Differential Equations
Order and Degree
Grade 12
Question:
<p>The differential equation representing the family of curves \(y^2 = 2c(x + \sqrt{c})\), where \(c > 0\), is a parameter, is of order and degree as follows:</p>
<p>order 1, degree 2</p>
<p>order 1, degree 1</p>
<p>order 1, degree 3</p>
<p>order 2, degree 2</p>
Step-by-Step Solution
Key Concept: Eliminate the parameter c from the family of curves by differentiating the equation and using substitution; the order equals the number of parameters to eliminate, and degree is the highest power of the highest derivative after clearing fractions and radicals.
<p><strong>Step 1:</strong> Start with the family of curves: $y^2 = 2c(x + \sqrt{c})$</p><p><strong>Step 2:</strong> Differentiate both sides with respect to $x$: $2y\frac{dy}{dx} = 2c$</p><p>This gives us: $y\frac{dy}{dx} = c$ ... (i)</p><p><strong>Step 3:</strong> From equation (i), we have $c = y\frac{dy}{dx}$. Substitute this back into the original equation:</p><p>$y^2 = 2y\frac{dy}{dx}\left(x + \sqrt{y\frac{dy}{dx}}\right)$</p><p><strong>Step 4:</strong> Simplify: $y = 2\frac{dy}{dx}\left(x + \sqrt{y\frac{dy}{dx}}\right)$</p><p>$y = 2x\frac{dy}{dx} + 2\sqrt{y\frac{dy}{dx}} \cdot \sqrt{\frac{dy}{dx}}$</p><p>$y = 2x\frac{dy}{dx} + 2\frac{dy}{dx}\sqrt{y\frac{dy}{dx}}$</p><p><strong>Step 5:</strong> Rearrange: $y - 2x\frac{dy}{dx} = 2\frac{dy}{dx}\sqrt{y\frac{dy}{dx}}$</p><p>Square both sides to eliminate the radical: $(y - 2x\frac{dy}{dx})^2 = 4\left(\frac{dy}{dx}\right)^2 \cdot y\frac{dy}{dx}$</p><p><strong>Step 6:</strong> The resulting differential equation is of <strong>Order 1</strong> (highest derivative is first order) and <strong>Degree 2</strong> (highest power of $\frac{dy}{dx}$ is 2 after squaring).</p><p>∴ Answer: C (Order 1, Degree 2)</p>
Correct Answer: C