In this section we need to take a look at the equation of a line in R 3 R 3 . As we saw in the previous section the equation y = m x + b y = m x + b does not describe a line in R 3 R 3 , instead it describes a plane. This doesn’t mean however that we can’t write down an equation for a line in 3-D space. We’re just going to need a new way of writing down the equation of a curve. So, before we get into the equations of lines we first need to briefly look at vector functions. We’re going to take a more in depth look at vector functions later. At this point all that we need to worry about is notational issues and how they can be used to give the equation of a curve. The best way to get an idea of what a vector function is and what its graph looks like is to look at an example. So, consider the following vector function. → r ( t ) = ⟨ t , 1 ⟩ r → ( t ) = ⟨ t , 1 ⟩ A vector function is a function that takes one or more variables, one in this case, and returns a...
We’ll start this chapter off with the material that most text books will cover in this chapter. We will take the material from the Second Order chapter and expand it out to
order linear differential equations. As we’ll see almost all of the 2nd order material will very naturally extend out to order with only a little bit of new material.
So, let’s start things off here with some basic concepts for order linear differential equations. The most general order linear differential equation is,
where you’ll hopefully recall that,
Many of the theorems and ideas for this material require that has a coefficient of 1 and so if we divide out by we get,
As we might suspect an IVP for an order differential equation will require the following initial conditions.
The following theorem tells us when we can expect there to be a unique solution to the IVP given by and .
Theorem 1
Suppose the functions and are all continuous in some open interval containing then there is a unique solution to the IVP given by and and the solution will exist for all in .
This theorem is a very natural extension of a similar theorem we saw in the 1st order material.
Next we need to move into a discussion of the order linear homogeneous differential equation,
Let’s suppose that we know are all solutions to then by the an extension of the Principle of Superposition we know that
will also be a solution to . The real question here is whether or not this will form a general solution to .
In order for this to be a general solution then we will have to be able to find constants for any choice of (in the interval from Theorem 1) and any choice of . Or, in other words we need to be able to find that will solve,
Just as we did for 2nd order differential equations, we can use Cramer’s Rule to solve this and the denominator of each the answers will be the following determinant of an x matrix.
As we did back with the 2nd order material we’ll define this to be the Wronskian and denote it by,
Now that we have the definition of the Wronskian out of the way we need to get back to the question at hand. Because the Wronskian is the denominator in the solution to each of the we can see that we’ll have a solution provided it is not zero for any value of that we chose to evaluate the Wronskian at. The following theorem summarizes all this up.
Theorem 2
Suppose the functions are all continuous on the open interval and further suppose that are all solutions to . If for every in then form a Fundamental Set of Solutions and the general solution to is,
Recall as well that if a set of solutions form a fundamental set of solutions then they will also be a set of linearly independent functions.
We’ll close this section off with a quick reminder of how we find solutions to the nonhomogeneous differential equation, . We first need the order version of a theorem we saw back in the 2nd order material.
Theorem 3
Suppose that and are two solutions to and that are a fundamental set of solutions to the homogeneous differential equation then,
is a solution to and it can be written as
is a solution to and it can be written as
Now, just as we did with the 2nd order material if we let be the general solution to and if we let be any solution to then using the result of this theorem we see that we must have,
where, is called the complementary solution and is called a particular solution.
Over the course of the next couple of sections we’ll discuss the differences in finding the complementary and particular solutions for order differential equations in relation to what we know about 2nd order differential equations. We’ll see that, for the most part, the methods are the same. The amount of work involved however will often be significantly more.
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