Linear Homogeneous Second-Order Ordinary Differential Equations Analysis and Visualization
Linear Homogeneous Second- Order Ordinary Differential Equations Analysis and Visualization
Show that for all real numbers and
is a solution to the ODE
hence is a solution to the ODE.
End of Example 10.6
The last example admits the following generalization.
If and are solutions to Eqn. (10.4)
is also a solution.
10.8 [Characteristic Polynomial]
is called the characteristic polynomial for the homogeneous ODE
The roots of the characteristic polynomial are called its characteristic roots.
The characteristic roots can be classified into one of three categories:
• Real distinct; i.e.
• Real identical; i.e.,
• Complex conjugate ; i.e., with
For instance. if and are real distinct, we see that
must be solutions to Eqn. (10.4). Even when the roots
and are identical or are
complex conjugate, we will determine
appropriate solutions. The next theorem provides these.
10.9 [General Solution]
Suppose the coefficients and are real numbers and that the characteristic polynomial has roots
and There are three possibilities for a general solution to
where and are arbitrary real numbers. Each of these solutions is defined at for all
Observe that we can set
and or and
in Example 10.6 to obtain the particular
In fact, we can do the same in Case 1 of Theorem 10.9 to get the particular solutions and These
particular solutions will play an important role in the theory of linear second -order ODEs with constant coefficients .
Determine the general solution to the ODE
to get complex conjugate roots Note that So according to Theorem 10.6, Case 3 prevails so that the general solution is
The reader should check this;, i.e., verify that
is a solution to the ODE.
End of Example 10.10
Observe that we can set in Example 10.9 to obtain the particular
and In fact, we can do the same in Case 3 of Theorem 10.9 to get the particular solutions and
In fact, the particular solutions obtained by setting in Theorem 10.6 are
distinguished by the role they play with regard solving an IVP for
Theorem 10.6 begs the question: Do these cases cover all possible solutions?
Given a set of initial conditions, say
can we be confident that one of the three cases provides a unique solution? The answer is YES as we shall soon
see. But first we examine how initial values affect solutions.
10.4 INITIAL CONDITIONS
Whereas a first-order ODE
generally admits a single solution that satisfies
second-order ODE can
have infinitely many solutions satisfying It is instructive to examine a simple ODE and see why this is so.
Consider the ODE
Calculate the characteristic roots by solving the quadratic equation
Thus we get the real repeated root
From Case 2 of Theorem 10.9, the general solution is
Now suppose we wish to determine the solution that satisfies Thus we must have
As no restriction is placed on we can only say that the solution is
Consequently, there are infinitely many solutions - one for each value of Figure 10.5(a) depicts four such solutions.
Observe how each of the solutions issues forth from the initial point Each
solution has a different slope at We
can calculate these from the corresponding solutions. Since then
For each value of we calculate the corresponding value of
the slope of the
corresponding solution at For
The following table lists these values.
Figure 10.5(b) displays the same graphs , this time labeled with the values of instead of
We see from the preceding analysis of the solutions to
that specification of a
value for and a value for
is sufficient to uniquely determine a solution. With this in mind we can define what we mean by initial conditions and
an IVP for a linear homogeneous second-order ODE.
Definition 10.11 (Initial Value Problem)
An initial value problem (IVP) for a linear homogeneous second-order ODE
consists of two things :
1. An ODE
2. Initial Conditions
where are given numbers and are called initial values or initial data.