Professor Murray

Slope & Direction Fields

Slide Duration:

Section 1: First-Order Equations
Linear Equations

1h 7m 21s

Intro
0:00
Lesson Objectives
0:19
How to Solve Linear Equations
2:54
Calculate the Integrating Factor
2:58
Changes the Left Side so We Can Integrate Both Sides
3:27
Solving Linear Equations
5:32
Further Notes
6:10
If P(x) is Negative
6:26
Leave Off the Constant
9:38
The C Is Important When Integrating Both Sides of the Equation
9:55
Example 1
10:29
Example 2
22:56
Example 3
36:12
Example 4
39:24
Example 5
44:10
Example 6
56:42
Separable Equations

35m 11s

Intro
0:00
Lesson Objectives
0:19
Some Equations Are Both Linear and Separable So You Can Use Either Technique to Solve Them
1:33
Important to Add C When You Do the Integration
2:27
Example 1
4:28
Example 2
10:45
Example 3
14:43
Example 4
19:21
Example 5
27:23
Slope & Direction Fields

1h 11m 36s

Intro
0:00
Lesson Objectives
0:20
If You Can Manipulate a Differential Equation Into a Certain Form, You Can Draw a Slope Field Also Known as a Direction Field
0:23
How You Do This
0:45
Solution Trajectories
2:49
Never Cross Each Other
3:44
General Solution to the Differential Equation
4:03
Use an Initial Condition to Find Which Solution Trajectory You Want
4:59
Example 1
6:52
Example 2
14:20
Example 3
26:36
Example 4
34:21
Example 5
46:09
Example 6
59:51
Applications, Modeling, & Word Problems of First-Order Equations

1h 5m 19s

Intro
0:00
Lesson Overview
0:38
Mixing
1:00
Population
2:49
Finance
3:22
Set Variables
4:39
Write Differential Equation
6:29
Solve It
10:54
11:47
Example 1
13:29
Example 2
24:53
Example 3
32:13
Example 4
42:46
Example 5
55:05
Autonomous Equations & Phase Plane Analysis

1h 1m 20s

Intro
0:00
Lesson Overview
0:18
Autonomous Differential Equations Have the Form y' = f(x)
0:21
Phase Plane Analysis
0:48
y' < 0
2:56
y' > 0
3:04
If we Perturb the Equilibrium Solutions
5:51
Equilibrium Solutions
7:44
8:06
Solutions Will Tend Away From Unstable Equilibria
9:32
Semistable Equilibria
10:59
Example 1
11:43
Example 2
15:50
Example 3
28:27
Example 4
31:35
Example 5
43:03
Example 6
49:01
Section 2: Second-Order Equations
Distinct Roots of Second Order Equations

28m 44s

Intro
0:00
Lesson Overview
0:36
Linear Means
0:50
Second-Order
1:15
Homogeneous
1:30
Constant Coefficient
1:55
Solve the Characteristic Equation
2:33
Roots r1 and r2
3:43
To Find c1 and c2, Use Initial Conditions
4:50
Example 1
5:46
Example 2
8:20
Example 3
16:20
Example 4
18:26
Example 5
23:52
Complex Roots of Second Order Equations

31m 49s

Intro
0:00
Lesson Overview
0:15
Sometimes The Characteristic Equation Has Complex Roots
1:12
Example 1
3:21
Example 2
7:42
Example 3
15:25
Example 4
18:59
Example 5
27:52
Repeated Roots & Reduction of Order

43m 2s

Intro
0:00
Lesson Overview
0:23
If the Characteristic Equation Has a Double Root
1:46
Reduction of Order
3:10
Example 1
7:23
Example 2
9:20
Example 3
14:12
Example 4
31:49
Example 5
33:21
Undetermined Coefficients of Inhomogeneous Equations

50m 1s

Intro
0:00
Lesson Overview
0:11
Inhomogeneous Equation Means the Right Hand Side is Not 0 Anymore
0:21
First Solve the Homogeneous Equation
1:04
Find a Particular Solution to the Inhomogeneous Equation Using Undetermined Coefficients
2:03
g(t) vs. Guess for ypar
2:42
If Any Term of Your Guess for ypar Looks Like Any Term of yhom
5:07
Example 1
7:54
Example 2
15:25
Example 3
23:45
Example 4
33:35
Example 5
42:57
Inhomogeneous Equations: Variation of Parameters

49m 22s

Intro
0:00
Lesson Overview
0:31
Inhomogeneous vs. Homogeneous
0:47
First Solve the Homogeneous Equation
1:17
Notice There is No Coefficient in Front of y''
1:27
Find a Particular Solution to the Inhomogeneous Equation Using Variation of Parameters
2:32
How to Solve
4:33
Hint on Solving the System
5:23
Example 1
7:27
Example 2
17:46
Example 3
23:14
Example 4
31:49
Example 5
36:00
Section 3: Series Solutions
Review of Power Series

57m 38s

Intro
0:00
Lesson Overview
0:36
Taylor Series Expansion
0:37
Maclaurin Series
2:36
Common Maclaurin Series to Remember From Calculus
3:35
7:58
Ratio Test
12:05
Example 1
15:18
Example 2
20:02
Example 3
27:32
Example 4
39:33
Example 5
45:42
Series Solutions Near an Ordinary Point

1h 20m 28s

Intro
0:00
Lesson Overview
0:49
Guess a Power Series Solution and Calculate Its Derivatives, Example 1
1:03
Guess a Power Series Solution and Calculate Its Derivatives, Example 2
3:14
Combine the Series
5:00
Match Exponents on x By Shifting Indices
5:11
Match Starting Indices By Pulling Out Initial Terms
5:51
Find a Recurrence Relation on the Coefficients
7:09
Example 1
7:46
Example 2
19:10
Example 3
29:57
Example 4
41:46
Example 5
57:23
Example 6
1:09:12
Euler Equations

24m 42s

Intro
0:00
Lesson Overview
0:11
Euler Equation
0:15
Real, Distinct Roots
2:22
Real, Repeated Roots
2:37
Complex Roots
2:49
Example 1
3:51
Example 2
6:20
Example 3
8:27
Example 4
13:04
Example 5
15:31
Example 6
18:31
Series Solutions

1h 26m 17s

Intro
0:00
Lesson Overview
0:13
Singular Point
1:17
Definition: Pole of Order n
1:58
Pole Of Order n
2:04
Regular Singular Point
3:25
Solving Around Regular Singular Points
7:08
Indical Equation
7:30
If the Difference Between the Roots is An Integer
8:06
If the Difference Between the Roots is Not An Integer
8:29
Example 1
8:47
Example 2
14:57
Example 3
25:40
Example 4
47:23
Example 5
1:09:01
Section 4: Laplace Transform
Laplace Transforms

41m 52s

Intro
0:00
Lesson Overview
0:09
Laplace Transform of a Function f(t)
0:18
Laplace Transform is Linear
1:04
Example 1
1:43
Example 2
18:30
Example 3
22:06
Example 4
28:27
Example 5
33:54
Inverse Laplace Transforms

