Differential Equations
Formation of Differential Equations
Grade 12
Question:
<p>The differential equation representing the family of ellipses having foci either on the <em>x</em>-axis or on the <em>y</em>-axis, centre at the origin and passing through the point (0, 3) is</p>
<p>\(xyy'' + x(y')^2 - yy' = 0\)</p>
<p>\(x + yy'' = 0\)</p>
<p>\(xyy' + y^2 - 9 = 0\)</p>
<p>\(xyy' - y^2 + 9 = 0\)</p>
Step-by-Step Solution
Key Concept: Since the ellipse passes through (0,3) and has center at origin, we know 3 is either the semi-major or semi-minor axis. Differentiate the general ellipse equation x²/a² + y²/b² = 1 to eliminate arbitrary constants a and b.
<p><strong>Step 1:</strong> The general ellipse equation centered at origin is: x²/a² + y²/b² = 1</p><p><strong>Step 2:</strong> Since (0,3) lies on the ellipse: 0/a² + 9/b² = 1, so b² = 9</p><p><strong>Step 3:</strong> The equation becomes: x²/a² + y²/9 = 1, containing one arbitrary constant a²</p><p><strong>Step 4:</strong> Differentiate with respect to x: 2x/a² + 2y(dy/dx)/9 = 0</p><p>This gives: x/a² = -y(dy/dx)/9, so a² = -9x/(y·dy/dx)</p><p><strong>Step 5:</strong> Substitute back into x²/a² + y²/9 = 1:</p><p>x²·y(dy/dx)/(-9x) + y²/9 = 1</p><p>-xy(dy/dx)/9 + y²/9 = 1</p><p><strong>Step 6:</strong> Multiply by 9: -xy(dy/dx) + y² = 9</p><p>∴ <strong>y² - xy(dy/dx) = 9</strong> or <strong>9 = y² - xy(dy/dx)</strong></p>
Correct Answer: D