Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>Let \(\dfrac{dy}{dx} + y = f(x)\) where \(y\) is a continuous function of \(x\) with \(y(0) = 1\) and \(f(x) = \begin{cases} e^{-x} & \text{if } 0 \leq x \leq 2 \\ e^{-2} & \text{if } x > 2 \end{cases}\). Which of the following hold(s) good?</p>
<p>(a) \(y(1) = 2e^{-1}\)</p>
<p>(b) \(y'(1) = -e^{-1}\)</p>
<p>(c) \(y'(3) = -2e^{-3}\)</p>
<p>(d) \(y'(3) = 2e^{-3}\)</p>
Step-by-Step Solution
Key Concept: This is a first-order linear ODE solved using integrating factor e^x. The key is recognizing that f(x) is piecewise, requiring separate solutions on [0,2] and (2,∞) with continuity matching at x=2.
<p><strong>Step 1: Solve for 0 ≤ x ≤ 2</strong></p><p>dy/dx + y = e^(-x). Integrating factor: μ(x) = e^x</p><p>d/dx(ye^x) = e^x · e^(-x) = 1</p><p>ye^x = x + C₁</p><p>y = (x + C₁)e^(-x)</p><p>Using y(0) = 1: 1 = C₁, so y = (x + 1)e^(-x) for 0 ≤ x ≤ 2</p><p><strong>Step 2: Find y(2⁻)</strong></p><p>y(2) = (2 + 1)e^(-2) = 3e^(-2)</p><p><strong>Step 3: Solve for x > 2</strong></p><p>dy/dx + y = e^(-2). Integrating factor: μ(x) = e^x</p><p>d/dx(ye^x) = e^x · e^(-2) = e^(-2) · e^x</p><p>ye^x = e^(-2) · e^x + C₂</p><p>y = e^(-2) + C₂e^(-x)</p><p>Using continuity at x = 2: 3e^(-2) = e^(-2) + C₂e^(-2)</p><p>C₂ = 2</p><p>y = e^(-2) + 2e^(-x) for x > 2</p><p><strong>Step 4: Verify properties</strong></p><p>• y(0) = 1 ✓</p><p>• y(2⁻) = y(2⁺) = 3e^(-2) (continuous) ✓</p><p>• dy/dx|_(x=2⁻) = -3e^(-2) + e^(-2) = -2e^(-2)</p><p>• dy/dx|_(x=2⁺) = -2e^(-2) (matches, derivative continuous) ✓</p><p>• As x→∞: y→e^(-2) ✓</p><p>∴ Answer: ABC</p>
Correct Answer: ABC