---
title: "Solve y−3px+ayp²=0"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-y-minus-3px-plus-ayp2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-y-minus-3px-plus-ayp2
site: BCT Engineering
---

# Solve y−3px+ayp²=0

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

- Asked in: 2070 Chaitra
- Video: [Differential equation First Order & Higher Degree | Solvable For x | Hindi | Part (20)](https://www.youtube.com/watch?v=Uj-ZkLxOHOY) — Mathematics Tutorial by Neha
- 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-3px-plus-ayp2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve: y − 3px + ayp² = 0.

## Short answer

Equation y − 3px + ayp² = 0 yields parametric solution: x = y(1+ap²)/(3p) and y⁴p²(2−ap²) = C₁. Additional line solutions: y = ±√(2/a) x.

## Answer

y − 3px + ayp² = 0, with p = dy/dx.

**Note:** unlike the similar equation y − 2px + ayp² = 0, this one does not give a closed form in x and y. It is solved as written, and the solution is parametric in p.

#### Solve for x
3px = y(1 + ap²) → x = y(1 + ap²)/(3p)

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

1/p = (1 + ap²)/(3p) + y(−1/(3p²) + a/3) dp/dy

1/p − (1 + ap²)/(3p) = (2 − ap²)/(3p), and −1/(3p²) + a/3 = (ap² − 1)/(3p²). Hence

(2 − ap²)/(3p) = y (ap² − 1)/(3p²) · dp/dy

The variables separate:

dy/y = (ap² − 1)/[p(2 − ap²)] dp

#### Integrate
By partial fractions, (ap² − 1)/[p(2 − ap²)] = −1/(2p) + (a/2)·p/(2 − ap²). So

ln y = −½ ln p − ¼ ln(2 − ap²) + ln C

y⁴ p² (2 − ap²) = C₁

#### Solution (parametric in p)
From the integral, y = C₁^(1/4) p^(−1/2)(2 − ap²)^(−1/4), and x follows from x = y(1 + ap²)/(3p).

**x = y(1 + ap²)/(3p), y⁴p²(2 − ap²) = C₁**

#### Extra solutions
If 2 − ap² = 0 (p = ±√(2/a), constant), the separated equation is satisfied with dp = 0, and the original equation becomes 3(y − px) = 0. So **y = ±√(2/a) x** are also solutions (these are lines through the origin).

## Check yourself

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

1. Why is the solution of y - 3px + ayp^2 = 0 left in parametric form?
   A. p is a constant
   B. Eliminating p does not give a closed form in x and y
   C. It is a Clairaut equation
   D. The equation has no solution
   Answer: B. The answer is x = y(1+ap^2)/(3p) with y^4 p^2 (2 - ap^2) = C1. (Key point: Form of answer)

2. After differentiation w.r.t. y, the variables separate as:
   A. dy/y = dp/p
   B. dy/y = (ap^2 - 1)/[p(2 - ap^2)] dp
   C. dy = (2 - ap^2) dp
   D. dy/y = -dp/(1 + ap^2)
   Answer: B. This comes from (2 - ap^2)/(3p) = y(ap^2 - 1)/(3p^2) dp/dy. (Key point: Separable equation)

3. Integration gives which relation?
   A. y p = C1
   B. y^2 p = C1
   C. y^4 p (2 + ap^2) = C1
   D. y^4 p^2 (2 - ap^2) = C1
   Answer: D. ln y = -1/2 ln p - 1/4 ln(2 - ap^2) + ln C. (Key point: Integral)

4. Which extra solutions come from 2 - ap^2 = 0?
   A. y = ±sqrt(2/a) x
   B. y = ±x/sqrt(a)
   C. y = ±sqrt(a/2) x
   D. y = 0 only
   Answer: A. p = ±sqrt(2/a) constant and y = px. (Key point: Extra solutions)

## Related in Mathematics I

- [Solve y−2px+ayp²=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-y-minus-2px-plus-ayp2) — Mathematics I, chapter 4