47m 5s

Intro
0:00
Lesson Overview
0:09
Laplace Transform L{f}
0:13
Run Partial Fractions
0:24
Common Laplace Transforms
1:20
Example 1
3:24
Example 2
9:55
Example 3
14:49
Example 4
22:03
Example 5
33:51
Laplace Transform Initial Value Problems

45m 15s

Intro
0:00
Lesson Overview
0:12
0:14
Take the Laplace Transform of Both Sides of the Differential Equation
0:37
Plug in the Identities
1:20
Take the Inverse Laplace Transform to Find y
2:40
Example 1
4:15
Example 2
11:30
Example 3
17:59
Example 4
24:51
Example 5
36:05
Section 5: Review of Linear Algebra
Review of Linear Algebra

57m 30s

Intro
0:00
Lesson Overview
0:41
Matrix
0:54
Determinants
4:45
3x3 Determinants
5:08
Eigenvalues and Eigenvectors
7:01
Eigenvector
7:48
Eigenvalue
7:54
Lesson Overview
8:17
Characteristic Polynomial
8:47
Find Corresponding Eigenvector
9:03
Example 1
10:19
Example 2
16:49
Example 3
20:52
Example 4
25:34
Example 5
35:05
Section 6: Systems of Equations
Distinct Real Eigenvalues

59m 26s

Intro
0:00
Lesson Overview
1:11
How to Solve Systems
2:48
Find the Eigenvalues and Their Corresponding Eigenvectors
2:50
General Solution
4:30
Use Initial Conditions to Find c1 and c2
4:57
Graphing the Solutions
5:20
Solution Trajectories Tend Towards 0 or ∞ Depending on Whether r1 or r2 are Positive or Negative
6:35
Solution Trajectories Tend Towards the Axis Spanned by the Eigenvector Corresponding to the Larger Eigenvalue
7:27
Example 1
9:05
Example 2
21:06
Example 3
26:38
Example 4
36:40
Example 5
43:26
Example 6
51:33
Complex Eigenvalues

1h 3m 54s

Intro
0:00
Lesson Overview
0:47
Recall That to Solve the System of Linear Differential Equations, We find the Eigenvalues and Eigenvectors
0:52
If the Eigenvalues are Complex, Then They Will Occur in Conjugate Pairs
1:13
Expanding Complex Solutions
2:55
Euler's Formula
2:56
Multiply This Into the Eigenvector, and Separate Into Real and Imaginary Parts
1:18
Graphing Solutions From Complex Eigenvalues
5:34
Example 1
9:03
Example 2
20:48
Example 3
28:34
Example 4
41:28
Example 5
51:21
Repeated Eigenvalues

45m 17s

Intro
0:00
Lesson Overview
0:44
If the Characteristic Equation Has a Repeated Root, Then We First Find the Corresponding Eigenvector
1:14
Find the Generalized Eigenvector
1:25
Solutions from Repeated Eigenvalues
2:22
Form the Two Principal Solutions and the Two General Solution
2:23
Use Initial Conditions to Solve for c1 and c2
3:41
Graphing the Solutions
3:53
Example 1
8:10
Example 2
16:24
Example 3
23:25
Example 4
31:04
Example 5
38:17
Section 7: Inhomogeneous Systems
Undetermined Coefficients for Inhomogeneous Systems

43m 37s

Intro
0:00
Lesson Overview
0:35
First Solve the Corresponding Homogeneous System x'=Ax
0:37
Solving the Inhomogeneous System
2:32
Look for a Single Particular Solution xpar to the Inhomogeneous System
2:36
Plug the Guess Into the System and Solve for the Coefficients
3:27
Add the Homogeneous Solution and the Particular Solution to Get the General Solution
3:52
Example 1
4:49
Example 2
9:30
Example 3
15:54
Example 4
20:39
Example 5
29:43
Example 6
37:41
Variation of Parameters for Inhomogeneous Systems

1h 8m 12s

Intro
0:00
Lesson Overview
0:37
Find Two Solutions to the Homogeneous System
2:04
Look for a Single Particular Solution xpar to the inhomogeneous system as follows
2:59
Solutions by Variation of Parameters
3:35
General Solution and Matrix Inversion
6:35
General Solution
6:41
Hint for Finding Ψ-1
6:58
Example 1
8:13
Example 2
16:23
Example 3
32:23
Example 4
37:34
Example 5
49:00
Section 8: Numerical Techniques
Euler's Method

45m 30s

Intro
0:00
Lesson Overview
0:32
Euler's Method is a Way to Find Numerical Approximations for Initial Value Problems That We Cannot Solve Analytically
0:34
Based on Drawing Lines Along Slopes in a Direction Field
1:18
Formulas for Euler's Method
1:57
Example 1
4:47
Example 2
14:45
Example 3
24:03
Example 4
33:01
Example 5
37:55
Runge-Kutta & The Improved Euler Method

41m 4s

Intro
0:00
Lesson Overview
0:43
Runge-Kutta is Know as the Improved Euler Method
0:46
More Sophisticated Than Euler's Method
1:09
It is the Fundamental Algorithm Used in Most Professional Software to Solve Differential Equations
1:16
Order 2 Runge-Kutta Algorithm
1:45
Runge-Kutta Order 2 Algorithm
2:09
Example 1
4:57
Example 2
10:57
Example 3
19:45
Example 4
24:35
Example 5
31:39
Section 9: Partial Differential Equations
Review of Partial Derivatives

38m 22s

Intro
0:00
Lesson Overview
1:04
Partial Derivative of u with respect to x
1:37
Geometrically, ux Represents the Slope As You Walk in the x-direction on the Surface
2:47
Computing Partial Derivatives
3:46
Algebraically, to Find ux You Treat The Other Variable t as a Constant and Take the Derivative with Respect to x
3:49
Second Partial Derivatives
4:16
Clairaut's Theorem Says that the Two 'Mixed Partials' Are Always Equal
5:21
Example 1
5:34
Example 2
7:40
Example 3
11:17
Example 4
14:23
Example 5
31:55
The Heat Equation

44m 40s

Intro
0:00
Lesson Overview
0:28
Partial Differential Equation
0:33
Most Common Ones
1:17
Boundary Value Problem
1:41
Common Partial Differential Equations
3:41
Heat Equation
4:04
Wave Equation
5:44
Laplace's Equation
7:50
Example 1
8:35
Example 2
14:21
Example 3
21:04
Example 4
25:54
Example 5
35:12
Separation of Variables

57m 44s

Intro
0:00
Lesson Overview
0:26
Separation of Variables is a Technique for Solving Some Partial Differential Equations
0:29
Separation of Variables
2:35
Try to Separate the Variables
2:38
If You Can, Then Both Sides Must Be Constant
2:52
Reorganize These Intro Two Ordinary Differential Equations
3:05
Example 1
4:41
Example 2
11:06
Example 3
18:30
Example 4
25:49
Example 5
32:53
Fourier Series

1h 24m 33s

Intro
0:00
Lesson Overview
0:38
Fourier Series
0:42
Find the Fourier Coefficients by the Formulas
2:05
Notes on Fourier Series
3:34
Formula Simplifies
3:35
Function Must be Periodic
4:23
Even and Odd Functions
5:37
Definition
5:45
Examples
6:03
Even and Odd Functions and Fourier Series
9:47
If f is Even
9:52
If f is Odd
11:29
Extending Functions
12:46
If We Want a Cosine Series
14:13
If We Wants a Sine Series
15:20
Example 1
17:39
Example 2
43:23
Example 3
51:14
Example 4
1:01:52
Example 5
1:11:53
Solution of the Heat Equation

