = \frac{π}{4}

Therefore:

lim_{x \to \infty} e^{x} y (x) = lim_{x \to \infty} (tan^{- 1} (e^{x}) + \frac{π}{4}) = \frac{π}{2} + \frac{π}{4} = \frac{3 π}{4}

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option