---
title: "Solve y=2px+p³y²"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-y-2px-plus-p3y2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-y-2px-plus-p3y2
site: BCT Engineering
---

# Solve y=2px+p³y²

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

- Asked in: 2073 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-y-2px-plus-p3y2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve: y = 2px + p³y², where p = dy/dx.

## Short answer

Solving y = 2px + p³y² for x and differentiating yields general solution y² = 2cx + c³. The envelope gives singular solution 27y⁴ + 32x³ = 0.

## Answer

y = 2px + p³y², with p = dy/dx. It is solvable for x.

#### Solve for x
x = (y − p³y²)/(2p) = y/(2p) − p²y²/2

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

1/p = 1/(2p) − (y/(2p²)) dp/dy − p²y − p y² dp/dy

1/(2p) + p²y = −(dp/dy)[y/(2p²) + py²]

Taking out common factors: (1 + 2p³y)/(2p) = −(dp/dy) · y(1 + 2p³y)/(2p²)

So (1 + 2p³y)[1/(2p) + (y/(2p²)) dp/dy] = 0.

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

Substituting in the given equation:

y = 2cx/y + c³y²/y³ = 2cx/y + c³/y → y² = 2cx + c³

**y² = 2cx + c³**

#### Singular solution
The factor 1 + 2p³y = 0 gives the envelope of this family. Differentiating y² = 2cx + c³ w.r.t. c: 0 = 2x + 3c², so c² = −2x/3. Then

y⁴ = c²(2x + c²)² = (−2x/3)(4x/3)² = −32x³/27

**27y⁴ + 32x³ = 0** (singular solution)

## Check yourself

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

1. y = 2px + p^3 y^2 is solved by expressing:
   A. y in terms of x and p
   B. Clairaut's form
   C. x in terms of y and p
   D. p in terms of y only
   Answer: C. x = y/(2p) - p^2 y^2/2. (Key point: Method)

2. The differentiated equation splits into factors:
   A. (1 - 2p^3 y) and p
   B. (p + y) and (p - y)
   C. (1 + p^3) and y
   D. (1 + 2p^3 y) and [1/(2p) + (y/(2p^2))dp/dy]
   Answer: D. Both sides contain (1 + 2p^3 y)/(2p). (Key point: Common factor)

3. From py = c, the general solution is:
   A. y^2 = 2cx + c^3
   B. y^2 = 2cx - c^3
   C. y^2 = cx + c^3
   D. y = 2cx + c^3
   Answer: A. y = 2cx/y + c^3/y. (Key point: General solution)

4. The singular solution is:
   A. 27y^4 - 32x^3 = 0
   B. 27y^2 + 32x^3 = 0
   C. 27y^4 + 32x^3 = 0
   D. y^4 = 27x^3
   Answer: C. Differentiating w.r.t. c gives c^2 = -2x/3, and eliminating gives y^4 = -32x^3/27. (Key point: Singular solution)

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