Differential Equations
Homogeneous Differential Equations
Grade 12
Question:
<p>The solution of the differential equation <span class="math">(x - y) dy - (x + y) dx = 0</span> is</p>
<p>(a) <span class="math">\tan^{-1} \frac{x}{y} + \frac{1}{2} \log (x^2 + y^2) = C</span></p>
<p>(b) <span class="math">\tan^{-1} \frac{y}{x} - \frac{1}{2} \log (x^2 + y^2) = C</span></p>
<p>(c) <span class="math">\cot^{-1} \frac{y}{x} + \frac{1}{2} \log (x^2 + y^2) = C</span></p>
Step-by-Step Solution
Key Concept: Recognize this as a homogeneous equation and use the substitution y = vx to convert it to a separable form involving inverse trigonometric and logarithmic functions.
<p><strong>Step 1:</strong> Rewrite the equation as <span class="math">\frac{dy}{dx} = \frac{x + y}{x - y}</span></p><p><strong>Step 2:</strong> This is a homogeneous equation. Let <span class="math">y = vx</span>, then <span class="math">\frac{dy}{dx} = v + x \frac{dv}{dx}</span></p><p><strong>Step 3:</strong> Substituting: <span class="math">v + x \frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}</span></p><p><strong>Step 4:</strong> Simplifying: <span class="math">x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v^2}{1 - v}</span></p><p><strong>Step 5:</strong> Separating variables: <span class="math">\frac{1 - v}{1 + v^2} dv = \frac{dx}{x}</span></p><p><strong>Step 6:</strong> Integrating both sides: <span class="math">\tan^{-1} v - \frac{1}{2} \log(1 + v^2) = \log x + C</span></p><p><strong>Step 7:</strong> Substituting back <span class="math">v = \frac{y}{x}</span>: <span class="math">\tan^{-1} \frac{y}{x} - \frac{1}{2} \log(x^2 + y^2) = C</span></p><p>∴ Answer is (b).</p>
Correct Answer: b