Limits, Continuity & Differentiability
Differentiation
Grade 12
Question:
<p>If \(2x = y^{1/5} + y^{-1/5}\) and \((x^2 - 1)\dfrac{d^2y}{dx^2} + \lambda x\dfrac{dy}{dx} + ky = 0\), then \(\lambda + k\) is equal to</p>
<p>\(-23\)</p>
<p>\(-24\)</p>
<p>\(26\)</p>
<p>\(-26\)</p>
Step-by-Step Solution
Key Concept: Differentiate the constraint 2x = y^(1/5) + y^(-1/5) twice to find relationships between dy/dx and d²y/dx², then match coefficients with the given differential equation to identify λ and k.
<p><strong>Step 1:</strong> Start with the constraint: 2x = y^(1/5) + y^(-1/5)</p><p><strong>Step 2:</strong> Differentiate both sides with respect to x:</p><p>2 = (1/5)y^(-4/5)·(dy/dx) - (1/5)y^(-6/5)·(dy/dx)</p><p>2 = (dy/dx)·(1/5)[y^(-4/5) - y^(-6/5)]</p><p>10 = (dy/dx)·(y^(-6/5))[y^(2/5) - 1]</p><p><strong>Step 3:</strong> Differentiate again to find d²y/dx²:</p><p>0 = d²y/dx² · (y^(-6/5))[y^(2/5) - 1] + (dy/dx)·[d/dx(y^(-6/5)[y^(2/5) - 1])]</p><p>After simplification using the constraint y^(2/5) - 1 = 5(x² - 1)y^(6/5):</p><p>0 = (x² - 1)·d²y/dx² + 5x·dy/dx + (1/5)y</p><p><strong>Step 4:</strong> Compare with (x² - 1)d²y/dx² + λx·dy/dx + ky = 0</p><p>λ = 5 and k = 1/5</p><p>∴ λ + k = 5 + 1/5 = <strong>26/5</strong></p>
Correct Answer: A