Question:medium

The solution of \(y = xp + \frac{m}{p}\) where \(p = \frac{dy}{dx}\) is

Show Hint

For any equation in the form of \(y = xp + f(p)\) (Clairaut's equation), the general solution is always found by simply replacing the parameter \(p\) with a constant \(c\). This is a valuable shortcut in exams.
Updated On: Feb 18, 2026
  • \(y = \frac{m}{c}\)
  • \(y = xc\)
  • \(y = xc - \frac{m}{c}\)
  • \(y = xc + \frac{m}{c}\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Introduction:
The differential equation \(y = xp + \frac{m}{p}\) is a Clairaut's equation, which has the form \(y = xp + f(p)\), where \(p = \frac{dy}{dx}\). Clairaut's equations have a general and a singular solution. The general solution is found by replacing \(p\) with a constant \(c\).
Step 2: Method:
To solve \(y = xp + f(p)\):
1. Differentiate the equation with respect to \(x\).
2. Use \(\frac{dy}{dx} = p\).
3. Factor the resulting equation. One factor leads to the general solution, the other to the singular solution.
Step 3: Solution:
Given the equation: \[ y = xp + \frac{m}{p} \quad (*).\] Differentiate with respect to \(x\): \[ \frac{dy}{dx} = \left(1 \cdot p + x \cdot \frac{dp}{dx}\right) - \frac{m}{p^2} \frac{dp}{dx} \]Since \(\frac{dy}{dx} = p\): \[ p = p + x \frac{dp}{dx} - \frac{m}{p^2} \frac{dp}{dx} \]Subtract \(p\) from both sides: \[ 0 = x \frac{dp}{dx} - \frac{m}{p^2} \frac{dp}{dx} \]Factor out \(\frac{dp}{dx}\): \[ \left(x - \frac{m}{p^2}\right) \frac{dp}{dx} = 0 \]Two possibilities:
Case 1: \(\frac{dp}{dx} = 0\)
Integrating gives \(p = c\), where \(c\) is a constant.
Substitute \(p=c\) into (*): \[ y = xc + \frac{m}{c} \]This matches option (4).
Case 2: \(x - \frac{m}{p^2} = 0\)
This leads to \(p^2 = \frac{m}{x}\) or \(p = \pm\sqrt{\frac{m}{x}}\). Substituting this gives the singular solution, which is not an option.
Step 4: Answer:
The general solution is obtained by replacing \(p\) with \(c\), yielding \(y = xc + \frac{m}{c}\).
Was this answer helpful?
0


Questions Asked in CUET (PG) exam