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

To solve the given differential equation and find the limit, we must first analyze the given information:

The differential equation is \((1 + e^{2x})\left(\frac{dy}{dx} + y\right) = 1\), and it passes through the point \((0, \frac{\pi}{2})\). We are tasked to find \(\lim_{{x \to \infty}} e^x y(x)\). 

1. Start by simplifying the differential equation. We have: \((1 + e^{2x})\left(\frac{dy}{dx} + y\right) = 1\)
2. Expanding and rearranging the terms, we get: \(\frac{dy}{dx} + y = \frac{1}{1 + e^{2x}}\)
3. This is a linear first-order differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = 1\) and \(Q(x) = \frac{1}{1 + e^{2x}}\).
4. To solve, we find the integrating factor \(e^{\int P(x) dx} = e^{\int 1 dx} = e^x\).
5. Multiply the entire equation by the integrating factor: \(e^x \frac{dy}{dx} + e^x y = \frac{e^x}{1 + e^{2x}}\)
6. The left-hand side is the derivative of \(e^x y\), which gives: \(\frac{d}{dx}(e^x y) = \frac{e^x}{1 + e^{2x}}\)
7. Integrating both sides with respect to \(x\), we have: \(e^x y = \int \frac{e^x}{1 + e^{2x}} dx + C\)
8. Let's solve the integral: \(\int \frac{e^x}{1 + e^{2x}} dx\)

Using substitution, let \(u = e^x\), then \(du = e^x dx\) or \(dx = \frac{du}{u}\), and \(\int \frac{du}{u(1 + u^2)} = \arctan(u) + C_1\)

1. Thus, the solution becomes: \(e^x y = \arctan(e^x) + C\)
2. Apply the initial condition \((0, \frac{\pi}{2})\): \(e^0 \cdot \frac{\pi}{2} = \arctan(1) + C\) \(\frac{\pi}{2} = \frac{\pi}{4} + C\) \(C = \frac{\pi}{4}\)
3. The particular solution is: \(e^x y = \arctan(e^x) + \frac{\pi}{4}\)
4. To find the limit, evaluate: \(\lim_{{x \to \infty}} e^x y = \lim_{{x \to \infty}} [\arctan(e^x) + \frac{\pi}{4}]\)
5. As \(x \to \infty\), \(e^x \to \infty\), and \(\arctan(e^x) \to \frac{\pi}{2}\).
6. Thus, the limit becomes: \(\frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}\).

Therefore, the correct answer is \(\frac{3\pi}{4}\).

Answer 2

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

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

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

\(⇒\)\(e^x y(x) = \tan^{-1}(e^x) + C\)
\(∵\) 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): \(\frac{3\pi}{4}\)