Differential Equations
Solution of differential equations
Grade 12
Question:
<p>The solution of \(x^2y_1^2 + xyy_1 - 6y^2 = 0\) are</p>
<p>(a) \(y = cx^2\)</p>
<p>(b) \(x^2y = c\)</p>
<p>(c) \(\dfrac{1}{2}\log y = c + \log x\)</p>
<p>(d) \(x^3y = c\)</p>
Step-by-Step Solution
Key Concept: Factor the differential equation as a quadratic in y₁ (dy/dx) to obtain two separable first-order ODEs, then solve each by separation of variables.
<p><strong>Step 1:</strong> Recognize the equation x²y₁² + xyy₁ - 6y² = 0 as quadratic in y₁. Factor as: (xy₁ - 2y)(xy₁ + 3y) = 0</p><p><strong>Step 2:</strong> This gives two cases:</p><p><strong>Case 1:</strong> xy₁ - 2y = 0 ⟹ x(dy/dx) = 2y ⟹ dy/y = 2dx/x ⟹ ln|y| = 2ln|x| + c₁ ⟹ <strong>y = c₁x²</strong></p><p><strong>Case 3:</strong> xy₁ + 3y = 0 ⟹ x(dy/dx) = -3y ⟹ dy/y = -3dx/x ⟹ ln|y| = -3ln|x| + c₂ ⟹ <strong>y = c₂/x³</strong></p><p><strong>Step 4:</strong> The complete solution is the family of curves: y = c₁x² and y = c₂x⁻³</p><p>∴ Answer: A</p>
Correct Answer: A