Differential Equations
Variable separable
Grade 12
Question:
<p>Given that the slope of the tangent to a curve \(y = y(x)\) at any point \((x, y)\) is \(\dfrac{2y}{x^2}\). If the curve passes through the centre of the circle \(x^2 + y^2 - 2x - 2y = 0\), then its equation is:</p>
<p>\(x \log_e |y| = 2(x - 1)\)</p>
<p>\(x \log_e |y| = -2(x - 1)\)</p>
<p>\(x^2 \log_e |y| = -2(x - 1)\)</p>
<p>\(x \log_e |y| = x - 1\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a separable differential equation where dy/dx = 2y/x². Separate variables and integrate, then use the initial condition (center of circle) to find the arbitrary constant.
<p><strong>Step 1:</strong> Set up the differential equation from the given slope:</p><p>dy/dx = 2y/x²</p><p><strong>Step 2:</strong> Separate variables:</p><p>dy/y = 2dx/x²</p><p><strong>Step 3:</strong> Integrate both sides:</p><p>∫dy/y = ∫2dx/x²</p><p>ln|y| = -2/x + C</p><p><strong>Step 4:</strong> Find the center of the circle x² + y² - 2x - 2y = 0:</p><p>(x-1)² + (y-1)² = 2</p><p>Center: (1, 1)</p><p><strong>Step 5:</strong> Apply initial condition y(1) = 1:</p><p>ln(1) = -2/1 + C</p><p>0 = -2 + C</p><p>C = 2</p><p><strong>Step 6:</strong> Write the solution:</p><p>ln|y| = -2/x + 2</p><p>|y| = e^(2 - 2/x) = e²·e^(-2/x)</p><p>y = e²·e^(-2/x) or y = Ae^(-2/x) where A = e²</p><p>∴ Answer: A</p>
Correct Answer: A