Question

Which of the following is the singular solution of \(p=\log(px-y)\), \(p=\dfrac{dy}{dx}\)?

• A. \(c=\log(cx-y)\)
• B. \(c=e^{(cx-y)}\)
• C. \(y=x(\log x-1)\)
• D. \(x=y(\log y-1)\)

Hint:

For singular solution of \(y=px+f(p)\), use \(\frac{\partial y}{\partial p}=0\) to eliminate \(p\).

Answer 1

The Correct Option is C

Concept:
The given equation is \[ p=\log(px-y) \] where \[ p=\frac{dy}{dx} \] This equation is of Clairaut-type form after rearrangement.

Step 1: Remove logarithm. Given, \[ p=\log(px-y) \] Taking exponential on both sides, \[ e^p=px-y \] Therefore, \[ y=px-e^p \]

Step 2: Treat \(p\) as parameter. This is in the form \[ y=px+f(p) \] where \[ f(p)=-e^p \] For singular solution, differentiate with respect to \(p\): \[ \frac{\partial y}{\partial p}=0 \]

Step 3: Differentiate with respect to \(p\). \[ y=px-e^p \] \[ \frac{\partial y}{\partial p}=x-e^p \] For singular solution, \[ x-e^p=0 \] \[ x=e^p \] Taking logarithm, \[ p=\log x \]

Step 4: Substitute \(p=\log x\) in \(y=px-e^p\). \[ y=x\log x-e^{\log x} \] Since \[ e^{\log x}=x \] we get \[ y=x\log x-x \] \[ y=x(\log x-1) \]

Step 5: Final answer. \[ \boxed{y=x(\log x-1)} \]