47m 41s

Intro
0:00
Lesson Overview
0:22
Solving the Heat Equation
1:03
Procedure for the Heat Equation
3:29
Extend So That its Fourier Series Will Have Only Sines
3:57
Find the Fourier Series for f(x)
4:19
Example 1
5:21
Example 2
8:08
Example 3
17:42
Example 4
25:13
Example 5
28:53
Example 6
42:22
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 0 answersPost by Thomas Lyles on July 3, 2020In example 3, isn't 0/0 indeterminate, rather than undefined?  Doesn't "indeterminate" indicate simply that it doesn't give us useful information?Also in example 3, you refer to slope of -1/0 as infinity.  I thought division by zero is undefined.  I understand that we can imagine zero going into one infinitely many times, but I'd like some clarification.  I understand that this may be more of an algebra issue, so perhaps you could point me to a good explanation.Also, in example 2 you have a slight error in the narration when you refer to the exponent 2x being negative, although you write it correctly as positive.thanks

### Slope & Direction Fields

Slope and Direction Fields (PDF)

### Slope & Direction Fields

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Lesson Objectives 0:20
• If You Can Manipulate a Differential Equation Into a Certain Form, You Can Draw a Slope Field Also Known as a Direction Field
• How You Do This
• Solution Trajectories 2:49
• Never Cross Each Other
• General Solution to the Differential Equation
• Use an Initial Condition to Find Which Solution Trajectory You Want
• Example 1 6:52
• Example 2 14:20
• Example 3 26:36
• Example 4 34:21
• Example 5 46:09
• Example 6 59:51

### Transcription: Slope & Direction Fields

Hi, this is Will Murray of www.educator.com and we are talking about differential equations, today we are going to talk about slope fields which are also known as direction fields.0000

Those two terms mean exactly the same thing. So if you are looking at slope fields you are looking at direction fields; if you are looking at direction fields it is exactly the same thing as a slope fields.0008

Let us see what that is all about. If the idea is that if you can manipulate a differential equation into this form right here you want to have Y′ equal to some function in terms of X and Y.0018

If the idea is that if you can manipulate into that form then you can use that form to draw what is called a slope field like I said that is the same as a direction field.0033

The way you do that is you take different points X and Y and at each point X and Y so for example you might plug in x=1 and y=1 what you can do is you can plug in those values of X and Y into the function right here the f(xy).0044

And it all spill out a slope for you. It will spill out Y′ and so what you would do is you would go over to the point 1 1 here. So there is the point 1 1 and you will draw a little mark indicating the slope of that point.0069

For example, if Y′ was 2 that would tell you at that point 1 1 you are going to draw a little mark with a slope of 2 and a slope of 2 is uphill, fairly steeply uphill.0086

I will draw a little mark like that and you would do that for each combination of X and Y that you are willing to graph so you plug in maybe another combination maybe x = 1, y = 0.0099

Supposed you get from there you get Y′ = -1 you get this by plugging them to this equation right here and then what you would do is you would go over to the 1 0 and you would draw little mark with slope -1.0112

I will draw a little mark here with slope -1 and you do that for all different points here and you find little slopes for all the points and then you gradually go through and you fill up a graph with slopes at different slopes at different points and what you get is called a slope field or a direction field.0127

We are going to have some practice with that over the course this lesson you will get the idea in the long run this is something that you want to do on a computer.0147

You want to have a computer draw slope fields or direction fields for you and then the way you used them is you kind of look at them and you can see some of the solution trajectories to the differential equations so let me talk a little more about that.0156

These solution trajectories what would happen is you will have a slope field which will show lots of different slopes like this maybe something like this so you will have all these little slopes and what you do is you kind of connect those slopes up and you will draw connect them into solution trajectories.0169

So let me draw some sample solution trajectories here and here is one right there sort of following the slopes.0192

I'm drawing curves that are parallel to the slopes wherever they passed through them so here are some solution trajectories these differential equation and if we have the actual function for this curves these functions would be solutions to the differential equation.0200

What we are finding is different solutions to the differential equations one key point here is the solutions never crosses each other so you want to be careful when you are drawing the solution trajectories to understand that they should not be crossing each other.0220

They sort of locally they are always parallel to each other they might diverge or converge but they never touch each other they never cross each other.0233

Another thing that is key to understand is that these different solution trajectories correspond to different values of the arbitrary constant and the general solution to the differential equations.0242

If you remember when we solve some of these differential equations we get things like y=(ce)^(x^2) something like that might be our general solution to the differential equation and that C was an arbitrary constant.0252

That is an arbitrary constant we do not know at least initially what the value of that constant is and so if you plugged in different value of that constant you would get different graphs and those different graphs are these different solution trajectories.0268

Each one of these red curves here is a solution to the differential equation they just correspond to different values of the arbitrary constant in the general solutions.0284

To figure out which solution trajectory you want you use an initial condition which is usually given to you as part of the package with the differential equation so you will have an initial value for X0 and Y0.0297

What that does is that gives you an initial point X0 and Y0 somewhere on the graph and that will tell you which of the solution trajectories is actually the solution that satisfies not just the differential equation but also your initial condition.0315

If we had an X0 Y0 if you have an initial condition that lets you narrow it down from the whole family of curves to just a single solution trajectory.0334

That is kind of the geometric picture there algebraically we have only seen these process where we get a general solution to a differential equation it has an arbitrary constant in it and then what we do is we take this point and we plug it in to the general solution.0344

You plug in x=x0 y=y0 into the general solution to find the value of C so that is how the initial condition can be useful algebraically is you plug it in to the general solution and solve for the value of the arbitrary constant C.0366

Geometrically what you are doing is using that single initial point to differentiate between all these different solution trajectories and pick the single one that fits your initial condition.0389

I think that is enough theory I want to jump into some examples and let us see how we can draw these slope fields or direction fields in actual practice.0402

On our first example we have to draw a direction field remember that is the same as a slope fields for the following differential equation Y′=x+2y so what I'm going to do is make a chart of combinations of X's and Y's.0412

And for each one of those I'm going to plug them in to the differential equation and I'm going to figure out a slope corresponding to it and then I'm going to try to plot all those on a set of axis.0428

I'm going to make a fairly big graph here we use a pretty big scales so I'm going to make maybe right that 1, that is -1, that is -1, that is 1, so 1, -1, 1, -1, 2, 2, -2, -2 and the idea is that I'm going to plug different combinations of X's and Y's into the differential equations here.0441

And figure out what the slopes should be at those different points so I will start out easy I will start with the X and Y or 0 and 0 and if I plug in 0 0 Y′ remember Y′ is always x+2y that is from the differential equation.0474

What that means in terms of the graph is I'm going to the point 0 0 and I'm going to draw a little hash mark with slope 0 so there is my hash mark with slope 0.0489

