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

# PDE by separation of variables

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2082 Kartik (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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Solve the PDE y²uₓ − x²u_y = 0 by the method of separation of variables.

## Short answer

Assume u = X(x)Y(y); separation gives X'/(x^2 X) = Y'/(y^2 Y) = k. Solving yields X = C1 e^(k x^3/3), Y = C2 e^(k y^3/3). General solution: u(x,y) = C exp(k(x^3 + y^3)/3) with arbitrary constants C and k.

## Answer

Assume u(x, y) = X(x)Y(y), where X depends on x only and Y on y only. Then uₓ = X′Y and u_y = XY′.

```text
y² X′ Y − x² X Y′ = 0
Divide by x² y² X Y:     X′/(x² X) = Y′/(y² Y) = k      (constant)
```

Each side depends on a different variable, so both equal a constant k.

- X′ = k x² X gives X = C₁ e^(kx³/3)
- Y′ = k y² Y gives Y = C₂ e^(ky³/3)

**Answer:** u(x, y) = C exp[ k(x³ + y³)/3 ], where C = C₁C₂ and k are arbitrary constants.

Check: uₓ = kx²u and u_y = ky²u, so y²uₓ − x²u_y = kx²y²u − kx²y²u = 0.

## Check yourself

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

1. Assume:
   A. u = X + Y
   B. u = X(x,y)
   C. u = XY' 
   D. u = X(x)Y(y)
   Answer: D. Product form. (Key point: Assumption)

2. Dividing by x^2 y^2 XY gives:
   A. X'/(x^2 X) = Y'/(y^2 Y) = k
   B. X'/X = Y'/Y = k
   C. X'/(xX) = Y'/(yY)
   D. X' = Y'
   Answer: A. Each side one variable. (Key point: Separation)

3. u =
   A. C exp[k(x + y)]
   B. C exp[k(x^3 + y^3)/3]
   C. C exp[k(x^2+y^2)/2]
   D. C(x^3+y^3)
   Answer: B. Integrate each ODE. (Key point: Result)

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