Differential Equations
First order differential equations
Grade 12
Question:
<p>Solve \(x\left(\frac{dy}{dx}\right)^2 + (y - x)\frac{dy}{dx} - y = 0\).</p>
<p>\((y - x - c)(xy - c) = 0\)</p>
<p>\((y + x - c)(xy + c) = 0\)</p>
<p>\((y - x + c)(xy - c) = 0\)</p>
<p>\((y - x - c)(xy + c) = 0\)</p>
Step-by-Step Solution
Key Concept: Treat this quadratic equation in dy/dx as a factorizable expression: factor it into two linear differential equations, then solve each separately using standard methods.
<p><strong>Step 1:</strong> Recognize the given equation as quadratic in dy/dx:</p><p>x(dy/dx)² + (y - x)(dy/dx) - y = 0</p><p><strong>Step 2:</strong> Factor by treating as ax² + bx + c = 0 form:</p><p>x(dy/dx)² + y(dy/dx) - x(dy/dx) - y = 0</p><p>x(dy/dx)[(dy/dx) - 1] + y[(dy/dx) - 1] = 0</p><p>[(dy/dx) - 1][x(dy/dx) + y] = 0</p><p><strong>Step 3:</strong> Solve first factor: dy/dx - 1 = 0</p><p>dy/dx = 1 ⟹ y = x + c₁</p><p><strong>Step 4:</strong> Solve second factor: x(dy/dx) + y = 0</p><p>x(dy/dx) = -y ⟹ dy/y = -dx/x</p><p>ln|y| = -ln|x| + c₂ ⟹ xy = c₃ (or y = c/x)</p><p><strong>Step 5:</strong> General solution is given by combining both families:</p><p>(y - x - c₁)(xy - c₂) = 0</p><p>∴ Answer: <strong>y = x + c₁</strong> and <strong>xy = c₂</strong> (two families of solutions)</p>
Correct Answer: A