Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>Consider the differential equation, \(y^2 dx + \left(x - \dfrac{1}{y}\right)dy = 0\). If value of \(y\) is 1 when \(x = 1\), then the value of \(x\) for which \(y = 2\), is:</p>
<p>\(\dfrac{5}{2} + \dfrac{1}{\sqrt{e}}\)</p>
<p>\(\dfrac{3}{2} - \dfrac{1}{\sqrt{e}}\)</p>
<p>\(\dfrac{1}{2} + \dfrac{1}{\sqrt{e}}\)</p>
<p>\(\dfrac{3}{2} - \sqrt{e}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as an exact differential equation by rearranging to M dx + N dy = 0 form, then verify ∂M/∂y = ∂N/∂x to find the solution using the potential function method.
<p><strong>Step 1:</strong> Rewrite the equation in standard form M dx + N dy = 0:</p><p>y² dx + (x - 1/y) dy = 0</p><p>Here M = y², N = x - 1/y</p><p><strong>Step 2:</strong> Check if exact: ∂M/∂y = 2y and ∂N/∂x = 1</p><p>This is NOT exact. Rewrite by dividing through by y²:</p><p>dx + (x/y² - 1/y³) dy = 0</p><p><strong>Step 3:</strong> Now M = 1, N = x/y² - 1/y³</p><p>Check: ∂M/∂y = 0 and ∂N/∂x = 1/y². Still not exact.</p><p><strong>Step 4:</strong> Try integrating factor μ = 1/y. Multiply original by 1/y:</p><p>y dx + (x/y - 1/y²) dy = 0</p><p>Now M = y, N = x/y - 1/y²</p><p>∂M/∂y = 1, ∂N/∂x = 1/y ✓ (Still checking...)</p><p><strong>Step 5:</strong> Return to original form and integrate directly:</p><p>d(xy) = y dx + x dy, so rearrange:</p><p>y² dx + x dy = y² dx + x dy - 1/y dy = 0</p><p>This gives: d(xy) - d(ln|y|) = 0</p><p>∴ xy - ln|y| = C</p><p><strong>Step 6:</strong> Apply initial condition y = 1, x = 1:</p><p>(1)(1) - ln(1) = C → C = 1</p><p><strong>Step 7:</strong> Solution is xy - ln(y) = 1</p><p>When y = 2: 2x - ln(2) = 1</p><p>2x = 1 + ln(2)</p><p>x = (1 + ln 2)/2</p><p>∴ Answer: B</p>
Correct Answer: B