Question

Let \(𝑦=𝑦(𝑥)\) be the solution of the differential equation \(𝑥\;𝑙𝑜𝑔_𝑒⁡𝑥\frac{𝑑𝑦}{ 𝑑𝑥} +𝑦=𝑥^2𝑙𝑜𝑔_𝑒⁡𝑥,(𝑥>1)\).  If \(𝑦(2)=2\), then \(𝑦(𝑒)\) is equal to 

• A. \(\frac{2+e^2}{2}\)
• B. \(\frac{1+e^2}{2}\)
• C. \(\frac{1+e^2}{4}\)
• D. \(\frac{4+e^2}{4}\)

Answer 1

The Correct Option is D

(A)Rewrite the given differential equation: \[ \frac{dy}{dx} + \frac{1}{x \log_e x} y = x. \] (B)This is a linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x), \] where \( P(x) = \frac{1}{x \log_e x} \) and \( Q(x) = x \). (C)The integrating factor (I.F.) is given by: \[ \text{I.F.} = e^{\int P(x) dx} = e^{\int \frac{1}{x \log_e x} dx}. \] Substituting \( u = \log_e x \), \( du = \frac{1}{x} dx \), we get: \[ \int \frac{1}{x \log_e x} dx = \int \frac{1}{u} du = \log_e u = \log_e(\log_e x). \] Therefore: \[ \text{I.F.} = e^{\log_e(\log_e x)} = \log_e x. \] (D)Multiply through the differential equation by \( \log_e x \): \[ (\log_e x) \frac{dy}{dx} + \frac{y}{x} = x \log_e x. \] (E)Recognize the left-hand side as a derivative: \[ \frac{d}{dx}(y \log_e x) = x \log_e x. \] (F) Integrate both sides: \[ y \log_e x = \int x \log_e x \, dx. \] (G) Use integration by parts for \( \int x \log_e x \, dx \), letting \( u = \log_e x \) and \( dv = x dx \): \[ u = \log_e x, \, du = \frac{1}{x} dx, \, v = \frac{x^2}{2}. \] Then: \[ \int x \log_e x \, dx = \frac{x^2}{2} \log_e x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \log_e x - \frac{x^2}{4}. \] Substituting back: \[ y \log_e x = \frac{x^2}{2} \log_e x - \frac{x^2}{4} + C. \] Divide through by \( \log_e x \): \[ y = \frac{x^2}{2} - \frac{x^2}{4 \log_e x} + \frac{C}{\log_e x}. \] Use the initial condition \( y(2) = 2 \) to find \( C \). When \( x = 2 \), \( \log_e 2 \neq 0 \): \[ 2 = \frac{4}{2} - \frac{4}{4 \log_e 2} + \frac{C}{\log_e 2}. \] Simplify: \[ 2 = 2 - \frac{1}{\log_e 2} + \frac{C}{\log_e 2}. \] Solving for \( C \): \[ C = 1. \] The solution becomes: \[ y = \frac{x^2}{2} - \frac{x^2}{4 \log_e x} + \frac{1}{\log_e x}. \]  (K) Evaluate \( y(e) \): \[ y(e) = \frac{e^2}{2} - \frac{e^2}{4 \log_e e} + \frac{1}{\log_e e}. \] Since \( \log_e e = 1 \): \[ y(e) = \frac{e^2}{2} - \frac{e^2}{4} + 1 = \frac{2e^2}{4} - \frac{e^2}{4} + 1 = \frac{e^2}{4} + 1. \] Simplify: \[ y(e) = \frac{4 + e^2}{4}. \]