<p>Let \(y = y(x)\) satisfies the differential equation \(y' = \ln(xy' - y)\). If \(y(1) = -1\) where \(y\) is twice differentiable and \(y''(x) \neq 0\), then \(y(e)\) equals:</p>
Step-by-Step Solution
Key Concept: Differentiate the given differential equation to create a separable ODE in terms of y', then use the initial condition to find the particular solution by recognizing that y' must be constant or follow a specific pattern.
<p><strong>Step 1:</strong> Start with y' = ln(xy' - y). Let p = y', so p = ln(xp - y).</p><p><strong>Step 2:</strong> Differentiate both sides with respect to x: p' = d/dx[ln(xp - y)] = (p + xp' - p)/(xp - y) = xp'/(xp - y).</p><p><strong>Step 3:</strong> From the original equation: xp - y = e^p, so p' = xp'/e^p, which gives e^p · p' = xp'.</p><p><strong>Step 4:</strong> Since y'' ≠ 0 (meaning p' ≠ 0), we can divide by p' to get e^p = x. Therefore y' = p = ln(x).</p><p><strong>Step 5:</strong> Integrate: y = ∫ln(x)dx = x·ln(x) - x + C.</p><p><strong>Step 6:</strong> Apply initial condition y(1) = -1: -1 = 1·ln(1) - 1 + C = 0 - 1 + C, so C = 0.</p><p><strong>Step 7:</strong> Thus y(x) = x·ln(x) - x, and y(e) = e·ln(e) - e = e·1 - e = 0.</p><p>∴ Answer: A (y(e) = 0)</p>
Correct Answer: A