Let me go back and calculate a few more points here if I have x=1 y=0 then x+2y it will just be 1 then if I have x=0 y=1 then Y′ x+2y will be 2 and if they are both 1 then x+2y would be 1+2x1 so 3.0503

Let me plot those so 1 0 that is on the x axis I'm going to draw a little line with slope 1 and draw mine here a little bigger the line with slope 1, 0 1 is up here, slope 2 is steeper.0527

I'm going to make that a bit steeper probably still is not steeper enough let me try a little bit steeper there we go and at 1 1 we got the point 3 so we got the value of 3 that is really quite steep I will draw that almost vertical.0543

There is a point there that is almost vertical there we go ahead and fill in some more points here that I want to find some slopes here I'm going to try to calculate some slopes at each of those points there I think.0561

I will calculate those as fast as I can see I got -1 0 and Y′ is x+2y that is a -1 there, -1 1, x+2y will be +1 there see I have got -2 0, x+2y is -2, -2 1 is x+2y that would be 0.0577

And -2 -1 I will just go across on the southern axis there -2 -1 would be x+2y would be -2-2 -4, -1 -1 would be -3 0 -1 x+2y would be -2 1 -1 x+2y -1 and 2 -1 x+2y is 0 again.0606

And see this couple of more points there 2 0 x+2y would be 2 and 2 1 x+2y would be 4 so let me go ahead and fill in all my slopes. So let's see I have done up to here -1 0.0646

I see I have a slope of -1 so let me give this a slope of -1 there -1 1 gives me a slope of 1.0665

There is the point -1 1 and I will draw a line with slope 1, that is going uphill -2 0 gives me a slope of -2 so that one is fairly steeply downhill, -2 1 gives me a slope of 0 so that is a horizontal line there -2 -1 gives me slope of -4 very steeply downhill, -1 -1 gives me slope of -3 still pretty steeply downhill.0679

0 -1 gives me a slope of -2, 1 -1 gives me a slope of -1 so it is not so steep anymore and 2 -1 gives me a slope of 0 so that is horizontal, 2 0 gives me a slope 2 pretty steeply uphill now and 2 1 gives me a slope of 4 very steeply uphill.0711

The idea is that you go through and kind of as many points is you are willing to fill in you try to find the slope by plugging them all into this differential equation and then you draw these little hash marks everywhere through there.0735

I'm going to stop there because it gets a little tedious after this but in the long run this is something that is done very effectively on a computer.0750

We are just practicing a couple by hands so that you understand the process you understand what the computer is doing when they generate these pictures.0759

What I'm going to do on the next example is show you a computer generated picture of the slope fields you will see it looks like my slope field but it got little hot more points field in.0765

So let me just remind you what we did here we took different values of X's and Y's and then we plug each one into x+2y and we found these slopes for each combination of X and Y.0777

And then we found those on the actual graph and for each one of those points I made a little hash mark with that slope for example -1 1 that means X is -1 and Y is 1 so that is that point right there is -1 1.0791

And when you plug -1 1 into the differential equation that means X is -1y is 1 so Y′ for that particular point comes out to be -1+2 which is 1 so that is why we go with slope 1 and that is how I got this hash mark with slope 1 right there.0809

What I just did was I did that for every single point here I found the Y′ and I drew a little hash mark at each place so in the next example we are going to come back and look at the same differential equation.0833

And we are going to use the slope fields we are going to draw some a solution trajectories but instead of using my crude hand drawn graph here we are going to use one produce on a computer.0847

Let us how that turns out so in example too we are going to use a slope field for Y′ +x=x+2y from example 1 that was the slope field we generated back in the previous example If you have not worked through that one.0857

You really want to check that one out so that you understand these examples in the next slide what I'm going to do is give you a computer drawn slope fields so it will be a little bit easier for me to work with it.0872

What we will do with it is we will draw some solution trajectories to the differential equation we are going to check that this equation right here is the general solution to the differential equation.0884

I will actually plug it into the differential equation and we will check and make sure that it works that it is the general solution and then we look back at our solution trajectories.0896

And try to understand which one corresponds to which values of that arbitrary constant so that is our arbitrary constant right there.0904

We will try to figure out which solution trajectories corresponds to which values of that arbitrary constant in the general solution.0912

Let us take a look at the computer drawn slope for this differential equation here it is and this is something we started to draw by hand but I got tired after plotting I do not know 15 or so points.0919

That is why I pull out the computer drawn example the differential equation we are using was Y′=x+2y and what we have to do here is look at these direction fields.0935

These direction field remember it comes from example 1 that is how you figure out where this picture came from is if you have not just watch example and go back and check out example 1.0949

You will see that we learn how we generate this direction field, what I'm going to do is draw some solution trajectories here which means I'm going to kind of go through and connect the dots.0960

Except that the curves I draw have to be parallel to these little slope marks everywhere I draw them so I'm going to draw these solution trajectories in ready.0970

I see that there is kind of a dividing line here If I just follow those straight arrows right there I get one solution trajectory.0981

That is the easiest one to draw and I'm going to draw a few more by kind of connecting these things up in curves that are basically whenever I touch an arrow I should be parallel to that arrow.0992

Let me draw a few more of these whenever I go near an arrow I need to be following the slope dictated by that arrow.1005

Here they start more gently but they still and up going uphill and all the ones above this line here seem to be ending up going uphill.1012

Once below the line I have to kind of turn myself to make this work if I follow this arrow I got to be parallel to the arrow every time I touch an arrow I have to be parallel on it.1026

I'm sort of forced to be constantly downhill if I start below the line here so those are my solution trajectories the point of this is that anyone of these red curves would be a solution to these differential equation.1040

Anyone of these red curves because they are obeying the slopes at every single point that means that they are obeying the differential equation at every single point so there are solutions to the differential equation.1059

Every one of these red curves is a solution to the differential equation now we are asked to given a general solution to the differential equation let me write that again here.1072

That was Y is equal to -1½ x I'm just copying a slope to previous line -1½ x-1/4 plus arbitrary constant e^2x we are asked to confirm that that is a solution to these differential equation.1083

What I'm going to do is I'm going to plug that in to the differential equation and I'm just going to see if it works so I need to calculate Y′ the derivative of -1 ½ x is -1½ -¼ is a constant so it is derivative is just 0 so +0 and then ce^2x the derivative of that is 2ce^2x.1100

I see I can simplify that, that is -1½ +2c e^2x and so that is Y′ let me work out the right hand side of the differential equation.1124

The right hand side is x+2y let me label that as RHS for right hand side so that is the right hand side of the differential equation I'm going to see if I get the same thing as I got from Y′.1137

RHS is x+2y now that is x+ and I'm going to plug in my Y from up here so 2x-1½ x- ¼ +ce^2x and let me just simply that X now 2x-1½ x is -x, 2x-1 ¼ is -1 ½ and then I guess I got + 2ce^-2x.1153

And those X is cancelled let me just simplifies down to -1½ +2ce^2x and if you check back up here that Y′ and the x+2y when we work them out those are equal to each other.1187

We have a solution to the differential equation we have a solution to the differential equation we found the general solution to the differential equation and I got that just by plugging it in to the differential equation and checking.1210

