---
title: "Linear independence of solutions of x²y'''−3xy''+3y'=0"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:linear-independence-solutions-x2y-3xy"
source: https://www.bctengineer.com/q/semester_2/mathematics/linear-independence-solutions-x2y-3xy
site: BCT Engineering
---

# Linear independence of solutions of x²y'''−3xy''+3y'=0

Mathematics · Chapter 4: Solution of Differential Equations in Series and Special Functions

- Asked in: 2069 Poush, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/linear-independence-solutions-x2y-3xy
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Show that the solutions of x²y''' − 3xy'' + 3y' = 0, (x > 0) are linearly independent.

## Short answer

The Cauchy-Euler equation has solutions y₁=1, y₂=x², y₃=x⁴. Their Wronskian W(x)=16x³≠0 for x>0 proves linear independence.

## Answer

Equation: x²y‴ − 3xy″ + 3y′ = 0 (x > 0). It is an equidimensional (Cauchy–Euler) equation, so try solutions of the form y = xᵐ.

#### Find the solutions
Because y itself does not appear, y = 1 (a constant) is a solution (m = 0). For the others, y′ = mxᵐ⁻¹, y″ = m(m−1)xᵐ⁻², y‴ = m(m−1)(m−2)xᵐ⁻³:

```text
m(m−1)(m−2) − 3m(m−1) + 3m = 0
m[(m−1)(m−2) − 3(m−1) + 3] = 0
m[m² − 6m + 8] = 0
m(m − 2)(m − 4) = 0   →   m = 0, 2, 4
```
Three solutions: y₁ = 1, y₂ = x², y₃ = x⁴.

Check, e.g. y = x⁴: x²(24x) − 3x(12x²) + 3(4x³) = 24x³ − 36x³ + 12x³ = 0 ✓.

#### Wronskian
```text
        | 1   x²   x⁴   |
W(x) =  | 0   2x   4x³  | = 1·(2x·12x² − 4x³·2) = 24x³ − 8x³ = 16x³
        | 0   2    12x² |
```
For x > 0, W(x) = 16x³ ≠ 0. A non-zero Wronskian means the solutions 1, x², x⁴ are linearly independent.

**General solution: y = c₁ + c₂x² + c₃x⁴.**

## Check yourself

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

1. The auxiliary equation gives m(m − 2)(m − 4) = 0. Which solutions follow?
   A. 1, x, x^2
   B. x^2, x^4, x^6
   C. 1, x^2, x^4
   D. e^x, e^(2x), e^(4x)
   Answer: C. m = 0, 2, 4 gives y = 1, x^2, x^4. (Key point: Roots)

2. What is the Wronskian of 1, x^2, x^4?
   A. 0
   B. 16x^3
   C. 8x^3
   D. 24x^3
   Answer: B. W = 2x·12x^2 − 4x^3·2 = 24x^3 − 8x^3 = 16x^3. (Key point: Wronskian)

3. Why are the solutions linearly independent for x > 0?
   A. W = 16x^3 is non-zero
   B. W is zero everywhere
   C. They have different powers of x, so W is irrelevant
   D. Because there are three of them
   Answer: A. A non-zero Wronskian proves linear independence. (Key point: Conclusion)

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