Differential Equations
Linear Differential Equation
Grade 12
Question:
<p>Given \(\dfrac{dy}{dx} + \left(\dfrac{2x+1}{x}\right)y = e^{-2x}\), and the curve passes through \(\left(1,\,\dfrac{1}{2}e^{-2}\right)\). Find \(y(\log_e 2)\).</p>
<p>\(\dfrac{1}{4}\log_e 2\)</p>
<p>\(\dfrac{1}{8}\log_e 2\)</p>
<p>\(\log_e 2\)</p>
<p>\(2\log_e 2\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a linear first-order ODE of the form dy/dx + P(x)y = Q(x). Find the integrating factor μ(x) = e^∫P(x)dx, then the solution becomes y·μ(x) = ∫Q(x)·μ(x)dx.
<p><strong>Step 1:</strong> Identify P(x) = (2x+1)/x and Q(x) = e^(-2x). Find the integrating factor:</p><p>μ(x) = e^∫(2x+1)/x dx = e^∫(2 + 1/x)dx = e^(2x + ln x) = x·e^(2x)</p><p><strong>Step 2:</strong> Multiply both sides by μ(x):</p><p>x·e^(2x)·dy/dx + (2x+1)e^(2x)·y = x·e^(-2x)·e^(2x) = x</p><p>The left side is d/dx[x·e^(2x)·y], so:</p><p>d/dx[x·e^(2x)·y] = x</p><p><strong>Step 3:</strong> Integrate both sides:</p><p>x·e^(2x)·y = ∫x dx = x²/2 + C</p><p><strong>Step 4:</strong> Apply initial condition (1, e^(-2)/2):</p><p>1·e²·(e^(-2)/2) = 1/2 + C</p><p>1/2 = 1/2 + C ⟹ C = 0</p><p><strong>Step 5:</strong> Therefore: y = x/(2e^(2x))</p><p><strong>Step 6:</strong> Find y(ln 2):</p><p>y(ln 2) = ln 2/(2e^(2ln 2)) = ln 2/(2·4) = (ln 2)/8</p><p>∴ Answer: B</p>
Correct Answer: B