Differential Equations
Linear Differential Equations
Grade None
Question:
<p>Given <br/> \(y^2\,dx + \left(x - \dfrac{1}{y}\right)dy = 0\)<br/>Find the solution of the differential equation.</p>
<p>\(x = \dfrac{1}{y} + 1 - e^{1/y}\)</p>
<p>\(x = \dfrac{1}{y} + 1 + ce^{1/y}\)</p>
<p>\(x = \dfrac{1}{y} + 1 + ce^{-1/y}\)</p>
<p>\(x = -\dfrac{1}{y} + 1 + ce^{1/y}\)</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 potential function F(x,y).
<p><strong>Step 1:</strong> Rewrite the equation in standard form M dx + N dy = 0:</p><p>M = y² and N = x - 1/y</p><p><strong>Step 2:</strong> Check exactness: ∂M/∂y = 2y and ∂N/∂x = 1</p><p>Since ∂M/∂y ≠ ∂N/∂x, this is NOT exact. Multiply by integrating factor μ = 1/y²:</p><p><strong>Step 3:</strong> Multiplying by 1/y²: dx + (x/y² - 1/y³)dy = 0</p><p>Now M = 1, N = x/y² - 1/y³</p><p>∂M/∂y = 0 and ∂N/∂x = 1/y² ... Still checking: Actually multiply by μ = y instead.</p><p><strong>Step 3 (Corrected):</strong> Multiply entire equation by y: y³ dx + (xy - 1/y)dy = 0</p><p>Better approach: Rearrange as y²dx + x dy = dy/y</p><p><strong>Step 4:</strong> Recognize that d(xy) = y dx + x dy, so: y² dx + x dy - dy/y = 0</p><p>This gives: d(xy) = dy/y, integrating both sides:</p><p>xy = ln|y| + C</p><p><strong>∴ Answer: xy - ln|y| = C or xy = ln|y| + C</strong></p>
Correct Answer: C