<p>The solution of differential equation \(\dfrac{dy}{dx} = \dfrac{x^2 + y^2 + 1}{2xy}\) satisfying \(y(1) = 0\) is given by:</p>
Step-by-Step Solution
Key Concept: Recognize this as a homogeneous-looking equation that becomes separable after the substitution y² = v. The equation structure suggests treating y² as a single variable to convert a non-linear ODE into a linear one.
<p><strong>Step 1:</strong> Rewrite the given equation:</p><p>2xy·dy = (x² + y² + 1)dx</p><p><strong>Step 2:</strong> Recognize this can be rewritten as:</p><p>2xy·dy - y²·dx = x²·dx + dx</p><p><strong>Step 3:</strong> Notice that d(xy²) = y²·dx + 2xy·dy, so:</p><p>d(xy²) - 2y²·dx = x²·dx + dx</p><p>Alternatively, rearrange as: 2xy·(dy/dx) - y² = x² + 1</p><p><strong>Step 4:</strong> Let v = y², then dv/dx = 2y·(dy/dx). The equation becomes:</p><p>x·(dv/dx) - v = x² + 1</p><p><strong>Step 5:</strong> This is a linear ODE: dv/dx - (v/x) = x + (1/x)</p><p>Integrating factor: μ(x) = e^(-∫dx/x) = 1/x</p><p><strong>Step 6:</strong> Multiply by integrating factor:</p><p>(1/x)·dv/dx - v/x² = 1 + 1/x²</p><p>d/dx(v/x) = 1 + 1/x²</p><p><strong>Step 7:</strong> Integrate both sides:</p><p>v/x = x - 1/x + C</p><p>y²/x = x - 1/x + C</p><p><strong>Step 8:</strong> Apply initial condition y(1) = 0:</p><p>0/1 = 1 - 1 + C → C = 0</p><p><strong>Step 9:</strong> Therefore: y²/x = x - 1/x</p><p>y² = x² - 1</p><p>∴ Answer: C</p>
Correct Answer: C