Differential Equations
Formation of Differential Equations
Grade 12
Question:
<p>The differential equation for the family of curves \(x^2 + y^2 - 2ay = 0\), where \(a\) is an arbitrary constant is</p>
<p>\(2(x^2 - y^2)y' = xy\)</p>
<p>\(2(x^2 + y^2)y' = xy\)</p>
<p>\((x^2 - y^2)y' = 2xy\)</p>
<p>\((x^2 + y^2)y' = 2xy\)</p>
Step-by-Step Solution
Key Concept: Differentiate the family of curves equation to eliminate the arbitrary constant 'a', then solve for the differential equation relating x and y.
<p><strong>Step 1:</strong> Start with the family of curves: <em>x</em><sup>2</sup> + <em>y</em><sup>2</sup> - 2<em>ay</em> = 0</p><p><strong>Step 2:</strong> Differentiate both sides with respect to <em>x</em>:</p><p>2<em>x</em> + 2<em>y</em>(d<em>y</em>/d<em>x</em>) - 2<em>a</em>(d<em>y</em>/d<em>x</em>) = 0</p><p><strong>Step 3:</strong> Simplify and factor out (d<em>y</em>/d<em>x</em>):</p><p>2<em>x</em> + (d<em>y</em>/d<em>x</em>)(2<em>y</em> - 2<em>a</em>) = 0</p><p><strong>Step 4:</strong> From Step 3: 2<em>a</em> = 2<em>y</em> + (d<em>x</em>/d<em>y</em>) · <em>x</em></p><p>From original equation: 2<em>a</em> = (<em>x</em><sup>2</sup> + <em>y</em><sup>2</sup>)/<em>y</em></p><p><strong>Step 5:</strong> Equate both expressions for 2<em>a</em>:</p><p>2<em>xy</em>(d<em>y</em>/d<em>x</em>) = <em>x</em><sup>2</sup> - <em>y</em><sup>2</sup></p><p><strong>Step 6:</strong> The differential equation is: <strong>d<em>y</em>/d<em>x</em> = (<em>x</em><sup>2</sup> - <em>y</em><sup>2</sup>)/(2<em>xy</em>)</strong></p><p>∴ Answer: C</p>
Correct Answer: C