Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>Let \(y = y(x)\) be the solution of the differential equation \(\frac{dy}{dx} + 2y = f(x)\), where \[f(x) = \begin{cases} 1, & x \in [0,1] \\ 0, & \text{otherwise} \end{cases}\] If \(y(0) = 0\), then \(y\!\left(\dfrac{3}{2}\right)\) is</p>
<p>\(\dfrac{e^2+1}{2e^4}\)</p>
<p>\(\dfrac{1}{2e}\)</p>
<p>\(\dfrac{e^2-1}{e^3}\)</p>
<p>\(\dfrac{e^2-1}{2e^3}\)</p>
Step-by-Step Solution
Key Concept: The solution requires solving the linear ODE in two regions separately: [0,1] where f(x)=1, and (1,∞) where f(x)=0. The value at x=3/2 uses the solution in the second region with initial condition from y(1).
<p><strong>Step 1: Solve for x ∈ [0,1] where f(x)=1</strong></p><p>The equation becomes: dy/dx + 2y = 1</p><p>Integrating factor: μ(x) = e^(2x)</p><p>Multiply by μ: d/dx[e^(2x)·y] = e^(2x)</p><p>Integrate: e^(2x)·y = (1/2)e^(2x) + C</p><p>So y = 1/2 + Ce^(-2x)</p><p>Using y(0) = 0: 0 = 1/2 + C, so C = -1/2</p><p>Therefore: y(x) = (1/2)(1 - e^(-2x)) for x ∈ [0,1]</p><p><strong>Step 2: Find y(1)</strong></p><p>y(1) = (1/2)(1 - e^(-2)) = (1 - e^(-2))/2</p><p><strong>Step 3: Solve for x > 1 where f(x)=0</strong></p><p>The equation becomes: dy/dx + 2y = 0</p><p>This gives: y(x) = Ae^(-2x)</p><p>Using initial condition y(1) = (1 - e^(-2))/2:</p><p>A·e^(-2) = (1 - e^(-2))/2</p><p>So A = (1 - e^(-2))/(2e^(-2))</p><p><strong>Step 4: Calculate y(3/2)</strong></p><p>y(3/2) = A·e^(-3) = [(1 - e^(-2))/(2e^(-2))]·e^(-3)</p><p>= (1 - e^(-2))/(2e^(-2))·e^(-3) = (1 - e^(-2))·e^(-1)/2</p><p>= (e^(-1) - e^(-3))/2</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D