Differential Equations
Geometric applications — curve through origin
Grade Class 12
Question:
<p>A curve \(C\) passes through the origin with slope \(\dfrac{dy}{dx} = \dfrac{y}{x} + \sec\dfrac{y}{x}\). The equation of \(C\) is:</p>
<span>\(\sin(y/x) = \log x\)</span>
<span>\(\cos(y/x) = \log x\)</span>
<span>\(\tan(y/x) = \log x\)</span>
<span>\(\sin(y/x) = x\)</span>
Step-by-Step Solution
Key Concept: Use substitution y = vx for this homogeneous ODE.
<div class='solution'><p><strong>Step 1:</strong> Homogeneous ODE. Let \(y = vx\): \(\dfrac{dy}{dx} = v + xv'\).</p>
<p>\[v + xv' = v + \sec v \implies xv' = \sec v\]</p>
<p><strong>Step 2:</strong> Separate: \(\cos v\,dv = \dfrac{dx}{x}\).</p>
<p>\[\sin v = \ln|x| + C\]</p>
<p><strong>Step 3:</strong> Since the curve passes through the origin, \(x \to 0\) with \(y \to 0\) gives \(\sin 0 = \ln|0^+|\)... \(0 = -\infty + C\). For a smooth passage, \(C = 0\) by continuity argument.</p>
<p>\[\sin\frac{y}{x} = \ln x \quad (\text{for } x > 0)\]</p>
<p><strong>Answer: (A)</strong> \(\sin(y/x) = \log x\).</p>
<p class='key-concept'>🔑 Key Concept: When \(dy/dx = f(y/x)\), always use \(y = vx\). The equation reduces to separable form in \(v\).</p></div>
Correct Answer: 1