Questions: Higher-Order Linear Differential Equations

Question 1

The characteristic polynomial of a 4th-order linear ODE factors as (r − 3)²(r² + 4). How many linearly independent homogeneous solutions exist, and what are they?

• A. 2 solutions: e^(3x) and cos(2x)
• B. 3 solutions: e^(3x), xe^(3x), and cos(2x)
• C. 4 solutions: e^(3x), xe^(3x), cos(2x), and sin(2x)
• D. 5 solutions: e^(3x), xe^(3x), x²e^(3x), cos(2x), and sin(2x)

Answer: (C)

A 4th-order ODE requires exactly 4 linearly independent homogeneous solutions. The double root r = 3 contributes two: e^(3x) and xe^(3x). The complex pair r = ±2i contributes two more: cos(2x) and sin(2x). Total: 4, matching the order. Option D would arise from a triple root at r = 3, which this polynomial does not have.

Question 2

A student claims that a 5th-order linear constant-coefficient homogeneous ODE must have exactly 5 linearly independent solutions. Is this correct?

• A. No — the number of independent solutions depends on whether the characteristic roots are real or complex
• B. No — repeated roots reduce the total number of independent solutions
• C. Yes — this is guaranteed by the structure of linear ODEs, and the multiplicity rules ensure exactly 5 independent solutions are produced from the characteristic polynomial
• D. Yes — but only if all 5 characteristic roots are distinct real numbers

Answer: (C)

An nth-order linear ODE always has exactly n linearly independent homogeneous solutions. The multiplicity rules are specifically designed to ensure this count works out: a root of multiplicity k contributes exactly k independent solutions (e^(rx), xe^(rx), ..., x^(k-1)e^(rx)), so the total across all roots equals the degree of the characteristic polynomial, which equals the order of the ODE.

Question 3

A repeated real root r of multiplicity k contributes exactly k linearly independent solutions: e^(rx), xe^(rx), x²e^(rx), ..., x^(k-1)e^(rx).

• T. True
• F. False

Answer: True

This is the multiplicity rule for repeated real roots. The factor x^j multiplied by e^(rx) for j = 0, 1, ..., k-1 produces k solutions that are linearly independent (can be verified by Wronskian). This rule is essential for ensuring the solution space has the correct dimension — equal to the order of the ODE.

Question 4

Solving higher-order linear ODEs requires fundamentally new methods beyond those developed for second-order equations.

• T. True
• F. False

Answer: False

Higher-order equations are a direct extension, not a new theory. The same principle applies: find the characteristic polynomial by substituting y = e^(rx), factor it, apply the same root rules (real, complex, repeated), collect exactly n independent homogeneous solutions, and add a particular solution. The only change is that the characteristic polynomial is degree n rather than 2. All existing techniques (undetermined coefficients, variation of parameters) extend directly.

Question 5

How does the degree of the characteristic equation relate to the order of the ODE, and why must you collect exactly that many linearly independent homogeneous solutions?

Model answer: The characteristic polynomial has the same degree as the order of the ODE — an nth-order equation yields a degree-n polynomial with exactly n roots (counting multiplicity, over ℂ). Each root contributes one or more linearly independent solutions according to its type and multiplicity, and the total always equals n. This is required because the solution space of an nth-order linear homogeneous ODE is an n-dimensional vector space — it has exactly n degrees of freedom, and the general solution must span that space.

This dimension argument is why the multiplicity rules exist: they ensure that even when roots repeat (which could naively suggest fewer solutions), we still extract the right number by multiplying by x, x², etc. Missing even one independent solution means the general solution is incomplete — it will fail to satisfy all possible initial conditions.