Differential Equations
Formation of Differential Equations
Grade 12
Question:
<p>The equation of family of circles passing through the origin is \((x-0)^2 + (y-a)^2 = a^2\). The differential equation of this family is:</p>
<p>\(\frac{dy}{dx} = \frac{2xy}{x^2 - y^2}\)</p>
<p>\(\frac{dy}{dx} = \frac{x^2 - y^2}{2xy}\)</p>
<p>\(\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\)</p>
<p>\(\frac{dy}{dx} = \frac{2xy}{x^2 + y^2}\)</p>
Step-by-Step Solution
Key Concept: Eliminate the parameter 'a' from the circle equation by differentiating it and using the relationship between x, y, and dy/dx to obtain a parameter-free differential equation.
<p><strong>Step 1:</strong> Start with the family equation: $(x-0)^2 + (y-a)^2 = a^2$</p><p>Expand: $x^2 + y^2 - 2ay + a^2 = a^2$</p><p>Simplify: $x^2 + y^2 - 2ay = 0$ ... (1)</p><p><strong>Step 2:</strong> Differentiate equation (1) with respect to x:</p><p>$2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0$</p><p>$2x + 2\frac{dy}{dx}(y - a) = 0$</p><p>$x + (y-a)\frac{dy}{dx} = 0$ ... (2)</p><p><strong>Step 3:</strong> From equation (2): $y - a = -\frac{x}{\frac{dy}{dx}}$</p><p><strong>Step 4:</strong> Substitute back into equation (1):</p><p>$x^2 + y^2 - 2\left(-\frac{x}{\frac{dy}{dx}}\right)y = 0$</p><p>$x^2 + y^2 + \frac{2xy}{\frac{dy}{dx}} = 0$</p><p>Multiply by $\frac{dy}{dx}$: $(x^2 + y^2)\frac{dy}{dx} + 2xy = 0$</p><p>∴ <strong>Answer: $(x^2 + y^2)\frac{dy}{dx} + 2xy = 0$</strong></p>
Correct Answer: C