Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>Given <br/> \(x\dfrac{dy}{dx} + 2y = x^2\) where \(x \neq 0\) and \(y(1) = 1\). Find \(y\).</p>
<p>\(y = \dfrac{x^2}{4} + \dfrac{3}{4x^2}\)</p>
<p>\(y = \dfrac{x^2}{4} + \dfrac{3}{4x^2}\)</p>
<p>\(y = \dfrac{x^4}{4} + \dfrac{3}{4}\)</p>
<p>\(y = \dfrac{x^2}{4} - \dfrac{3}{4x^2}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a first-order linear differential equation in standard form dy/dx + P(x)y = Q(x). Divide by x first, then use the integrating factor method with μ(x) = e^∫P(x)dx.
<p><strong>Step 1: Rewrite in standard form</strong></p><p>Divide the equation by x: <br/>dy/dx + (2/x)y = x</p><p><strong>Step 2: Find the integrating factor</strong></p><p>P(x) = 2/x, so μ(x) = e^∫(2/x)dx = e^(2ln|x|) = x²</p><p><strong>Step 3: Multiply both sides by integrating factor</strong></p><p>x²·dy/dx + 2xy = x³</p><p>The left side is d/dx(x²y)</p><p><strong>Step 4: Integrate both sides</strong></p><p>d/dx(x²y) = x³</p><p>x²y = ∫x³dx = x⁴/4 + C</p><p>y = x²/4 + C/x²</p><p><strong>Step 5: Apply initial condition y(1) = 1</strong></p><p>1 = 1/4 + C</p><p>C = 3/4</p><p><strong>Step 6: Final solution</strong></p><p>y = x²/4 + 3/(4x²)</p><p>Or: y = (x⁴ + 3)/(4x²)</p><p>∴ Answer: A</p>
Correct Answer: A