Page 65 - Advanced engineering mathematics

2.1 The Linear Second-Order Equation 45

We single out exactly one of these curves if we specify its slope at (0,3). For example, if we
specify that y (0) =−1, then

y (0) = c =−1,

3
so y(x) = 2x − x + 3. This is the only solution passing through (0,3) with slope −1.
To sum up, in this example, we obtain a unique solution by specifying a point that the graph
must pass through, together with the slope this solution must have at this point.
This leads us to define the initial value problem for the linear second-order differential
equation as the problem

y + p(x)y + q(x)y = f (x); y(x 0 ) = A, y (x 0 ) = B

in which x 0 , A, and B are given. We will state, without proof, an existence theorem for this initial
value problem.
THEOREM 2.1 Existence of Solutions

Let p,q, and f be continuous on an open interval I. Then the initial value problem

y + p(x)y + q(x)y = f (x); y(x 0 ) = A, y (x 0 ) = B,
has a unique solution on I.
We now have an idea of the kind of problem we will be solving and of some conditions
under which we are guaranteed a solution. Now we want to develop a strategy to follow to solve
linear equations and initial value problems. This strategy will be in two steps, beginning with the
case that f (x) is identically zero.
The Structure of Solutions

The second-order linear homogeneous equation has the form

y + p(x)y + q(x)y = 0. (2.2)

If y 1 and y 2 are solutions and c 1 and c 2 are numbers, we call c 1 y 1 +c 2 y 2 a linear combination
of y 1 and y 2 . It is an important property of the homogeneous linear equation (2.2) that a linear
combination of solutions is again a solution.

THEOREM 2.2

Every linear combination of solutions of the homogeneous linear equation (2.2) is also a
solution.

Proof Let y 1 and y 2 be solutions, and let c 1 and c 2 be numbers. Substitute c 1 y 1 + c 2 y 2 into the
differential equation:

(c 1 y 1 + c 2 y 2 ) + p(x)(c 1 y 1 + c 2 y 2 ) + q(x)(c 1 y 1 + c 2 y 2 )

= c 1 y + c 2 y + c 1 p(x)y + c 2 p(x)y + c 1 q(x)y 1 + c 2 q(x)y 2
1
1
2
2

= c 1 y + p(x)y + q(x)y 1 + c 2 y + p(x)y + q(x)y 2
1
2
1
2
= c 1 (0) + c 2 (0) = 0
because of the assumption that y 1 and y 2 are both solutions.
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October 14, 2010 14:12 THM/NEIL Page-45 27410_02_ch02_p43-76