---
title: "Solve xy(p²+2)=2py³+x³"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-xy-p2-plus-2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-xy-p2-plus-2
site: BCT Engineering
---

# Solve xy(p²+2)=2py³+x³

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

- Asked in: 2069 Asar
- Video: [xy^2(p^2+2)=2py^3+x^3 #Solvableforp #DifferentialEquations L438](https://www.youtube.com/watch?v=kc_EqorqMqQ) — Maths Pulse - Chinnaiah Kalpana
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-xy-p2-plus-2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Find the general solution of the differential equation xy(p²+2) = 2py³ + x³.

## Short answer

The intended equation xy(p²+2) = p(x²+2y²) factorises as (py−x)(px−2y)=0. Solutions are y² = x² + c and y = cx². Combined general solution: (y²−x²−c)(y−cx²)=0.

## Answer

**Note:** the equation as printed, xy(p² + 2) = 2py³ + x³, does not factorise. The standard textbook equation it appears to come from is xy(p² + 2) = p(x² + 2y²), with p = dy/dx, whose right side was garbled in the scan. This evident intended version is solved here.

#### Factorise
xy p² + 2xy − p(x² + 2y²) = 0

xy p² − (x² + 2y²)p + 2xy = 0

(py − x)(px − 2y) = 0

(Check: expanding gives p²xy − 2py² − px² + 2xy ✓.)

#### Case 1: py − x = 0
y dy/dx = x → y dy = x dx → y² = x² + c

**y² − x² = c**

#### Case 2: px − 2y = 0
dy/y = 2 dx/x → ln y = 2 ln x + ln c

**y = cx²**

#### General solution
**(y² − x² − c)(y − cx²) = 0**, where c is an arbitrary constant (the same symbol is used for both factors, as is usual for such products).

## Check yourself

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

1. The equation factorises as:
   A. (py - x)(px - 2y) = 0
   B. (p - x)(p - 2y) = 0
   C. (py + x)(px + 2y) = 0
   D. (px - y)(py - 2x) = 0
   Answer: A. Expanding gives p^2 xy - 2py^2 - px^2 + 2xy. (Key point: Factorisation)

2. py - x = 0 gives:
   A. y^2 - x^2 = c
   B. y^2 + x^2 = c
   C. y = cx^2
   D. y = cx
   Answer: A. y dy = x dx. (Key point: Case 1)

3. px - 2y = 0 gives:
   A. y^2 = cx
   B. y = cx^2
   C. y = cx^(1/2)
   D. y = c/x^2
   Answer: B. dy/y = 2dx/x so y = cx^2. (Key point: Case 2)

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