Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>Let \(y = y(x)\) be the solution of the differential equation, \(x\frac{dy}{dx} + y = x\log_e x,\ (x > 1)\). If \(2y(2) = \log_e 4 - 1\), then \(y(e)\) is equal to:</p>
<p>\(-\dfrac{e}{2}\)</p>
<p>\(-\dfrac{e^2}{2}\)</p>
<p>\(\dfrac{e}{4}\)</p>
<p>\(\dfrac{e^2}{4}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a first-order linear ODE of the form dy/dx + P(x)y = Q(x). Use integrating factor μ(x) = e^(∫P(x)dx) to convert the left side into a perfect derivative d/dx[μ(x)y].
<p><strong>Step 1:</strong> Rewrite the equation in standard form by dividing by x:</p><p>dy/dx + (y/x) = log_e x</p><p><strong>Step 2:</strong> Identify integrating factor μ(x) = e^(∫1/x dx) = e^(log_e x) = x</p><p><strong>Step 3:</strong> Multiply both sides by μ(x) = x:</p><p>x(dy/dx) + y = x·log_e x</p><p>This gives: d/dx(xy) = x·log_e x</p><p><strong>Step 4:</strong> Integrate both sides:</p><p>xy = ∫ x·log_e x dx</p><p>Using integration by parts with u = log_e x, dv = x dx:</p><p>xy = (x²/2)·log_e x - ∫(x²/2)·(1/x) dx</p><p>xy = (x²/2)·log_e x - x²/4 + C</p><p><strong>Step 5:</strong> Apply initial condition 2y(2) = log_e 4 - 1, so y(2) = (log_e 4 - 1)/2:</p><p>2·(log_e 4 - 1)/2 = (4/2)·log_e 2 - 4/4 + C</p><p>log_e 4 - 1 = 2·log_e 2 - 1 + C</p><p>Since log_e 4 = 2·log_e 2: C = 0</p><p><strong>Step 6:</strong> Find y(e):</p><p>e·y(e) = (e²/2)·log_e e - e²/4</p><p>e·y(e) = e²/2 - e²/4 = e²/4</p><p>y(e) = e/4</p><p>∴ Answer: e/4</p>
Correct Answer: C