1550078481-Ordinary_Differential_Equations__Roberts_

N-th Order Linear Differential Equations 177

zero function. Hence,

y(x) = k1y1(x) + kzyz(x) + · · · + knYn(x) = 0 for all x EI.

That is, the functions y 1 ( x), Y2 ( x), ... , Yn ( x) are linearly dependent on the

interval I , which is a contradiction.

In effect, Theorem 4.3 says, as illustrated by the proof, "If y 1 (x ), yz(x ), ... ,

Yn ( x) are solutions on the interval I of the n-th order homogeneous linear
differential equation (15), then either

(1) W(y 1 ,y 2 , ... ,yn,x) = 0 for all x EI and the solutions are linearly
dependent on I
or

(2) W(y 1 ,yz, ... ,yn,x)-=/= 0 for all x E I and the solutions are linearly

independent on I."

That is , if y 1 ( x), Y2 ( x), ... , Yn ( x) are solutions on an interval I of the same

n-th order homogeneous linear differential equation, it is not possible for their
Wronskian to be zero at one point in I and to be nonzero at another point in
I. Hence, to check a set of n solutions to (15) on an interval I to see if they
are linearly dependent on I or linearly independent on I, all we need to do is
to evaluate the Wronskian of the solutions at some convenient point in I and
see if it is zero or not.

EXAMPLE 6 Determination of Linear Dependence

or Linear Independence

The functions ex, xex, and x^2 ex are solutions on (-oo, oo) of the third-order
homogeneous linear differential equation

y(3) - 3y(2) + 3y(l) - y = 0.

Determine if they are linearly dependent or linearly independent on ( -oo, oo).

SOLUTION

By definition

xex

(x + l)ex

(x + 2)ex

x2ex

(x^2 + 2x)ex

(x^2 + 4x + 2)ex

Computing this Wronskian directly is tedious at best. However, evaluating

the Wronskian at x = 0 E ( -oo, oo) and computing, we find easily that