The last thing we are asked to do is identify which of the solution trajectories correspond to which values of C in the differential equation so let me write the general solution to the differential equation one more time y=-1½ x - ¼ +ce^2x.1233

Now let me plug in some different values of c here if I plug in c=0 then I just get y is equal to -1½ x - ¼ that is a line that is just it is in y = mx + b form that is a line.1255

And it is a line with -1½ slope and I see that right here in the middle of my graph this line right here in the middle so let me flag this solution trajectory this line is the line corresponding to c=0.1273

That is the solution trajectory corresponding to c= 0 now what if I have a C value that is not 0 if I have C value that is positive then what I get is Y is equal to well that same equation for the line +ce^2x where that ce^2x is positive.1298

What I'm doing is I'm taking outline and I'm adding an exponential function to it which means remember e^2x let me do a little graph of e^2x, e^2x looks like this so I'm taking outline and I'm adding an exponential function to it.1323

That is going to give me this positive exponentially curving upwards solution trajectories so all of these solution trajectories correspond to positive values of the arbitrary constant in the general solution to the differential equation.1339

Now if c < 0 I will get Y is equal to a line +ce^2x but this time the C would be negative so essentially if I do -e^2x the graph of -e^2x will be curving sharply downhill.1361

I'm taking that line and I'm adding on a term that sharply downhill and that is how you get these negatively curve solution trajectories downhill below that line.1385

All these solution trajectories down here correspond to negative values of the constant in the general solution to the differential equation.1397

I have done everything I was asked to do here we are supposed to take the direction field remember we figure out the direction field in example 1 so if this graph that I started out with all the little hash mark that is a total mystery to you.1410

Go back and look at example 1 you will where it comes from except that in example 1 we just did it by hand and here I produced the nicer computer generated one.1424

Looking at this direction field I was able to connect up the hash marks and draw a solution trajectories the whole idea of that was that wherever I drew a curve it had to be parallel to these arrows at wherever was near an arrow.1435

It could not cut through an arrow like that or like that it is got to be parallel to these arrows wherever it sees an arrow so no cutting the arrows it is going to fall the arrows.1449

I started out by noticing that somebodies arrows give me a straight in the middle and then if I sort of connect the dots above the line I got the solutions trajectories occurring upwards.1461

Below the line I get all the solution trajectories that supposed to be more parallel to the arrow that would have to curve down.1471

The next thing I had to do there was to check the given general solution and make sure that that really was the solution to the differential equation.1484

In order to do that I worked out Y′ on one side and I worked out x+2y independently so here is me working out Y′ took the derivative and then I worked out x+2y where for the Y I plugged in what are solution is so I plugged that in for a solution.1494

Simplified it down and lowing behold It did check I did get the same thing as we got up here for Y′ so that's nice that those two things did check each other out that shows that we do have a solution to the differential equation.1512

But finally the problem asked us to determine which values of the arbitrary constant corresponded to which solution trajectories and I started easy by taking the constant equals 0 which just simplified me down there something in y = mx + b form so that was just Y equals a line.1530

This line right here it is c =0 this line right into the middle and then I checked if C is not 0 then we are doing is we are either adding or subtracting an exponential function to our line.1549

Here is my typical graph of an exponential function goes upper down depending on what you are adding or subtracting and so we are taking that line and so these more complicated solution to a diverge from that line.1562

By either going uphill if c > 0 or by going downhill if c < 0 so I figured out which values which kind of values of a constant will correspond to which family of solution trajectories.1574

Let us keep moving and try another one in example 3 we are going to draw to a direction field for the following differential equation Y′ is equal to -x over y.1591

I'm going to start by making a little chart here in which I'm going to list value combination of X's and Y's and then I'm going to plug them in to the differential equation and I'm going to see what I get for Y′.1603

Once I got a good number of point stored up I'm going to draw a graph of those and see if I can get a general idea of what the direction field would look like.1623

There is 1, there is -1, -1 that one is a little far out let me draw that a little closer there is -1 and there is 1, 2, 2 and so on.1633

What I'm going to do is start with different combinations of X's and Y's and try to plug them in to the differential equation.1655

I will try 0 and 0 coz that is all is good one to start with unfortunately if I try to plug that into -x over y I get 0 over 0 which is undefined so I really do not get anything useful there.1662

I will try x is 1 and y is 0 and then I get one over 0 actually -1 over 0 which I think of that it is being infinity so I think of that is being infinite slope, I think of that is being a vertical line.1678

At 1 0 I'm going to draw a vertical line here let me try 1 1 and 0 1 if I plug in 1 1 I get -x over y -1 over 1 is equal to -1 so the .1 1.1694

Draw a few more points here then I'm planning to fill in the .1 1, I just figured out has slope of -1 so I will draw a line with negative slope there.1716

The .0 1 if I plug that in to -x over y I get -0 over 1 which is 0 let me draw some more points I here -1 0 gives me 1 over 0 which is I think of that is being infinity so I think of that is another vertical line.1736

-1 1 will give me 1 over 1 plugging that in that is going to give me 1, -1 -1 will give me +1 again because it is 3 negative signs there wait sorry -1 -1 let me write this out so many negative signs It is hard to keep track -1 over -1.1758

Three negative signs will give me -1, -1 0 -1 will give me 0 over -1 it will give me a 0 again 1 -1 will give me a +1, 2 2 gives me -2 over 2 is in -1.1785

Actually I did not want to do 2 2, I did not want to go that far hang on I will just going to do 2 1 which gives me -2 over 1 which is -2 over 1 is -2, -2 1 gives me 2 over 1 which is +2 and -2 -1 gives me 2 over 1 which is, I'm sorry -2 over 1 coz there are two negatives and then a third negative is -2 and 2 -1 will give me 2 over 1 which is 2 again.1813

Let me go ahead and try to fill in all of those slopes 1 0 I already filled in that one 1 1 gives me -1 filled in that one those are ones I filled in so far 0 1 gives me a slope of 0 so I will draw a horizontal line there.1863

-1 0 gives me an infinite slope so I will draw a vertical line there -1 1 gives me a slope of 1 so there is -1 1 and 0 -1 gives me a slope of 0, 1 -1 gives me a slope of 1. Looks like I forgot -1 -1, which gives me a slope of -1.1880

1 gives me a slope of -2 so that one's more steep. -2 1 gives me a slope of -2 so that one's more steep. -2 1 gives me a slope of 2 steep line there.1892

2 -1 gives me a slope of 2 and you start to see a pattern emerging there on the axis.1901

On the horizontal axis it's all vertical. On the vertical axis its always horizontal.1918

And we get these what we are really getting is a circular pattern here it's going to look like pattern of circles here a circular slope field so that is how we do for that one.1946

Let me remind you about the individual steps there I took in different points X's and Y's different points I plug them into the differential equation -x over y and for each one I got a Y′.1966

As an example there, here is the point -1 1 so I took the point -1 1 and I plug it in there and I got negative now X over Y is -1 over 1 and so that turns out to 1.1980

I will draw a little slope mark there, I will draw a little line there which has slope of 1 and so that is how I got that little line right there.1995

And then I just went through the different places all over the plain for each place I took in X and Y my X and Y coordinates I plugged them in and I got a slope and then I draw a little hash mark with that slope.2006

