Page 67 - Advanced engineering mathematics

2.1 The Linear Second-Order Equation 47

In many cases, it is obvious whether two functions are linearly independent or dependent.
However, the Wronskian test is important, as we will see shortly (for example, in Section 2.3.1
and in the proof of Theorem 2.4).
We are now ready to determine what is needed to find all solutions of the homo-
geneous linear equation y + p(x)y + q(x)y = 0. We claim that, if we can find two

linearly independent solutions, then every solution must be a linear combination of these two
solutions.

THEOREM 2.4

Let y 1 and y 2 be linearly independent solutions of y + py + qy = 0 on an open interval I. Then

every solution on I is a linear combination of y 1 and y 2 .

This provides a strategy for finding all solutions of the homogeneous linear equation on an
open interval.
1. Find two linearly independent solutions y 1 and y 2 .
2. The linear combination

y = c 1 y 1 + c 2 y 2
contains all possible solutions.
For this reason, we call two linearly independent solutions y 1 and y 2 a fundamen-
tal set of solutions on I, and we call c 1 y 1 + c 2 y 2 the general solution of the differential
equation on I.
Once we have the general solution c 1 y 1 + c 2 y 2 of the differential equation, we can
find the unique solution of an initial value problem by using the initial conditions to
determine c 1 and c 2 .

Proof Let ϕ be any solution of y + py +qy =0on I. We want to show that there are numbers

c 1 and c 2 such that ϕ = c 1 y 1 + c 2 y 2 .

Choose any x 0 in I.Let ϕ(x 0 )= A and ϕ (x 0 )= B. Then ϕ is the unique solution of the initial
value problem
y + py + qy = 0; y(x 0 ) = A, y (x 0 ) = B.

Now consider the two algebraic equations in the two unknowns c 1 and c 2 :
y 1 (x 0 )c 1 + y 2 (x 0 )c 2 = A

y (x 0 )c 1 + y (x 0 )c 2 = B.
1
Because y 1 and y 2 are linearly independent, their Wronskian W(x) is nonzero. These two
algebraic equations therefore yield

Ay (x 0 ) − By 2 (x 0 ) By 1 (x 0 ) − Ay (x 0 )

c 1 = 2 and c 2 = 1 .
W(x 0 ) W(x 0 )
With these choices of c 1 and c 2 , c 1 y 1 + c 2 y 2 is also a solution of the initial value problem.
Since this problem has the unique solution ϕ, then ϕ = c 1 y 1 + c 2 y 2 , as we wanted to
show.

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October 14, 2010 14:12 THM/NEIL Page-47 27410_02_ch02_p43-76