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Calculus III - 3-Dimensional Space: Equations of Lines

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 vector. Note as we

Differential Equations - Higher Order: Basic Concepts


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 
nth order linear differential equations. As we’ll see almost all of the 2nd order material will very naturally extend out to nth order with only a little bit of new material.
So, let’s start things off here with some basic concepts for nth order linear differential equations. The most general nthorder linear differential equation is,
(1)Pn(t)y(n)+Pn1(t)y(n1)++P1(t)y+P0(t)y=G(t)
where you’ll hopefully recall that,
y(m)=dmydxm
Many of the theorems and ideas for this material require that y(n) has a coefficient of 1 and so if we divide out by Pn(t)we get,
(2)y(n)+pn1(t)y(n1)++p1(t)y+p0(t)y=g(t)
As we might suspect an IVP for an nth order differential equation will require the following n initial conditions.
(3)y(t0)=y¯0,y(t0)=y¯1,,y(n1)(t0)=y¯n1
The following theorem tells us when we can expect there to be a unique solution to the IVP given by (2) and (3).


Theorem 1

Suppose the functions p0,p1,,pn1 and g(t) are all continuous in some open interval I containing t0 then there is a unique solution to the IVP given by (2) and (3) and the solution will exist for all t in I.
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 nth order linear homogeneous differential equation,
(4)y(n)+pn1(t)y(n1)++p1(t)y+p0(t)y=0
Let’s suppose that we know y1(t),y2(t),,yn(t) are all solutions to (4) then by the an extension of the Principle of Superposition we know that
y(t)=c1y1(t)+c2y2(t)++cnyn(t)
will also be a solution to (4). The real question here is whether or not this will form a general solution to (4).
In order for this to be a general solution then we will have to be able to find constants c1,c2,,cn for any choice of t0 (in the interval I from Theorem 1) and any choice of y¯1,y¯2,,y¯n. Or, in other words we need to be able to find c1,c2,,cn that will solve,
c1y1(t0)+c2y2(t0)++cnyn(t0)=y¯0c1y1(t0)+c2y2(t0)++cnyn(t0)=y¯1c1y1(n1)(t0)+c2y2(n1)(t0)++cnyn(n1)(t0)=y¯n1
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 n x n matrix.
|y1y2yny1y2yny1(n1)y2(n1)yn(n1)|As we did back with the 2nd order material we’ll define this to be the Wronskian and denote it by,
W(y1,y2,yn)(t)=|y1y2yny1y2yny1(n1)y2(n1)yn(n1)|
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 ci we can see that we’ll have a solution provided it is not zero for any value of t=t0 that we chose to evaluate the Wronskian at. The following theorem summarizes all this up.


Theorem 2

Suppose the functions p0,p1,,pn1 are all continuous on the open interval I and further suppose that y1(t),y2(t),yn(t) are all solutions to (4). If W(y1,y2,yn)(t)0 for every t in I then y1(t),y2(t),yn(t)form a Fundamental Set of Solutions and the general solution to (4) is,
y(t)=c1y1(t)+c2y2(t)++cnyn(t)
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, (2). We first need the nth order version of a theorem we saw back in the 2nd order material.


Theorem 3

Suppose that Y1(t) and Y2(t) are two solutions to (2) and that y1(t),y2(t),yn(t) are a fundamental set of solutions to the homogeneous differential equation (4) then,
Y1(t)Y2(t)is a solution to (4) and it can be written as
Y1(t)Y2(t)=c1y1(t)+c2y2(t)++cnyn(t)
Now, just as we did with the 2nd order material if we let Y(t) be the general solution to (2) and if we let YP(t) be any solution to (2) then using the result of this theorem we see that we must have,
Y(t)=c1y1(t)+c2y2(t)++cnyn(t)+YP(t)=yc(t)+YP(t)
where, yc(t)=c1y1(t)+c2y2(t)++cnyn(t) is called the complementary solution and YP(t) 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 nth 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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