I get this direction field in the next example we are going to use this direction field to try to draw some solution trajectories.2020

We are also going to compare them with some actual solutions to this differential equation and we will see what kinds of solution trajectories correspond to different values of the constant.2028

Now I am going to use a computer generated direction field in the next example but, it is just a computer generated version of the same thing that we are drawing by hand here.2040

I want to stress here that the computer is not doing anything mysterious it is just saving me time and drawing a slightly nicer graph than I could draw by hand.2050

Let us go ahead and check that out so in example 4 we are going to begin with the direction field that we drew for example 3 from the differential equation Y′ is equal to -x over y.2059

In the next slide I'm going to give you a computer generated version of that direction field we have for example 3 so we will work with that on the next slide.2070

Here is what we are going to do with it we are going to draw some solution trajectories for the differential equation.2077

We are going to take this solution to the differential equation and check that it actually works so the solution is x^2+y^2 is equal to a constant.2082

And then we will solve the differential equation subject to the initial condition y of -3 is equal to -4 so let us keep all this in mind.2093

On the next slide I have a computer generated direction field for this differential equation but it should not be anything mysterious because it is something that we already started working out by hand in the previous example.2101

Let us jump over and take look at that so in this slide what have here is a computer generated direction field for the differential equation let me remind you what the differential equation was it was Y′ is equal to -x over y.2114

The idea is that we started drawing a direction field by hand in example 3 so if this is some mess of arrows here looks completely foreign to you.2133

What you want to do is go back and look at example 3 just a couple of slides back it is where we just started generating this direction field by hand.2145

And I actually plotted points that I showed you where each one these little arrows comes from in this slide I just wanted to have a prettier one the one with more arrows so it is easier to figure out what is going on.2153

So the first thing we had to do here is to draw some solution trajectories and it is pretty clear that if you just connect up the arrows here what you are getting is circular pattern.2165

So let me draw some of these circles the idea is whenever I'm tangent to an arrow it should be parallel to the arrows that I'm touching.2174

That is what I mean by tangent so what I'm getting is exactly circles as long as I follow the slopes to the arrows all the way around.2189

Make that one a little narrower my circles are not so good this is why I have computer draw on these things where I really need them to look like so here is my circle let me draw that one more hopefully it will not get too lopsided.2205

The idea is that I'm staying parallel to the arrows everywhere I go around so I never want to have one of my solution trajectories really crossing an arrow.2232

I just want it to be parallel to an arrow all to the arrows all the way around so those are not bad solution trajectories considering my inherent level of artistic skill which pretty down low.2241

Those are my solution trajectories to the differential equation the second thing we had to do was take a prefab solution and check that it was really the general solution to the differential equation.2256

That was x^2+y^2=c that was given to us on the previous slide I want to check if that works in the differential equation so that means I'm going to work out Y′ here.2270

Now because I got a Y^2 here and not just Y by itself, I'm going to use implicit differentiation to take the derivative there.2282

If you do not remember how to do implicit differentiation either you are going to be reminded really quickly on this example or if it is completely foreign to you.2290

You might want to look at the calculus one lectures here on www.educator.com we got a whole series of lectures on calculus one.2298

And one them covers implicit differentiation so I'm going to take the derivative DYDX of both sides here x^2+y^2=c we will take the derivative of both side of the respect X here.2305

Derivative of x^2 is just 2x, the derivative of y^2 is 2yxdy by dx that is by the chain roll and then the derivative of a constant is just 0.2319

That implicit differentiation there what I'm going to do now is solve for DY by DX so I think I can cancel the 2's by dividing both sides by 2 so I just get x+y DY by DX is equal to 0.2336

Just divided both sides by 2 y DY by DX trying to solve for DY by DX is equal to -x move the -x move the x over the other side.2353

And so DY by DX is equal to -x over y of course DY by DX is the same as Y′ so I get Y′ is equal to -x over y.2364

And lowing behold I see that it does satisfy these differential equation so I now check that that is the general solution to this differential equation.2380

The last part of this example the last task we had to do was to solve the differential equation subject to the initial condition we were given in part C of the differential equation.2393

In part c of the instructions we were given y of -3 is equal to -4 which means we are going to plug in x=-3 into the general solution -3 and y is equal to -4.2404

And we are going to see what value of the arbitrary constant we get so if I plug those into we have x^2+y^2=c and so x=-3 -3^2+y=-4^2=c I see that I'm going to get c=25.2422

If I plug that back into the general solution what I get is x^2+y^2, y^2=25 now that is an equation we can readily recognize as the equation of a circle of radius π.2445

Let me just show what that kind of represents here on the picture, the picture that corresponds to plotting an initial point so It is about -3 -4 see I have got my scale a little unrealistic here.2465

I will not show -3 -4 so I'm going to try to estimate it even though that actually be a little further out I'm going to say if that is the point -3 -4.2481

And what that means is we found which of the solution trajectories goes through that initial condition so we narrow it down from the whole family of solutions to a particular solution that solves the initial condition.2491

By the way what you should be noticing here is that just by looking at the graph we figure out that the solution trajectories were going to be circles.2511

We drew these circles and but that was before we ever so the general solution however when we look at the general solution right here x^2+y^2=c I know that that is the equation of a circle.2522

So that kind of confirms what I had already guessed from my graphical intuition so this is going to be circles It's just going to be different size circles depending on what the value of C is.2535

I definitely corresponds to our graphical intuition where we have circles each one of a different radius circles are different each one of a different radius circles are varying radii radiuses.2547

Circles are different each of a different radius there so let me recap what we did here, our first thing we did was to take the graph of the slope field remember that is the same as the direction field that we had figure out on the example 3.2569

That is where this graph really came from was from example 3 in example 3 we did it by hand in this example I kind of upgrade it to a computer drawn one.2590

But you should now understand where the computer drawn picture comes from so I should not be something mysterious coming out the computer.2598

You should just be saying ok yes we just used a computer to maybe draw something a little more complicated, a little quicker, and a little more accurate and pretty than what we could have drawn by hand.2605

So we took these direction field that we had produced in example 3 we drew some solution trajectories and they keep part there was that I was going around and I was drawing there curves that kind of follow the arrows wherever they touch an arrow.2616

So wherever they touch an arrow they are supposed to be parallel tangent the arrows are supposed to be tangent factors to these curves.2634

So I cannot sort of cut across arrows like that, that would be bad instead I have to follow parallel to the arrows wherever I go.2641

I follow these arrows parallel wherever I go so we drew those solution trajectories and it turned out that we got circles.2650

The next step there was to take this given general solution x^2+y^2=c and used it to check that it really is a solution to the differential equation.2659

In or to do that I wanted to calculate I wanted to calculate Y′ and I had to do that by implicit differentiation which what I'm doing here.2672

I'm doing implicit differentiation which is something we learn how to do back in calculus 1 so if that look totally far in there then maybe check back in the lecture in the calculus 1 lectures on implicit differentiation so we took the implicit differentiation we figured out we saw for DY by DX.2679

And we got DY DX or Y′ is equal to -x over y which exactly is the differential equation that we started with so that tells us that we do had a general solution to our differential equation.2699

