Differential Equations
Reducible to Separable Form
Grade 12
Question:
<p>The solution of the differential equation <span class="math">\(\frac{dy}{dx} = (4x + y + 1)^2\)</span> is:</p>
<p>(a) <span class="math">\(4x + y + 1 = 2\tan(2x + y + C)\)</span></p>
<p>(b) <span class="math">\(4x + y + 1 = 2\tan(x + 2y + C)\)</span></p>
<p>(c) <span class="math">\(4x + y + 1 = 2\tan(2y + C)\)</span></p>
<p>(d) <span class="math">\(4x + y + 1 = 2\tan(2x + C)\)</span></p>
Step-by-Step Solution
Key Concept: Recognize that this is a differential equation in the form dy/dx = f(ax + by + c). Use substitution v = 4x + y + 1 to reduce it to a separable equation, then integrate using standard trigonometric substitution.
<p><strong>Step 1: Make the substitution v = 4x + y + 1</strong></p><p>Differentiating both sides with respect to x: dv/dx = 4 + dy/dx</p><p>Therefore: dy/dx = dv/dx - 4</p><p></p><p><strong>Step 2: Substitute into the original equation</strong></p><p>The given equation is: dy/dx = (4x + y + 1)²</p><p>Substituting: dv/dx - 4 = v²</p><p>Rearranging: dv/dx = v² + 4</p><p></p><p><strong>Step 3: Separate variables</strong></p><p>dv/(v² + 4) = dx</p><p></p><p><strong>Step 4: Integrate both sides</strong></p><p>∫ dv/(v² + 4) = ∫ dx</p><p>Using the standard formula ∫ du/(u² + a²) = (1/a)tan⁻¹(u/a) + C:</p><p>∫ dv/(v² + 4) = (1/2)tan⁻¹(v/2) + K</p><p>The right side integrates to: x + C'</p><p></p><p><strong>Step 5: Solve for v</strong></p><p>(1/2)tan⁻¹(v/2) = x + C' (combining constants)</p><p>tan⁻¹(v/2) = 2x + K (where K = 2C')</p><p>v/2 = tan(2x + K)</p><p>v = 2tan(2x + K)</p><p></p><p><strong>Step 6: Back-substitute v = 4x + y + 1</strong></p><p>4x + y + 1 = 2tan(2x + C)</p><p>where C is an arbitrary constant.</p><p></p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a