---
title: "PDE by separation of variables"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:pde-by-separation-of-variables-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/pde-by-separation-of-variables-two
site: BCT Engineering
---

# PDE by separation of variables

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2081 Chaitra (new course), 2 marks
- Video: [Separation of Variables Method | Partial Differential Equation | Example  & Concepts  by GP Sir](https://www.youtube.com/watch?v=DgvQzfLpaTY) — Dr.Gajendra Purohit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/pde-by-separation-of-variables-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find the solution of the equation 3uₓ + 2u_y = 0 by the method of separation of the variables.

## Short answer

Using separation u=XY, we get 3X'/X = -2Y'/Y = constant k, leading to u(x,y) = C e^(k(x - 3y/2)) = C e^(k(2x-3y)/2) with arbitrary constants C and k.

## Answer

Let u = X(x)Y(y). Then uₓ = X′Y, u_y = XY′, and 3uₓ + 2u_y = 0 becomes

3X′Y + 2XY′ = 0 → 3X′/X = −2Y′/Y = 3k  (a constant)

- X′/X = k gives X = C₁ e^(kx)
- Y′/Y = −3k/2 gives Y = C₂ e^(−3ky/2)

**Answer:** u(x, y) = C exp[ k(x − 3y/2) ] = C e^(k(2x − 3y)/2), with C and k arbitrary constants.

Check: uₓ = ku and u_y = −(3k/2)u, so 3uₓ + 2u_y = 3ku − 3ku = 0.

## Check yourself

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

1. With u = X(x)Y(y), 3u_x + 2u_y = 0 becomes which separated form?
   A. X'/Y' = 3/2
   B. 3X'/X = -2Y'/Y
   C. 3X'/X = 2Y'/Y
   D. 3X'Y = 2XY' with no division
   Answer: B. Divide 3X'Y + 2XY' = 0 by XY and move the Y term across: 3X'/X = -2Y'/Y, both equal to a constant. (Key point: Separation step)

2. Why can both sides of 3X'/X = -2Y'/Y be set equal to a constant?
   A. Both sides are always zero
   B. One side depends only on x and the other only on y
   C. X and Y are both constants
   D. The PDE is linear
   Answer: B. A function of x alone equals a function of y alone only if both are constant. (Key point: Why a constant)

3. With 3X'/X = 3k, what is Y?
   A. Y = C2 e^(-2ky/3)
   B. Y = C2 e^(-3ky/2)
   C. Y = C2 e^(3ky/2)
   D. Y = C2 e^(-ky)
   Answer: B. -2Y'/Y = 3k gives Y'/Y = -3k/2, so Y = C2 e^(-3ky/2). (Key point: Solution form)

## Related in Mathematics III

- [PDE by separation of variables](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Apde-by-separation-of-variables) — Mathematics III, chapter 4

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