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Linear Homogeneous SecondOrder Ordinary Differential Equations Analysis and Visualization
Linear Homogeneous SecondOrder Ordinary Differential Equations Analysis and Visualization
Example
10.6
Show that for all real numbers and
is a solution to the ODE
Solution: Calculate
Then
hence is a solution to the ODE.
End of Example 10.6
The last example admits the following generalization.
Theorem
10.7 [Linearity]
If and
are solutions to Eqn. (10.4)
then for all real numbers and the linear combination
is also a solution.
Definition
10.8 [Characteristic Polynomial]
The 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.
Theorem
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
Proof:
Observe that we can set
and or and
in Example 10.6 to obtain the particular
solutions and
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 .
Example
10.10:
Determine the general solution to the ODE
Solution: The characteristic equation is
As this polynomial doesn't factor
readily, use the quadratic formula to calculate the
characteristic roots
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
solutions
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 firstorder ODE
generally admits a single solution that satisfies
a
secondorder 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
which implies
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
instance, when
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 secondorder ODE.
Definition 10.11 (Initial Value Problem)
An initial value problem (IVP) for a linear homogeneous secondorder ODE
consists of two things :
1. An ODE
and
2. Initial Conditions
where are given numbers and are called initial values or initial data.
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