---
title: "Solve (x−a)p²+(x−y)p−y=0"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-x-minus-a-p2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-x-minus-a-p2
site: BCT Engineering
---

# Solve (x−a)p²+(x−y)p−y=0

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

- Asked in: 2072 Chaitra
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-x-minus-a-p2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve: (x−a)p² + (x−y)p − y = 0, where p = dy/dx.

## Short answer

Substitute u = x+y to obtain Clairaut form w = XP − a/P with w = x+y−2a, X = x−a. General solution: x+y−2a = c(x−a) − a/c. Singular: (x+y−2a)² = −4a(x−a).

## Answer

(x − a)p² + (x − y)p − y = 0, with p = dy/dx.

#### Reduce to Clairaut's form
Put u = x + y, so du/dx = 1 + p, p = u' − 1. Substituting:

(x − a)(u' − 1)² + (2x − u)(u' − 1) − (u − x) = 0

Expanding and simplifying gives

(x − a)u'² + (2a − u)u' − a = 0

Divide by u' and rearrange:

u = (x − a)u' + 2a − a/u'

Let X = x − a and w = u − 2a = x + y − 2a (so dw/dX = u' = P). Then

**w = XP − a/P**, with P = dw/dX

This is Clairaut's equation.

#### General solution
Replace P by c: w = cX − a/c, that is

**x + y − 2a = c(x − a) − a/c**

(Equivalently y = (c − 1)x − a(c − 1)²/c, a family of straight lines. Check: with p = m, the original equation holds for y = mx − am²/(m + 1).)

#### Singular solution
Differentiate w = cX − a/c with respect to c: 0 = X + a/c², so c² = −a/X. Then a/c = −cX, w = 2cX and w² = 4c²X² = −4aX:

**(x + y − 2a)² = −4a(x − a)**

This parabola is the envelope of the family.

## Check yourself

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

1. Which substitution reduces the equation to Clairaut's form?
   A. u = x + y
   B. u = x - y
   C. u = xy
   D. u = y/x
   Answer: A. With u = x + y, p = u' - 1. (Key point: Substitution)

2. With X = x - a and w = x + y - 2a, the equation becomes:
   A. w = XP + a/P
   B. w = XP - aP
   C. w = X/P - a
   D. w = XP - a/P
   Answer: D. From u = (x-a)u' + 2a - a/u'. (Key point: Clairaut form)

3. General solution of a Clairaut equation is obtained by:
   A. Replacing P by X
   B. Integrating P w.r.t. X
   C. Replacing P by an arbitrary constant c
   D. Setting X = 0
   Answer: C. x + y - 2a = c(x - a) - a/c. (Key point: General solution)

4. The singular solution is:
   A. (x + y - 2a)^2 = -4a(x - a)
   B. (x + y - 2a)^2 = 4a(x - a)
   C. x + y - 2a = 0
   D. (x + y)^2 = -4ax
   Answer: A. Eliminating c from w = cX - a/c and X + a/c^2 = 0 gives w^2 = -4aX. (Key point: Singular solution)

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