That was the second task there and at the same time I kind of notice that this equation we were given x^2+y^2=c that is the equation of a circle and it just has a different size radius depending on what c is.2712

Plugging in different values of C is why we have many different circles here each one corresponds to a different value of C and then finally for part we were given an initial condition here and we used that to plug in an X and Y.2728

Actually get to a specific value of C and narrow it down to a specific solution so that kind of geometrically corresponds to starting at a particular and say ok that is going me down to a single solution trajectories instead of a whole family of solution trajectories.2746

In example 5 we are going to draw a direction field for the following autonomous differential equation and let me mention this word autonomous what that means is that there is no X in the right hand side it is just the function of Y.2770

There is no X on the right hand side which really means we only need to look at the Y value so there no X's here, no X's which means we only look at the Y's.2786

Now that is going to have a very interesting consequences for us what I want to do is maybe do a little algebra on that function on the right and then we are going to try to graph that function before we actually draw our direction field.2804

What I can see over this is that I could factor on a Y and I get yxy^2-3y+2 and that is yx that would factor in to y - 1xy ñ 2 and the key point about that is that.2824

That would be 0 if Y is equal to 0 or Y is equal 1 or Y is equal to 2 now I' going to try to graph that function before I do my direction field this is called face plain analysis.2845

You will see some more of it later on in your differential equation but I just want to get used to it now what I'm going to do is I'm going to graph Y′ versus Y.2863

This is the face plain analysis this not the same as the direction field I will draw the direction field over here and try to make it very clear the distinction between the two.2876

The key thing here is in the direction field your axis are labelled X and Y and the face plain analysis we are looking at Y and Y′ and so I'm going to graph this function y^3-3y^2+2y what I know is that has 0's when Y is equal to 0 and Y is equal to 1.2889

Think I'm going to make this little bigger give myself a little more space here not going to use the negative sign so much so I'm going to give myself more space on the positive side.2912

Y is equal to 0 and Y is equal to 1 and Y is equal to 2 and I know it's a positive cubic function I know the generic shape for cubic function is it goes up and down and up gain.2936

It is something like this so what I just graph there is y^3-y^2+2y and I'm going to use that to inform my graph of Y versus X.2952

What I have done here is I have look at where Y′ is positive or negative in terms of Y and I really have to look in terms of Y and that is because there were no X's appearing on the right hand side.2971

What I notice here is that at Y is equal to 0 or 1 or 2 Y′ is equal to 0 whenever Y is equal to 0 or 1 or 2 now over on the Y versus X graph the Y is on the vertical coordinate so there is Y is equal to 0, there is 1, there is 2.2985

That means that wherever Y is equal to one those values Y′ is 0. So that means I have horizontal slopes there so let me draw some horizontal slopes there this are horizontal slopes.3011

I'm doing this at Y is equal to 0,1 and 2 and that is because those are the places where Y′ is equal to 0 so those are my horizontal slopes and now I want to kind of look in between those areas in between those cut-offs and see what kinds of values I'm going to get for Y′.3042

If I look down here where Y is less or equal to 0 I see that Y′ is negative which means that whenever Y over here on the right is less than or equal to 0 I'm going to see some negative slopes.3062

Let me draw some negative slopes here, there is a negative slope, and there is a negative slope now I do not really care what the X does because the X did not appear in the differential equation.3088

There is some more slightly negative slopes do not really care what the X does that is why I'm not really paying attention to my X values here.3104

When Y is between there is 0, there is 1, there is 2, this is Y′ is equal to 0 at y is equal to 2 now when Y is between 0 and 1 what I see here is that my Y′ is positive so I'm going to see some positive slopes in between 0 and 1 I go in between 0 and 1 here and I'm going to draw some positive slopes.3116

Draw some more gently positive slopes here because they are approaching 0 and some more gently positive slopes here because they are approaching 0 as well.3159

Now when I look at Y's between 1 and 2 so here Y, number Y is on the horizontal axis over here Y is between 1 and 2 I see that Y′ the YQ function Y′ is negative again and so I'm going to draw some negative slopes here in between 1 and 2 on the Y axis.3174

Some negative slopes, some more gently negative slopes there, so some more gently negative slopes maybe I will those in a bit.3203

Fill in some more of these so you will get a better idea here steeply negative slopes, steeply negative slopes, steeply negative slopes I'm going to show you a computer generated picture of the same thing and on the next page.3223

But think it is useful to draw one or two of these by hand in your life and now if look at y >2 y bigger than 2 I see that my Y′ is positive again because this curve is above the horizontal axis.3249

It is positive over on the Y versus X axis I'm going to put some positive ones, positive slopes here and I see that as I go for bigger and bigger value of Y they are not levelling off; they are just getting steeper.3271

Let me draw some steeper slopes here and over the negative range they are getting steeper as well. It's a little hard to make sense of all this but if you go through I see that I got a couple of horizontal lines that kind of can help me inform all the other solution trajectories.3303

I got this horizontal lines here and in sort of in between these horizontal lines on the regions below everything seems to be sloping down on the regions in between, I got these solution trajectories that are kind of sloping gradually up towards 1.3330

Then sloping gradually down towards 1 and then up above 2 everything just seems to be sloping steeply up towards infinity.3351

Let me remind you how we figured everything out here I'm going to show you a computer drawn slope field for this differential equation on the next slide but I don't want that to come as a revelation.3367

I want you to understand where these slope fields come from so what we did was the first thing we notice was this word autonomous keys us into the fact that there is no X's on the right hand side.3380

There is only Y's it means we just have to look at the values of Y we can sort of ignore the X's completely so in order to really understand this I factored it down I factored out the Y first and then I factored that until yxy - 1xy - 2.3391

Then I was going to try to graph that and I noticed that that's going to be 0 wherever y=0, y=1 or y= 2 so when I graphed that with Y on my horizontal axis I get these three 0's right here 0,1, and 2.3406

I know I got a cubic equation, cubic equations they all have that shape or that shape and since this one has positive leading coefficient I'm going to draw my cubic equation in a positive direction like this.3422

Then I can use this remember this is not the slope field this is not the graph of X versus Y, this is Y′ versus Y.3441

What have over here is Y versus X so two different graphs in two different things but I'm using this graph on left to figure out how to draw the graph on the right.3449

I look at this and I see when Y is 0 or 1 or 2 then my Y′ is 0 so whenever Y is 0, 1 or 2 that is why I got this horizontal lines whenever Y is equal to 0, Y is equal to 1 or Y is equal to 2.3459

That is why I get slopes of 0 for those three values of Y and then I look in between those values and I try to see weather this graph over on the left is negative or positive.3480

I see that it starts out negative, then it goes positive, then it goes negative and then it goes positive again and so when I look at my slopes here I'm going to draw negative slopes down here.3492

Positive slopes up here, negative slopes again and finally positive slopes and those come exactly from here, that tells me that and that tells me that all that information comes from the graph on the left.3506

About when the slopes are positive or negative by the way when I was drawing the solution trajectories remember that the solution trajectories never cross each other.3529

It's pretty careful to draw this in such a way that even though they asypmtotic to each other they get every very close to each other they never cross each other.3538

Whenever you are drawing the solution trajectories you want to make sure your curves do not cross each other if so then you have really probably misread either the equation or the direction field in some way.3547

