---
title: "Solve y−2px+ayp²=0"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-y-minus-2px-plus-ayp2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-y-minus-2px-plus-ayp2
site: BCT Engineering
---

# Solve y−2px+ayp²=0

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

- Asked in: 2072 Kartik
- Video: [203. y=2px-ayp^2| Solve for x | Engineering Mathematics | B.A. BSc. B.E & B.Ed](https://www.youtube.com/watch?v=toAL6NRvwoo) — The Classroom Nepal🇳🇵
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-y-minus-2px-plus-ayp2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve: y − 2px + ayp² = 0, where p = dy/dx.

## Short answer

Equation y − 2px + ayp² = 0 is solvable for x. General solution from py = c gives y² = 2cx − ac². Singular solution from 1 − ap² = 0 gives y² = x²/a.

## Answer

The equation y − 2px + ayp² = 0 (p = dy/dx) is solvable for x.

#### Solve for x
2px = y + ayp² → x = y(1 + ap²)/(2p) = y(1/(2p) + ap/2)

#### Differentiate w.r.t. y
dx/dy = 1/p, so

1/p = (1 + ap²)/(2p) + y(−1/(2p²) + a/2) dp/dy

1/p − (1 + ap²)/(2p) = (1 − ap²)/(2p), and −1/(2p²) + a/2 = (ap² − 1)/(2p²). So

(1 − ap²)/(2p) = y (ap² − 1)/(2p²) · dp/dy

(1 − ap²)[1/(2p) + y/(2p²) · dp/dy] = 0

#### General solution
From the second factor: 1/p + (y/p²) dp/dy = 0 → dp/p = −dy/y, so py = c, p = c/y.

Substitute in the given equation:

y − 2(c/y)x + ay(c²/y²) = 0 → y² − 2cx + ac² = 0

**y² = 2cx − ac²**

#### Singular solution
The first factor gives ap² = 1, p = ±1/√a. Then the equation reduces to 2y = 2px, y = px = ±x/√a, so

**y² = x²/a**

This is also the envelope of the family: ∂/∂c gives 2x − 2ac = 0, c = x/a, and y² = 2x²/a − x²/a = x²/a.

## Check yourself

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

1. The equation y - 2px + ayp^2 = 0 is solved by writing:
   A. y in terms of x and p
   B. Clairaut's form
   C. p in terms of x
   D. x in terms of y and p, differentiating w.r.t. y
   Answer: D. x = y(1 + ap^2)/(2p). (Key point: Approach)

2. The differentiated equation factorises into:
   A. (p - a)(...) = 0
   B. (1 + ap^2)[...] = 0
   C. (1 - ap^2)[1/(2p) + (y/(2p^2))dp/dy] = 0
   D. p(1 - ap^2) = 0
   Answer: C. The common factor is (1 - ap^2). (Key point: Factorisation)

3. From py = c, substitution gives:
   A. y^2 = cx - ac
   B. y^2 = 2cx - ac^2
   C. y = 2cx - ac
   D. y^2 = 2cx + ac^2
   Answer: B. y - 2cx/y + ac^2/y = 0, so y^2 = 2cx - ac^2. (Key point: General solution)

4. The factor 1 - ap^2 = 0 gives the singular solution:
   A. y^2 = x^2/a
   B. y = ax
   C. y^2 = ax^2
   D. y^2 = 4ax
   Answer: A. p = ±1/sqrt(a) and y = px, so y^2 = x^2/a. (Key point: Singular solution)

## Related in Mathematics I

- [Solve (1+y²)+(x−e^(tan⁻¹y))dy/dx=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-1-plus-y2-x-minus-e-atan-y) — Mathematics I, chapter 4, asked 6×
- [Solve dy/dx = y/x + tan(y/x)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-dy-dx-y-over-x-tan) — Mathematics I, chapter 4, asked 6×
- [Solve dy/dx − 2y tan x = y² tan x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-bernoulli-2y-tan-x) — Mathematics I, chapter 4, asked 6×
- [Solve (1+x²)dy/dx+y=e^(tan⁻¹x)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-1-plus-x2-dy-dx-etan) — Mathematics I, chapter 4, asked 6×
- [Solve x dy/dx + y log y = xy eˣ](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x-dy-dx-y-log-y) — Mathematics I, chapter 4, asked 4×
- [Solve x log x dy/dx + y = 2 log x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x-log-x-dy-dx) — Mathematics I, chapter 4, asked 4×

## Related in other subjects

- [RK-4: dy/dx = (1+x²)y](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Ark4-y-1-x2) — Numerical Method, chapter 6
- [Polar coordinates: ∫∫(x²+y²) dy dx](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apolar-coordinates-x2-y2-dy) — Mathematics, chapter 2, asked 5×
- [Evaluate ∫∫ e^(y/x) dy dx](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-y-x-dy) — Mathematics, chapter 2
- [Evaluate ∫∫ e^(x²) dx dy](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-x2-dx-dy) — Mathematics, chapter 2