---
title: "Solve a partial differential equation"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:solve-a-partial-differential-equation-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-a-partial-differential-equation-two
site: BCT Engineering
---

# Solve a partial differential equation

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2081 Chaitra (new course), 2 marks
- Video: [Oxford Calculus: Solving Simple PDEs](https://www.youtube.com/watch?v=uztjxrGY6Jw) — Tom Rocks Maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-a-partial-differential-equation-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Solve the partial differential equation ∂²u/∂x² = 12x²(t + 1) subject to u(0, t) = cos 2t and uₓ(0, t) = sin t.

## Short answer

Integrate twice: u = x⁴(t+1) + x sin t + cos 2t. This satisfies the given PDE and initial conditions.

## Answer

Integrate ∂²u/∂x² = 12x²(t + 1) twice with respect to x. The "constants" of integration are functions of t.

```text
∂u/∂x = 4x³(t + 1) + φ(t)
u     = x⁴(t + 1) + x φ(t) + ψ(t)
```

**Using the conditions**

- uₓ(0, t) = φ(t) = sin t
- u(0, t) = ψ(t) = cos 2t

**Answer:** u(x, t) = x⁴(t + 1) + x sin t + cos 2t

Check: u_xx = 12x²(t + 1), u(0, t) = cos 2t and uₓ(0, t) = sin t.

## Check yourself

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

1. When integrating u_xx = 12x^2(t+1) twice with respect to x, what are the 'constants' of integration?
   A. Arbitrary functions of t
   B. Arbitrary functions of x
   C. Arbitrary functions of both x and t
   D. Plain numbers only
   Answer: A. Only x is integrated, so t acts like a constant and the integration 'constants' can be functions of t. (Key point: Integrating with respect to x)

2. Which condition fixes the function phi(t) multiplying x in u = x^4(t+1) + x phi(t) + psi(t)?
   A. u_xx = 12x^2(t+1)
   B. u(x,0) = 0
   C. u(0,t) = cos 2t
   D. u_x(0,t) = sin t
   Answer: D. Differentiating gives u_x = 4x^3(t+1) + phi(t); at x = 0 this equals phi(t), so phi = sin t. (Key point: Using u_x condition)

3. What does u(0,t) = cos 2t determine?
   A. psi(t) = cos 2t
   B. psi(t) = sin t
   C. The x^4 coefficient
   D. phi(t) = cos 2t
   Answer: A. At x = 0 the first two terms vanish, leaving u(0,t) = psi(t) = cos 2t. (Key point: Using u condition)

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- [Solve a partial differential equation](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-a-partial-differential-equation) — Mathematics III, chapter 4
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