Question

If the general solution of the differential equation \(y' = \frac{y}{x} + \Phi\left(\frac{x}{y}\right)\), for some function \(\Phi\), is given by \(y\ln|cx| = x\), where c is an arbitrary constant, then \(\Phi(2)\) is equal to:

• A. \(4\)
• B. \(\frac{1}{4}\)
• C. \(-4\)
• D. \(-\frac{1}{4}\)

Hint:

From the given solution, \(\ln|cx| = x/y\); differentiate to get \(y'\) in terms of \(y/x\) and read off \(\Phi\).

Answer 1

The Correct Option is D

Given \(y\ln|cx| = x\), so \(\ln|cx| = \frac{x}{y}\).
Differentiate \(y\ln|cx| = x\): \(y'\ln|cx| + \frac{y}{x} = 1\).
Substitute \(\ln|cx| = \frac{x}{y}\): \(y'\cdot\frac{x}{y} + \frac{y}{x} = 1\), so \(y' = \frac{y}{x} - \frac{y^2}{x^2}\).
Comparing with \(y' = \frac{y}{x} + \Phi\left(\frac{x}{y}\right)\): \(\Phi\left(\frac{x}{y}\right) = -\frac{y^2}{x^2} = -\frac{1}{(x/y)^2}\), so \(\Phi(t) = -\frac{1}{t^2}\).
\(\Phi(2) = -\frac{1}{4}\).
Answer: (4) \(-\frac{1}{4}\)