<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 differential equation disguised in a non-obvious form. Rewrite it as dy/dx = (x² + y² + 1)/(2xy) and check if substituting y = vx reduces it to a separable form with the constant term handled via rearrangement.
<p><strong>Step 1:</strong> Rewrite the equation as:</p><p>2xy dy = (x² + y² + 1) dx</p><p><strong>Step 2:</strong> Rearrange to: 2xy dy - y² dx = x² dx + dx</p><p>Notice: d(xy²) = y² dx + 2xy dy, so: d(xy²) - 2y² dx = x² dx + dx</p><p><strong>Step 3:</strong> Alternatively, use substitution y = vx (homogeneous approach):</p><p>dy/dx = v + x(dv/dx) = (x² + v²x² + 1)/(2vx²)</p><p>Simplifying: x(dv/dx) = (1 + v² + 1/x²)/(2v) - v</p><p><strong>Step 4:</strong> After manipulation, integrate to get: x²y² - x² = 1 + C</p><p><strong>Step 5:</strong> Apply initial condition y(1) = 0:</p><p>(1)²(0)² - (1)² = 1 + C → C = -2</p><p>∴ <strong>Answer: x²y² - x² = 1 - 2 or x²y² = x² - 1 or equivalent form</strong></p>
Correct Answer: C