Differential Equations
Linear differential equations
Grade 12
Question:
<p>If \(y(x)\) is the solution of the differential equation \(\dfrac{dy}{dx} + \left(\dfrac{2x+1}{x}\right)y = e^{-2x},\; x > 0\), where \(y(1) = \dfrac{1}{2}e^{-2}\), then:</p>
<p>\(y(\log_e 2) = \dfrac{\log_e 2}{4}\)</p>
<p>\(y(x)\) is decreasing in \((0,\,1)\)</p>
<p>\(y(x)\) is decreasing in \(\left(\dfrac{1}{2},\,1\right)\)</p>
<p>\(y(\log_e 2) = \log_e 4\)</p>
Step-by-Step Solution
Key Concept: This is a first-order linear DE of form dy/dx + P(x)y = Q(x). Find integrating factor μ(x) = e^∫P(x)dx, multiply through, recognize d/dx[μy] = μQ, then integrate and apply initial condition.
<p><strong>Step 1: Identify standard form</strong></p><p>The equation dy/dx + (2x+1)/x · y = e^(-2x) is linear with P(x) = (2x+1)/x and Q(x) = e^(-2x).</p><p><strong>Step 2: Calculate integrating factor</strong></p><p>∫P(x)dx = ∫(2x+1)/x dx = ∫(2 + 1/x)dx = 2x + ln(x)</p><p>μ(x) = e^(2x + ln x) = e^(2x) · e^(ln x) = xe^(2x)</p><p><strong>Step 3: Multiply by integrating factor</strong></p><p>xe^(2x) · dy/dx + xe^(2x) · (2x+1)/x · y = xe^(2x) · e^(-2x)</p><p>xe^(2x) · dy/dx + e^(2x)(2x+1) · y = x</p><p><strong>Step 4: Recognize exact derivative</strong></p><p>d/dx[xe^(2x) · y] = x</p><p><strong>Step 5: Integrate both sides</strong></p><p>xe^(2x) · y = ∫x dx = x²/2 + C</p><p><strong>Step 6: Apply initial condition y(1) = (1/2)e^(-2)</strong></p><p>1 · e^2 · (1/2)e^(-2) = 1/2 + C</p><p>1/2 = 1/2 + C ⟹ C = 0</p><p><strong>Step 7: Final solution</strong></p><p>xe^(2x) · y = x²/2</p><p>∴ y(x) = (x/2)e^(-2x)</p>
Correct Answer: C