---
title: "Classify a second order PDE"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:classify-a-second-order-pde-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/classify-a-second-order-pde-two
site: BCT Engineering
---

# Classify a second order PDE

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2082 Kartik (new course), 2 marks
- Video: [Classification of Second-Order PDE | Partial Differential Equation Example  & Concepts  by GP Sir](https://www.youtube.com/watch?v=2DmqG66fVhg) — 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/classify-a-second-order-pde-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Classify the PDE: x² ∂²u/∂x² + 3 ∂²u/∂x∂y + x ∂²u/∂y² + 17 ∂u/∂t = 100u.

## Short answer

Coefficients: A = x^2, B = 3, C = x; discriminant B^2 − 4AC = 9 − 4x^3. The PDE is hyperbolic for x^3 < 9/4, parabolic on x^3 = 9/4, and elliptic for x^3 > 9/4.

## Answer

For A u_xx + B u_xy + C u_yy + (lower-order terms) = 0, the sign of B² − 4AC decides the type: negative is elliptic, zero is parabolic and positive is hyperbolic.

In the given equation the second order terms are x² u_xx + 3 u_xy + x u_yy. The terms 17 ∂u/∂t and −100u are of lower order (first order and zero order), so they do not affect the classification; the principal part involves x and y.

A = x², B = 3, C = x

B² − 4AC = 9 − 4x³

- 9 − 4x³ > 0, that is x³ < 9/4 (x < 1.31): **hyperbolic**
- 9 − 4x³ = 0, that is x³ = 9/4 (x ≈ 1.31): **parabolic**
- 9 − 4x³ < 0, that is x³ > 9/4 (x > 1.31): **elliptic**

So the type changes with the region: hyperbolic for x < (9/4)^(1/3), parabolic on x = (9/4)^(1/3), elliptic for x > (9/4)^(1/3).

## Check yourself

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

1. Which terms decide the type?
   A. Second-order terms only
   B. All terms
   C. First-order terms
   D. Zero-order terms
   Answer: A. Lower-order terms are ignored. (Key point: Principal part)

2. With A = x^2, B = 3, C = x:
   A. 9 − 4x^2
   B. B^2 − 4AC = 9 − 4x^3
   C. 3 − 4x^3
   D. 9 + 4x^3
   Answer: B. 4·x^2·x = 4x^3. (Key point: Discriminant)

3. For x > (9/4)^(1/3) the equation is:
   A. hyperbolic
   B. parabolic
   C. elliptic
   D. unclassified
   Answer: C. Discriminant negative. (Key point: Region)

## Related in Mathematics III

- [Classify a second order PDE](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aclassify-a-second-order-pde) — Mathematics III, chapter 4

## Related in other subjects

- [Second order constant delay filter](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aconstant-delay-second-order) — Filter Design, chapter 2, asked 4×
- [Second order maxima in single slit](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asingle-slit-second-order-maxima-ratio) — Engineering Physics, chapter 4, asked 2×
- [First and second order finite difference schemes](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Afd-first-second-order-schemes) — Computational Techniques in Civil Engineering, chapter 5, asked 2×
- [Saint-Venant, second-order explicit scheme](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asaint-venant-second-order-explicit) — Computational Techniques in Civil Engineering, chapter 5, asked 2×