Differential Equations
First order differential equations solvable for x
Grade 12
Question:
<p>Solve \(1 + \left(\frac{dy}{dx}\right)^2 = x\frac{dy}{dx}\).</p>
<p>\(y = c + \frac{p^2}{2} - \log p\)</p>
<p>\(y = c - \frac{p^2}{2} + \log p\)</p>
<p>\(y = c + \frac{p^2}{2} + \log p\)</p>
<p>\(y = c - \frac{p^2}{2} - \log p\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a Clairaut-type equation by treating dy/dx as a parameter p, then differentiate with respect to x to obtain a separable or solvable form. The solution consists of a general solution (family of lines) and a singular solution (envelope).
<p><strong>Step 1:</strong> Let p = dy/dx. The equation becomes: 1 + p² = xp</p><p><strong>Step 2:</strong> Rearrange: p² - xp + 1 = 0</p><p><strong>Step 3:</strong> Differentiate both sides with respect to x (treating p as a function of x):</p><p>2p(dp/dx) - p - x(dp/dx) + 0 = 0</p><p>(dp/dx)(2p - x) = p</p><p><strong>Step 4:</strong> Either dp/dx = 0 or (2p - x) = p/0</p><p>Case 1: If dp/dx = 0, then p = c (constant)</p><p>Substituting back: 1 + c² = xc</p><p>Therefore: <strong>y = cx - (1 + c²)/c</strong> or <strong>cy = c²x - (1 + c²)</strong> [General Solution - family of straight lines]</p><p><strong>Step 5:</strong> For singular solution, eliminate c from: 1 + c² = xc and the differentiation equation</p><p>From 1 + c² = xc: c² - xc + 1 = 0, so c = (x ± √(x² - 4))/2</p><p>Eliminating c yields: <strong>4y = x² - 4</strong> or <strong>y = (x² - 4)/4</strong> [Singular Solution - envelope]</p><p>∴ Answer: A</p>
Correct Answer: A