For more information, please see full course syllabus of Differential Equations

For more information, please see full course syllabus of Differential Equations

### 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
- Lesson Objectives
- 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
- Never Cross Each Other
- General Solution to the Differential Equation
- Use an Initial Condition to Find Which Solution Trajectory You Want
- Example 1
- Example 2
- Example 3
- Example 4
- Example 5
- Example 6

- 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

### Differential Equations Online Course

### 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 X _{0} and Y_{0}.*0297

*What that does is that gives you an initial point X _{0} and Y_{0} 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 X _{0} Y_{0} 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=x _{0} y=y_{0} 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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