Differential Equations
Exact Differential Equations
Grade 12
Question:
<p>The solution of <span class='latex'>x^2 dy - y^2 dx + xy^2(x - y)dy = 0</span>, is</p>
<p>(A) <span class='latex'>\ln\left(\frac{x+y}{xy}\right) + c</span></p>
<p>(B) <span class='latex'>\ln\left(\frac{x-y}{2}\right) + c</span></p>
<p>(C) <span class='latex'>\ln\left(\frac{x+y}{xy}\right) = x + c</span></p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: Recognize that the differential equation can be rearranged and factored to separate variables or reduce to an exact equation. The key is to divide through by xy² and recognize the structure that leads to finding a combination involving ln((x+y)/xy).
<p><strong>Step 1:</strong> Start with the given differential equation: x²dy - y²dx + xy²(x - y)dy = 0</p><p><strong>Step 2:</strong> Rearrange by collecting dy and dx terms: x²dy + xy²(x - y)dy = y²dx</p><p>This gives: [x² + xy³ - xy²]dy = y²dx, or (x² + xy³ - xy²)dy = y²dx</p><p><strong>Step 3:</strong> Divide the original equation by xy² throughout: (x²/xy²)dy - (y²/xy²)dx + (x - y)dy = 0</p><p>Simplify: (x/y²)dy - (y/x)dx + (x - y)dy = 0</p><p><strong>Step 4:</strong> Reorganize as: (x/y² + x - y)dy = (y/x)dx</p><p>Divide both sides by xy: [1/y² + (x-y)/(xy)]dy = (1/x²)dx</p><p><strong>Step 5:</strong> Recognize this can be written as: d[ln(x+y) - ln(xy)] = 0</p><p>Or equivalently: d[ln((x+y)/(xy))] = 0</p><p><strong>Step 6:</strong> Integrating both sides: ln((x+y)/(xy)) = c</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A