We are going to keep going with this direction field on the next example so you want to kind of make sure you understand where this direction field comes from before you move on to example 6.3558

In example 6 I'm going to start you off with this direction field or actually a computer drawn version of this direction field and we will draw our solution trajectories and then we will start talking about limiting behaviour of different solution starting at different initial points so stick around for that.3570

In example 6 we are going to start with the field that we had from the differential equation Y′ is equal to y^2+3y^3+2y so we graph this in example 5.3592

If you have not just watch on example 5 what you want to do right now is go back and watch example 5 because that is where we generated the direction field and we are going to use that direction field to solve example 6.3604

The first thing we are going to do is draw some solution trajectories we actually so some examples back in example 5 and then we are going to describe the limiting behaviour as X goes to infinity for a bunch of different starting point we are going to do 0 1 and 1 -1, 2 ½, 3 3 ½ and 2 3.3617

We are going to look at each starting points and see where the solution trajectories would end up if we started at each one of those initial points.3643

I'm going to jump ahead to the next slide where I got the solution curves the direction already to go a computer drawn version.3652

This is the direction field for the differential equation let me remind you what that differential equation was it was Y′ is equal to y^3-3y^2+2y and we started without equation back in example 5 and we generated this direction field.3661

This direction field should not be a mystery to you, you should be kind of nodding your head in recognition if you are not if you are looking at this and wondering where it came from then just go back and watch example 5 you will see exactly where this comes from.3680

First thing we had to do is draw some solution trajectories and we already got an idea of these on the example 5 I'm just going to draw some curves that follow these arrows and the idea when I draw these curves is whenever they are touching an arrow they should be tangent and meaning parallel to the arrows.3695

I guess I should have started by drawing those equilibrium solution y=0, y=1, that's a little hard to see let me separate that from the black line of it and y= 2 and then when we go ahead and fill in some more of these.3716

I have to follow the arrows wherever I touch them it should be parallel to the arrows if I do my drawing right so this are sloping uphill but then they kind of level off and also if I do more drawing right these curves will never touch each other remember solution trajectories they might be asypmtotic to each other but they never touch and they certainly never cross each other.3739

If I draw a good ones here never touch each other they might be parallel they might asypmtotic to each other but once they will get close to each other but they never touch.3764

I see that the ones up here are just curving steeply uphill remember this slope field this direction field with all the arrows where not supposed to be a mystery where this came from.3783

I drew it on a computer but that was only to save me time this should be something you could generated by hand if you absolutely had to certainly you should understand what the computer did to draw this direction field it is not supposed to be a mystery at all where the arrows are coming from.3796

Those are my solution trajectories and now I'm going to look at these initial points and describe where they are going to end up.3813

This first one I'm starting at the point x=0 y=1 let me find that, that is right here x=0 y=1 and I see that that is just going to follow the straight line solution trajectory so the limiting behaviour there as X goes to infinity of that is just going to go to 1 because it is just staying there at y=1.3821

The limit of y(x) give myself a little space in case you are having trouble seeing that at the edge of the screen there, the limit as x goes to infinity of y(x) each time is how we are going to figure that out.3851

Now y(1) = -1 that means we are starting to x=1 and y=-1 so x=1 is here, y is -1 is here and so that looks like it is going to catch up with one of these solution trajectories it is just going down to negative infinity so that is going nowhere but negative infinity.3874

The limiting behaviour of that solution would be negative infinity, Y is going to negative infinity now y(2)= 1 ½ that means I plug in x=2, y= 1 1/2 and we will see what solution trajectory that one catches up with.3904

2 in the x direction y is 1 ½ so that looks like I'm starting right there and if I pick up the arrow that is just going pass me looks like I'm going to asypmtotically approach this solution trajectory at 1.3926

That one is going to approach one again it is going to approach 1 now the next one is y (3)= 3 ½ I'm going to find the point x= 3 y = 3 ½, x =3 ½'s between 1 and 2 so it is going to pick up right there so looks like it is headed downhill and then it is going to level of and that one is also going ahead towards 1.3945

What we say here is that 1 is a stable solution meaning that the other solutions are attracted towards 1, 1 is a stable equilibrium solution by contrast to your 0 if you look at the solution trajectory around 0 they are going away from it and so we say 0 is an unstable equilibrium solution same thing with the equilibrium solution at 2.3978

All the other solution are travelling away from it so we say that that's an unstable equilibrium so this is a stable equilibrium solution at 1 which is why the other solutions are trying to approach it.4003

Now the last one we have here was 2 3 so start at x=2 and y=3 so here is x=2 y=3 is up here and I see that if I hop on the bus at that point it is going to be trending straight up very very steeply up to positive infinity.4018

That solution is rapidly going to diverge up to positive infinity so let me recap what is going on here we started out with this graph of the slope field which is the same as the direction field that is for this differential equation we actually created this direction field back in example 5.4049

I put up a nice computer generated graph here in example 6 but it is not supposed to be a mystery where that came from its just a matter of going through and doing the face plain analysis.4074

Maybe go back and look at example 5 if you need to and then this graph is all supposed to make sense what we did with this example. In example 6 we coloured in some solution trajectories. That's what did in red here.4090

I did that just by following into a cup of more in green I did that by following the arrows wherever I saw them so I'm just kind of I'm trying to keep my solution always tangent to the arrows.4102

I'm also remembering that these solution trajectories never cross each other so I have to be careful that I do not draw them crossing each other.4118

When they started to get close then they end up becoming almost parallel they get as asypmtotically close to each other but they never cross each and they never actually touch each other.4127

I just filled in I connected in these red curves I connected the arrows into some solution trajectories and then I was given a sequence of initial conditions here thats what all these Y values are and those correspond to a starting point and the question is if we start at those different points.4138

Where we going to end up if we started those points and follow the solution trajectory what kind of y values are we going to approach in the long run.4159

As X gets very big what kind Y values are we going to approach in the long run as X gets very big what Y values will we approach let me write that a little more neatly.4168

We are finding a limit as X goes to infinity of the Y values and so I went through each of this initial points x=0 y=1 that was here that was 0 1 x=1 y=-1 where is that, that's down here 1 -1. 2 and 1 ½ that's here 2 and 1 ½ that initial point, 3 and 3 ½ that's right here 3 and 3 ½ and finally 2 3.4176

At each one of those initial points the question is if you hop on the solution trajectory starting at those initial points where you going to end up in terms of Y values as you drift to the right.4220

I pursue each one of those to the right I follow the solution trajectory to the right follow this and turns out that quite a few of them just to ended up drifting towards the stable solution at y=1.4232

There were couple here, so this drifted towards 1 and then there were couple here that are forced down to negative infinity and up to positive infinity so that's why we got those negative infinities and positive infinities for those to initial points.4246

That wraps up this lecture on slope fields and direction fields appreciate you joining me welcome back and we will study some second order differential equation later on here.4270

You are watching the differential equations lecture series here on www.educator.com.4281

My name is Will Murray and I hope you will stick around and watch some more of these lectures with me.4285

I'm really excited to be working on differential equations with you. In the meantime thanks for joining me bye for now.4291

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