---
title: "Lagrange's method for a PDE"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:lagranges-method-for-a-pde-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/lagranges-method-for-a-pde-two
site: BCT Engineering
---

# Lagrange's method for a PDE

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2082 Kartik (new course), 4 marks
- Video: [Lagrange's Method to solve Partial Differential Equation | Msc Mathematics #mathematics](https://www.youtube.com/watch?v=MYSqmQ03l7A) — Spectrum of Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/lagranges-method-for-a-pde-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Apply Lagrange's method to solve the PDE y²p − xyq = x(z − 2y), where the symbols have their usual meanings.

## Short answer

Auxiliary equations dx/y^2 = dy/(−xy) = dz/[x(z−2y)]. First integral from −x dx = y dy gives x^2 + y^2 = a. Second from dy/(−y) = dz/(z−2y) yields yz − y^2 = b. General solution: φ(x^2 + y^2, yz − y^2) = 0.

## Answer

For Pp + Qq = R the auxiliary equations are dx/P = dy/Q = dz/R. Here P = y², Q = −xy, R = x(z − 2y).

dx/y² = dy/(−xy) = dz/[x(z − 2y)]

**First integral.** Take the first two ratios: dx/y² = dy/(−xy), so −x dx = y dy.

x dx + y dy = 0 → x² + y² = a

**Second integral.** Take the last two ratios (cancel x from the third): dy/(−y) = dz/(z − 2y).

```text
dz/dy = −(z − 2y)/y      →     dz/dy + z/y = 2
Multiply by y:           y dz/dy + z = 2y   →   d(yz)/dy = 2y
Integrate:               yz = y² + b        →   yz − y² = b
```

**General solution**

**φ( x² + y², yz − y² ) = 0**

where φ is an arbitrary function. Equivalently, yz − y² = f(x² + y²).

## Check yourself

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

1. P, Q, R are:
   A. y^2, xy, x(z−2y)
   B. y^2, −xy, x(z − 2y)
   C. x, y, z
   D. y, −x, z
   Answer: B. Read from the PDE. (Key point: P,Q,R)

2. First integral:
   A. x − y = a
   B. xy = a
   C. x^2 + y^2 = a
   D. x^2 − y^2 = a
   Answer: C. x dx + y dy = 0. (Key point: First)

3. Second integral:
   A. yz + y^2 = b
   B. z − y = b
   C. yz = b
   D. yz − y^2 = b
   Answer: D. d(yz) = 2y dy. (Key point: Second)

4. General solution:
   A. φ(x^2 + y^2, yz − y^2) = 0
   B. φ(x,y)=0
   C. φ(x^2+y^2, z) = 0
   D. z = x^2
   Answer: A. Arbitrary function. (Key point: General)

## Close variants set in other exams

- [Charpit's method for a PDE](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Acharpits-method-for-a-pde-two) — Mathematics III, chapter 4, asked 3×
- [Charpit's method for a PDE](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Acharpits-method-for-a-pde) — Mathematics III, chapter 4, asked 3×

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