Differential Equations
Homogeneous Differential Equations
Grade 12
Question:
<p>If the general solution of the differential equation \(y' = \dfrac{y}{x} + \Phi\!\left(\dfrac{x}{y}\right)\), for some function \(\Phi\), is given by \(y\ln|cx| = x\), where \(c\) is an arbitrary constant, then \(\Phi(2)\) is equal to</p>
<p>4</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(-4\)</p>
<p>\(-\dfrac{1}{4}\)</p>
Step-by-Step Solution
Key Concept: Differentiate the given solution y·ln|cx| = x implicitly to extract the form of Φ, then match coefficients with the original DE to identify Φ(2).
<p><strong>Step 1:</strong> Differentiate the solution y·ln|cx| = x with respect to x using the product rule:</p><p>y'·ln|cx| + y·(1/x) = 1</p><p><strong>Step 2:</strong> Solve for y':</p><p>y' = (1 - y/x)/ln|cx|</p><p><strong>Step 3:</strong> Rewrite the original DE as y' = y/x + Φ(x/y). Rearrange our result:</p><p>y' = y/x + (1 - y/x)/ln|cx| = y/x + [1/ln|cx| - y/(x·ln|cx|)]</p><p><strong>Step 4:</strong> From the solution y·ln|cx| = x, we have ln|cx| = x/y. Substitute this:</p><p>Φ(x/y) = 1/ln|cx| - y/(x·ln|cx|) = (y/x)·(1/ln|cx|) - y/(x·ln|cx|) = y/(x·ln|cx|)·[(y/x) - 1]</p><p>Since ln|cx| = x/y, we get: Φ(x/y) = (y/x)·(y/x) - (y/x)·1 = (y/x)² - (y/x)</p><p><strong>Step 5:</strong> Let t = x/y. Then Φ(t) = 1/t² - 1/t</p><p>Therefore: Φ(2) = 1/4 - 1/2 = <strong>-1/4</strong></p><p>∴ Answer: D</p>
Correct Answer: D