Differential Equations
Formation of Differential Equations
Grade 12
Question:
<p>The general equation of circles touching the y-axis at the origin is \((x+g)^2 + y^2 = g^2\). The differential equation of such circles is:</p>
<p>\(y^2 = x^2 + 2xy\dfrac{dy}{dx}\)</p>
<p>\(y^2 = x^2 - 2xy\dfrac{dy}{dx}\)</p>
<p>\(x^2 = y^2 + 2xy\dfrac{dy}{dx}\)</p>
<p>\(x^2 = y^2 - 2xy\dfrac{dy}{dx}\)</p>
Step-by-Step Solution
Key Concept: A circle touching the y-axis at origin has center at (−g, 0), so eliminating the parameter g from the circle equation (x+g)² + y² = g² by differentiating gives the differential equation. The key is that g is the family parameter, not a constant for a single circle.
<p><strong>Step 1:</strong> Start with the given circle equation: $(x+g)^2 + y^2 = g^2$</p><p><strong>Step 2:</strong> Expand: $x^2 + 2gx + g^2 + y^2 = g^2$</p><p>This simplifies to: $x^2 + 2gx + y^2 = 0$ ... (1)</p><p><strong>Step 3:</strong> Differentiate equation (1) with respect to x:</p><p>$2x + 2g + 2y\frac{dy}{dx} = 0$</p><p>$x + g + y\frac{dy}{dx} = 0$</p><p>Therefore: $g = -x - y\frac{dy}{dx}$ ... (2)</p><p><strong>Step 4:</strong> Substitute (2) into (1):</p><p>$x^2 + 2x\left(-x - y\frac{dy}{dx}\right) + y^2 = 0$</p><p>$x^2 - 2x^2 - 2xy\frac{dy}{dx} + y^2 = 0$</p><p>$-x^2 - 2xy\frac{dy}{dx} + y^2 = 0$</p><p>$2xy\frac{dy}{dx} = y^2 - x^2$</p><p><strong>∴ Answer:</strong> $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ (or equivalent form depending on option A)</p>
Correct Answer: A