Homogeneous Linear ODEs — Mathematics Knowledge System

Overview

A linear ODE is homogeneous when its forcing term is zero: aₙ(x)y⁽ⁿ⁾ + ⋯ + a₁(x)y′ + a₀(x)y = 0. The solutions of an n-th order homogeneous linear ODE form an n-dimensional vector space, so any solution is a linear combination of n linearly independent 'basis' solutions — a fact that follows directly from linearity of the differential operator. For constant-coefficient equations, the basis solutions come from the roots of the characteristic polynomial, exactly as in the second-order case but generalized to any order. Linear independence of candidate solutions is checked with the Wronskian, a determinant that is nonzero exactly when the solutions are independent.

Intuition

Because the equation is linear and homogeneous, if y₁ and y₂ are both solutions, so is any combination C₁y₁+C₂y₂ — plugging a sum into a linear operator just adds the results, and the zero right-hand side stays zero. This 'superposition' means you never have to find every solution individually: find enough independent basis solutions (matching the order of the equation) and every other solution is automatically some blend of them, just as every vector in 3D space is a blend of three independent basis vectors.

Formal Definition

Definition

The general n-th order linear homogeneous ODE with constant coefficients and its characteristic polynomial:

a_n y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0

General homogeneous linear ODE

a_n r^n + a_{n-1}r^{n-1} + \cdots + a_1 r + a_0 = 0

Characteristic equation (trial y=e^{rx})

y = C_1 y_1 + C_2 y_2 + \cdots + C_n y_n

General solution as a linear combination

W(y_1,\ldots,y_n)(x) \ne 0 \iff y_1,\ldots,y_n \text{ linearly independent}

Wronskian independence test

Worked Examples

Form the characteristic cubic from y=e^{rx}.
$r^3 - 6r^2 + 11r - 6 = 0$
Factor by testing small integer roots: r=1 works.
$(r-1)(r^2-5r+6) = (r-1)(r-2)(r-3) = 0$
Three distinct real roots give three independent exponential solutions.
$r_1=1,\ r_2=2,\ r_3=3$

Answer: y = C₁eˣ + C₂e^(2x) + C₃e^(3x) (verify one term: for y=e^x, y‴-6y″+11y′-6y = e^x(1-6+11-6)=0 ✓; similarly for e^{2x}: 8-24+22-6=0 ✓, and e^{3x}: 27-54+33-6=0 ✓).

Practice Problems

Difficulty 6/10

Solve y‴ − y′ = 0.

Difficulty 6/10

Solve y″ + 4y = 0 with y(0)=0, y′(0)=6.

Difficulty 7/10

A simplified beam-deflection model gives y⁗ − y = 0. Show that y = e^x is a solution and state the full characteristic-root structure.

Quiz

Summary

A homogeneous linear ODE has zero forcing term, and its solution set is an n-dimensional vector space by superposition.
For constant coefficients, characteristic-polynomial roots (real distinct, repeated, or complex) generate a basis of n independent solutions.
The Wronskian determinant confirms whether a candidate set of solutions is actually linearly independent.

References

WebsiteWikipedia — Linear differential equation