---
title: "Solve y = px − √(m²+p²)"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-clairaut-y-px-sqrt-m2-p2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-clairaut-y-px-sqrt-m2-p2
site: BCT Engineering
---

# Solve y = px − √(m²+p²)

Mathematics I · Chapter 4: Ordinary Differential Equations and their Applications

- Asked in: 2074 Asoj; 2069 Chaitra (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-clairaut-y-px-sqrt-m2-p2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve: y = px − √(m²+p²), where p = dy/dx.

## Short answer

Clairaut’s equation. Differentiation gives (dp/dx)[x − p/√(m²+p²)]=0. dp/dx=0 yields general solution y = cx − √(m²+c²). The other factor yields singular solution m²x² + y² = m².

## Answer

y = px − √(m² + p²) is Clairaut's equation, y = px + f(p), with f(p) = −√(m² + p²).

#### General solution
Differentiate with respect to x (p = dy/dx):

p = p + x dp/dx − [p/√(m² + p²)] dp/dx

(dp/dx)[x − p/√(m² + p²)] = 0

The factor dp/dx = 0 gives p = c (constant). Substituting in the equation:

**y = cx − √(m² + c²)** (general solution, a family of straight lines)

#### Singular solution
The other factor gives x = p/√(m² + p²). Hence

x²(m² + p²) = p² → p²(1 − x²) = m²x² → p = mx/√(1 − x²) (taking this sign)

Then √(m² + p²) = m/√(1 − x²), and

y = px − √(m² + p²) = m x²/√(1 − x²) − m/√(1 − x²) = −m√(1 − x²)

Squaring: y² = m²(1 − x²), that is

**m²x² + y² = m²** (singular solution)

This ellipse x² + y²/m² = 1 is the envelope of the family of lines. (It is also found by eliminating c between y = cx − √(m² + c²) and 0 = x − c/√(m² + c²).)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. y = px - sqrt(m^2 + p^2) is which type of equation?
   A. Exact equation
   B. Clairaut's equation
   C. Bernoulli's equation
   D. Homogeneous equation
   Answer: B. It has the form y = px + f(p). (Key point: Recognising the type)

2. How is the general solution obtained?
   A. Replace p by x
   B. Replace p by y/x
   C. Replace p by 1/x
   D. Replace p by a constant c
   Answer: D. From dp/dx = 0, p = c, so y = cx - sqrt(m^2 + c^2), a family of straight lines. (Key point: General solution)

3. Which factor produces the singular solution?
   A. x - p/sqrt(m^2 + p^2) = 0
   B. p = 0
   C. dp/dx = 0
   D. x = 0
   Answer: A. The second factor gives x = p/sqrt(m^2+p^2), which leads to the envelope. (Key point: Singular solution source)

4. The singular solution is:
   A. x^2 + y^2 = m^2
   B. y^2 = m^2 x^2
   C. m^2 x^2 - y^2 = m^2
   D. m^2 x^2 + y^2 = m^2
   Answer: D. Eliminating p gives y^2 = m^2(1 - x^2), an ellipse that is the envelope of the lines. (Key point: Singular solution)

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