Page 66 - Advanced engineering mathematics

46 CHAPTER 2 Linear Second-Order Equations

The point to taking linear combinations c 1 y 1 + c 2 y 2 is to generate new solutions from y 1 and
y 2 . However, if y 2 is itself a constant multiple of y 1 ,say y 2 = ky 2 , then the linear combination
c 1 y 1 + c 2 y 2 = c 1 y 1 + c 2 ky 1 = (c 1 + c 2 k)y 1
is just a constant multiple of y 1 ,so y 2 does not contribute any new information. This leads to the
following definition.

Two functions are linearly independent on an open interval I (which can be the entire real
line) if neither function is a constant multiple of the other for all x in the interval. If one
function is a constant multiple of the other on the entire interval, then these functions are
called linearly dependent.

EXAMPLE 2.1

cos(x) and sin(x) are solutions of y + y =0 over the entire real line. These solutions are linearly
independent, because there is no number k such that cos(x) = k sin(x) or sin(x) = k cos(x) for
all x. Because these solutions are linearly independent, linear combinations c 1 cos(x) + c 2 sin(x)
give us new solutions, not just constant multiples of one of the known solutions.

There is a simple test to determine whether two solutions of equation (2.2) are linearly
independent or dependent on an open interval I. Define the Wronskian W(y 1 , y 2 ) of two solutions
y 1 and y 2 to be the 2 × 2 determinant

y 1 y 2

W(y 1 , y 2 ) = = y 1 y − y 2 y .
y
y 2 1
1 2
Often we denote this Wronskian as just W(x).
THEOREM 2.3 Properties of the Wronskian
Suppose y 1 and y 2 are solutions of equation (2.2) on an open interval I.

1. Either W(x) = 0 for all x in I,or W(x) = 0 for all x in I.
2. y 1 and y 2 are linearly independent on I if and only if W(x) = 0on I.

Conclusion (2) is called the Wronskian test for linear independence. Two solutions are lin-
early independent on I exactly when their Wronskian is nonzero on I. In view of conclusion (1),
we need only check the Wronskian at a single point of I, since the Wronskian must be either
identically zero on the entire interval or nonzero on all of I. It cannot vanish for some x and be
nonzero for others in I.

EXAMPLE 2.2
2x
2x

Check by substitution that y 1 (x) = e and y 2 (x) = xe are solutions of y − 4y + 4y = 0 for all
x. The Wronskian is
2x 2x
e xe
4x 4x 4x 4x
W(x) = 2x 2x = e + 2xe − 2xe = e ,
2e e + 2xe 2x

and this is never zero, so y 1 and y 2 are linearly independent solutions.
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October 14, 2010 14:12 THM/NEIL Page-46 27410_02_ch02_p43-76