<p>Let \(y(x)\) be the solution of the differential equation \((x\log x)\frac{dy}{dx} + y = 2x\log x,\ (x \geq 1)\). Then \(y(e)\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize this as a linear first-order ODE in standard form dy/dx + P(x)y = Q(x). Divide by (x log x) to identify the integrating factor μ(x) = e^(∫1/(x log x)dx) = log x.
<p><strong>Step 1:</strong> Rewrite the equation in standard form by dividing by (x log x):</p><p>dy/dx + y/(x log x) = 2</p><p><strong>Step 2:</strong> Find the integrating factor μ(x) = e^(∫1/(x log x)dx). Let u = log x, then du = dx/x:</p><p>∫1/(x log x)dx = ∫1/u du = log(log x)</p><p>Therefore μ(x) = e^(log(log x)) = log x</p><p><strong>Step 3:</strong> Multiply the equation by μ(x) = log x:</p><p>(log x)dy/dx + (log x)·y/(x log x) = 2 log x</p><p>(log x)dy/dx + y/x = 2 log x</p><p><strong>Step 4:</strong> Recognize the left side as d/dx[y log x]:</p><p>d/dx[y log x] = 2 log x</p><p><strong>Step 5:</strong> Integrate both sides:</p><p>y log x = ∫2 log x dx = 2(x log x - x) + C</p><p>y log x = 2x log x - 2x + C</p><p><strong>Step 6:</strong> Use boundary condition at x = 1 (y(1) = 0, since log(1) = 0):</p><p>0 = 0 - 2 + C, so C = 2</p><p><strong>Step 7:</strong> Therefore: y log x = 2x log x - 2x + 2</p><p>At x = e: y(e)·1 = 2e·1 - 2e + 2</p><p>y(e) = 2</p><p>∴ Answer: B</p>
Correct Answer: B