Differential Equations
Linear differential equations
Grade 12
Question:
<p>If \(y = y(x)\) is the solution of the differential equation \(x\dfrac{dy}{dx} + 2y = x^2\) satisfying \(y(1) = 1\), then \(y\!\left(\dfrac{1}{2}\right)\) is equal to</p>
<p>\(\dfrac{7}{64}\)</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{49}{16}\)</p>
<p>\(\dfrac{13}{16}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a linear first-order DE. Divide by x to get standard form dy/dx + (2/x)y = x, then use integrating factor e^(∫2/x dx) = e^(2ln|x|) = x² to convert to d/dx[x²y] = x³.
<p><strong>Step 1:</strong> Rewrite the equation in standard form by dividing by x:</p><p>$$\frac{dy}{dx} + \frac{2y}{x} = x$$</p><p><strong>Step 2:</strong> Find integrating factor: $\mu(x) = e^{\int \frac{2}{x}dx} = e^{2\ln|x|} = x^2$</p><p><strong>Step 3:</strong> Multiply both sides by $x^2$:</p><p>$$x^2\frac{dy}{dx} + 2xy = x^3$$</p><p>This simplifies to: $\frac{d}{dx}[x^2 y] = x^3$</p><p><strong>Step 4:</strong> Integrate both sides:</p><p>$$x^2 y = \int x^3 dx = \frac{x^4}{4} + C$$</p><p>$$y = \frac{x^2}{4} + \frac{C}{x^2}$$</p><p><strong>Step 5:</strong> Apply initial condition $y(1) = 1$:</p><p>$$1 = \frac{1}{4} + C \implies C = \frac{3}{4}$$</p><p><strong>Step 6:</strong> Therefore: $y(x) = \frac{x^2}{4} + \frac{3}{4x^2}$</p><p><strong>Step 7:</strong> Calculate $y\left(\frac{1}{2}\right)$:</p><p>$$y\left(\frac{1}{2}\right) = \frac{(1/2)^2}{4} + \frac{3}{4(1/4)} = \frac{1}{16} + 3 = \frac{49}{16}$$</p><p>∴ Answer: D</p>
Correct Answer: D