Question

For a ∈ ℝ, let ya(x) be the solution of the differential equation
\(\frac{dy}{dx}+2y=\frac{1}{1+x^2}\) for x ∈ \(\R\)
satisfying y(0) = a. Then, which one of the following is TRUE ?

• A. \(\lim\limits_{x\rightarrow \infin}y_a(x)=0\) for every a ∈ \(\R\)
• B. \(\lim\limits_{x\rightarrow \infin}y_a(x)=1\) for every a ∈ \(\R\)
• C. There exists an a ∈ ℝ such that \(\lim\limits_{x \rightarrow \infin}y_a(x)\) exists but its value is different from 0 and 1
• D. There exists an a ∈ ℝ for which \(\lim \limits_{x \rightarrow \infin}y_a(x)\) does not exist

Answer 1

The Correct Option is A

We are given the differential equation: \[ \frac{dy}{dx} + 2y = \frac{1}{1 + x^2}. \] This is a first-order linear differential equation, and we can solve it using the integrating factor method. The integrating factor is: \[ \mu(x) = e^{\int 2 \, dx} = e^{2x}. \] Multiplying both sides of the differential equation by the integrating factor \( e^{2x} \), we get: \[ e^{2x} \frac{dy}{dx} + 2e^{2x} y = \frac{e^{2x}}{1 + x^2}. \] The left-hand side is now the derivative of \( e^{2x} y \), so we can integrate both sides: \[ \frac{d}{dx} \left( e^{2x} y \right) = \frac{e^{2x}}{1 + x^2}. \] Integrating the right-hand side with respect to \( x \), we get: \[ e^{2x} y = \int \frac{e^{2x}}{1 + x^2} dx + C, \] where \( C \) is the constant of integration. As \( x \to \infty \), the term \( \frac{e^{2x}}{1 + x^2} \) approaches 0, because the exponential growth in the numerator is not enough to overcome the growth of \( x^2 \) in the denominator. Therefore, for large \( x \), \( y_{\alpha}(x) \to 0 \) regardless of the initial condition \( \alpha \). Thus, the correct answer is (A): \( \lim_{x \to \infty} y_{\alpha}(x) = 0 \text{ for every } \alpha \in \mathbb{R} \).