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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...
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Differential Equations - Systems: Real Eigenvalues - ii



Example 3 Find the solution to the following system.x′1=x1+2x2x1(0)=0x′2=3x1+2x2x2(0)=−4

We first need to convert this into matrix form. This is easy enough. Here is the matrix form of the system.
x→′=(1232)x→,x→(0)=(0−4)This is just the system from the first example and so we’ve already got the solution to this system. Here it is.
x→(t)=−85e−t(−11)−45e4t(23)Now, since we want the solution to the system not in matrix form let’s go one step farther here. Let’s multiply the constants and exponentials into the vectors and then add up the two vectors.
x→(t)=(85e−t−85e−t)−(85e4t125e4t)=(85e−t−85e4t−85e−t−125e4t)Now, recall,
x→(t)=(x1(t)x2(t))So, the solution to the system is then,
x1(t)=85e−t−85e4tx2(t)=−85e−t−125e4t
Let’s work another example.


Example 4 Solve the following IVP.x→′=(−514−2)x→,x→(0)=(12)

So, the first thing that we need to do is find the eigenvalues for the matrix.
det(A−λI)=|−5−λ14−2−λ|=λ2+7λ+6=(λ+1)(λ+6)⇒λ1=−1,λ2=−6Now let’s find the eigenvectors for each of these.
λ1=−1 :
We’ll need to solve,
(−414−1)(η1η2)=(00)⇒−4η1+η2=0⇒η2=4η1The eigenvector in this case is,
η→=(η14η1)⇒η→(1)=(14),η1=1λ2=−6 :
We’ll need to solve,
(1144)(η1η2)=(00)⇒η1+η2=0⇒η1=−η2The eigenvector in this case is,
η→=(−η2η2)⇒η→(2)=(−11),η2=1Then general solution is then,
x→(t)=c1e−t(14)+c2e−6t(−11)Now, we need to find the constants. To do this we simply need to apply the initial conditions.
(12)=x→(0)=c1(14)+c2(−11)Now solve the system for the constants.
c1−c2=14c1+c2=2}⇒c1=35,c2=−25The solution is then,
x→(t)=35e−t(14)−25e−6t(−11)
Now let’s find the phase portrait for this system.


Example 5 Sketch the phase portrait for the following system.x→′=(−514−2)x→

From the last example we know that the eigenvalues and eigenvectors for this system are,
λ1=−1η→(1)=(14)λ2=−6η→(2)=(−11)This one is a little different from the first one. However, it starts in the same way. We’ll first sketch the trajectories corresponding to the eigenvectors. Notice as well that both of the eigenvalues are negative and so trajectories for these will move in towards the origin as t increases. When we sketch the trajectories we’ll add in arrows to denote the direction they take as t increases. Here is the sketch of these trajectories.
Now, here is where the slight difference from the first phase portrait comes up. All of the trajectories will move in towards the origin as t increases since both of the eigenvalues are negative. The issue that we need to decide upon is just how they do this. This is actually easier than it might appear to be at first.
The second eigenvalue is larger than the first. For large and positive t’s this means that the solution for this eigenvalue will be smaller than the solution for the first eigenvalue. Therefore, as t increases the trajectory will move in towards the origin and do so parallel to η→(1). Likewise, since the second eigenvalue is larger than the first this solution will dominate for large and negative t’s. Therefore, as we decrease t the trajectory will move away from the origin and do so parallel to η→(2).
Adding in some trajectories gives the following sketch.
In these cases we call the equilibrium solution (0,0) a node and it is asymptotically stable. Equilibrium solutions are asymptotically stable if all the trajectories move in towards it as t increases.

Note that nodes can also be unstable. In the last example if both of the eigenvalues had been positive all the trajectories would have moved away from the origin and in this case the equilibrium solution would have been unstable.
Before moving on to the next section we need to do one more example. When we first started talking about systems it was mentioned that we can convert a higher order differential equation into a system. We need to do an example like this so we can see how to solve higher order differential equations using systems.



Example 6 Convert the following differential equation into a system, solve the system and use this solution to get the solution to the original differential equation.2y″+5y′−3y=0,y(0)=−4y′(0)=9

So, we first need to convert this into a system. Here’s the change of variables,
x1=yx′1=y′=x2x2=y′x′2=y″=32y−52y′=32x1−52x2The system is then,
x→′=(0132−52)x→x→(0)=(−49)where,
x→(t)=(x1(t)x2(t))=(y(t)y′(t))Now we need to find the eigenvalues for the matrix.
det(A−λI)=|−λ132−52−λ|=λ2+52λ−32=12(λ+3)(2λ−1)λ1=−3,λ2=12Now let’s find the eigenvectors.
λ1=−3 :
We’ll need to solve,
(313212)(η1η2)=(00)⇒3η1+η2=0⇒η2=−3η1The eigenvector in this case is,
η→=(η1−3η1)⇒η→(1)=(1−3),η1=1λ2=12:
We’ll need to solve,
(−12132−3)(η1η2)=(00)⇒−12η1+η2=0⇒η2=12η1The eigenvector in this case is,
η→=(η112η1)⇒η→(2)=(21),η1=2The general solution is then,
x→(t)=c1e−3t(1−3)+c2et2(21)Apply the initial condition.
(−49)=x→(0)=c1(1−3)+c2(21)This gives the system of equations that we can solve for the constants.
c1+2c2=−4−3c1+c2=9}⇒c1=−227,c2=−37The actual solution to the system is then,
x→(t)=−227e−3t(1−3)−37et2(21)Now recalling that,
x→(t)=(y(t)y′(t))we can see that the solution to the original differential equation is just the top row of the solution to the matrix system. The solution to the original differential equation is then,
y(t)=−227e−3t−67et2Notice that as a check, in this case, the bottom row should be the derivative of the top row.

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