Question

Let the solution curve \(y = y(x)\) of the differential equation \((1 + e^{2x})\left(\frac{dy}{dx} + y\right) = 1\) pass through the point \((0, \frac{\pi}{2})\). Then, \(\lim_{{x \to \infty}} e^x y(x)\) is equal to

• A. \(\frac{\pi}{4}\)
• B. \(\frac{3\pi}{4}\)
• C. \(\frac{\pi}{2}\)
• D. \(\frac{3\pi}{2}\)

Answer 1

The Correct Option is B

\(D.E (1 + e^{2x})\frac{dy}{dx} + y = 1\)
\(⇒\) \(\frac{dy}{dx} + y = \frac{1}{1+e^{2x}}\)

\(\text{I.F.} = e^{\int 1 \,dx} = e^x\)

\(∴\)\(e^x y(x) = \int \frac{e^x}{1 + e^{2x}} \,dx\)

\(⇒\)\(e^x y(x) = \tan^{-1}(e^x) + C\)
\(∵\) It passes through
\((0, \frac{\pi}{2}), \quad C = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}\)

\(∴\) \(\lim_{{x \to \infty}} e^x y(x) = \lim_{{x \to \infty}} \tan^{-1}(e^x) + \frac{\pi}{4}\)
\(= \frac{3\pi}{4}\)
So, the correct option is (B): \(\frac{3\pi}{4}\)