Differential Equations
Formation of Differential Equations
Grade 12
Question:
<p>The differential equation of all circles passing through the origin and having their centres on the <em>x</em>-axis is</p>
<p>\(x^2 = y^2 + xy\dfrac{dy}{dx}\)</p>
<p>\(x^2 = y^2 + 3xy\dfrac{dy}{dx}\)</p>
<p>\(y^2 = x^2 + 2xy\dfrac{dy}{dx}\)</p>
<p>\(y^2 = x^2 - 2xy\dfrac{dy}{dx}\)</p>
Step-by-Step Solution
Key Concept: A circle with center on the x-axis passing through the origin has center (a, 0) and radius a, giving equation (x-a)² + y² = a². Differentiate to eliminate the parameter a and obtain the DE without involving the arbitrary constant.
<p><strong>Step 1:</strong> Family of circles with centers on x-axis passing through origin has center (a, 0) and radius = a (distance from center to origin).</p><p><strong>Step 2:</strong> Equation of circle: (x − a)² + y² = a²</p><p>Expanding: x² − 2ax + a² + y² = a²</p><p>∴ x² + y² − 2ax = 0 ... (1)</p><p><strong>Step 3:</strong> Differentiate equation (1) with respect to x:</p><p>2x + 2y(dy/dx) − 2a = 0</p><p>∴ a = x + y(dy/dx) ... (2)</p><p><strong>Step 4:</strong> Substitute (2) into (1):</p><p>x² + y² − 2x[x + y(dy/dx)] = 0</p><p>x² + y² − 2x² − 2xy(dy/dx) = 0</p><p>y² − x² − 2xy(dy/dx) = 0</p><p><strong>∴ DE: 2xy(dy/dx) = y² − x²</strong> or equivalently <strong>(y² − x²) − 2xy(dy/dx) = 0</strong></p>
Correct Answer: C