Differential Equations
Formation of Differential Equations
Grade 12
Question:
<p>The differential equation corresponding to the family of curves \(x^2 + y^2 - 2ay = 0\), where \(a\) is an arbitrary constant, is:</p>
<p>\((x^2 - y^2)y' = 2xy\)</p>
<p>\(x^2 y' - y^2 y' = 2xy\)</p>
<p>\((x^2 + y^2)y' = 2xy\)</p>
<p>\(x^2 y' + y^2 y' = 2xy\)</p>
Step-by-Step Solution
Key Concept: Eliminate the arbitrary constant 'a' by differentiating the curve equation and solving the system of the original equation and its derivative. This yields a differential equation containing only x, y, and dy/dx.
<p><strong>Step 1:</strong> Start with the family of curves: <strong>x² + y² - 2ay = 0</strong></p><p><strong>Step 2:</strong> Differentiate both sides with respect to x:<br>2x + 2y(dy/dx) - 2a(dy/dx) = 0</p><p><strong>Step 3:</strong> Simplify and solve for 'a':<br>2x + 2y(dy/dx) - 2a(dy/dx) = 0<br>2a(dy/dx) = 2x + 2y(dy/dx)<br>a = [x + y(dy/dx)]/(dy/dx)</p><p><strong>Step 4:</strong> Substitute this expression for 'a' back into the original equation:<br>x² + y² - 2y · [x + y(dy/dx)]/(dy/dx) = 0<br>x² + y² - 2y[x + y(dy/dx)]/(dy/dx) = 0</p><p><strong>Step 5:</strong> Multiply through by (dy/dx):<br>(x² + y²)(dy/dx) - 2xy - 2y²(dy/dx) = 0<br>(x² - y²)(dy/dx) = 2xy</p><p><strong>Step 6:</strong> Final differential equation:<br><strong>(x² - y²)dy/dx = 2xy</strong> or equivalently <strong>(x² - y²)dy = 2xy·dx</strong></p><p>∴ Answer: A</p>
Correct Answer: A