Differential Equations
Equations Reducible to Variable Separable Form
Grade 12
Question:
<p>The solution of the differential equation <span class="math">\frac{dy}{dx} = \sin(x + y) + \cos(x + y)</span> is</p>
<p>(a) <span class="math">\log \tan \frac{x + y}{2} + 1 = x + C</span></p>
<p>(b) <span class="math">\log \cot \frac{x + y}{2} + 1 = x + C</span></p>
<p>(c) <span class="math">\log \tan \frac{x - y}{2} - 1 = C</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use substitution z = x + y to reduce the equation to variable separable form, then apply trigonometric identities to simplify and integrate.
<p><strong>Step 1:</strong> Let <span class="math">z = x + y \Rightarrow \frac{dz}{dx} = 1 + \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{dz}{dx} - 1</span></p><p><strong>Step 2:</strong> Substituting into the given equation: <span class="math">\frac{dz}{dx} - 1 = \sin z + \cos z</span></p><p><strong>Step 3:</strong> This simplifies to <span class="math">\frac{dz}{dx} = 1 + \sin z + \cos z</span></p><p><strong>Step 4:</strong> Using the identity <span class="math">1 + \sin z + \cos z = 2\cos^2 \frac{z}{2} + 2\sin \frac{z}{2} \cos \frac{z}{2} = 2\cos \frac{z}{2}(\cos \frac{z}{2} + \sin \frac{z}{2})</span></p><p><strong>Step 5:</strong> Separating variables and integrating: <span class="math">\int \frac{dz}{2\cos \frac{z}{2}(\cos \frac{z}{2} + \sin \frac{z}{2})} = \int dx</span></p><p><strong>Step 6:</strong> After integration, we get <span class="math">\log \tan \frac{x+y}{2} + 1 = x + C</span></p><p>∴ Answer is (a).</p>
Correct Answer: a