In this video we're talking about how to find equilibrium solutions of a differential equation and how
to determine whether those equilibrium solutions are stable or unstable solutions.
And this particular problem we've been given the differential equation D-y over is equal to y minus
Y cubed.
So the first thing we want to realize is that why is a function of t.
So when we say D-y over d it's the derivative of Y with respect to t sorts y prime of t or just y prime
and then we have the original function over here on the right hand side.
Why and why cube.
So we have to think of this as why of T minus y of T cubed.
So what we want to do to find equilibrium solutions we want to set the derivative equal to zero.
Or remember that this is the derivative D-y over d t is the same thing as y prime of to or just y prime.
So we could really rewrite this as.
Y prime equal to y minus Y cubed and we want to set the derivative equal to zero just as if we were
finding critical points.
Since this is the derivative we can set this equal to zero and then we want to solve for y.
So we're dealing now with the equation y minus Y cubed is equal to zero.
If we factor out a y we get Y times the quantity 1 minus Y squared is equal to zero.
If we factor the one minus Y squared we get Y times 1 plus Y times 1 minus y equal 0.
Then we can say is that we have the critical points if we take each factor and set it equal to zero
individually we would have the critical points y equal to zero y equal to negative 1 from this factor
and positive one from this factor so 0 and positive negative 1.
So we have three critical points.
Now we need to set those aside for a second and find the second derivative.
Why.
Double prime to find the second derivative remember that we need to treat Y as a function.
So when we take the derivative of y prime we're going to have y double prime and then on the right hand
side here we have y minus Y cubed but y is a function.
So when we take the derivative of y we say the derivative of y is 1 but because Y is a function in terms
of t.
We then have to multiply by D-y over d t.
In other words whenever you take the derivative of y Just remember that you have to multiply by d y
over d t.
So then the derivative of negative y cubed is going to be negative three wives squared.
But because then we took the derivative of why we have to multiply by d y over d t.
Remember then that d y overdye T is the same thing as y prime.
So what we want to do is take y prime and plug it in for d y over d t.
So we end up with y double prime is equal to one time y over ditty or 1 times Y minus Y cubed 1 times
Y minus Y cubed is just y minus Y cubed.
Then we'll have minus three y squared multiplied by y minus Y cubed.
When we simplify here we're going to have y minus Y cubed.
When we distribute the negative 3 Y squared we're going to get minus 3 y cubed and then we've a negative
sign and a negative sign.
So we get a positive 3 y to the Fifth.
If we consolidate like terms we have our 3 y to the fifth which we'll put first because it's the highest
degree variable.
Then we have a minus Y cubed and a minus 3 y cubed which is going to give us a minus for y cubed.
And then we have a plus Y.
Now if we factor this we can pull out a Y.
And that's going to leave us with 3 y to the fourth minus for Y squared plus 1.
And then we can factor what's inside the parentheses and we get 3 Y squared here Y squared here because
multiplying those together will give us a 3 Y squared.
Then we're going to end up with a minus one and a minus one because that will give us minus 3 Y squared
and a minus Y squared which will give us the negative for Y squared and the negative one times negative
one is a positive one.
And then we could go even one step further and say why we'll take this Y squared minus one and factor
that into Y plus 1 times Y minus 1 and then we have our 3 y squared minus 1.
Now what we want to do is take our second derivative and set it equal to zero in order to find inflection
points.
I remember setting the first derivative equal to zero and solving for y gives critical points setting
the second derivative equal to zero and solving for y gives inflection points.
So setting this equal to zero and then taking each factor equal to zero individually we can say that
our inflection points are going to be y equal to zero.
From this factor right here from these next two factors y plus 1 and y minus 1 we get positive and negative
1.
And from our last factor 3 Y squared minus one equals zero.
If we added 1 to both sides we would end up with 3 Y squared equals 1.
Dividing both sides by 3 we'd end up with Y squared is equal to one third and then taking the square
root of sides we would end up with Y is equal to positive or negative square root of one third which
we could rewrite as positive or negative.
Square root of three over three.
Once we have our critical points and our inflection points we want to plot them on a simple number line.
So I've got a number line here.
I first want to plot my critical points and will plot critical points in orange.
So we have zero positive and negative one.
So we'll go out and plot those on the top side of the number line and here will say 0 we'll say negative
1 and we'll say positive one up here.
Then we want to take our critical points which will plot in green and we want to plot them on the underside
of the number line.
So we have zero.
We have positive and negative 1 so negative one positive one.
And then we have positive and negative squared of three over three which is about positive or negative
zero point five.
So if we just plot those let's say right about here and we'll call this negative square root 3 over
3 and right about here we'll call this positive square root 3 over 3.
Our next step is going to be to determine the behavior of the differential equation over each of these
intervals.
So what we want to do is test the function to see if it's increasing or decreasing.
On each of these intervals and to see if it's concave up or concave down on each of these intervals.
So first we'll test to see where the function is increasing and decreasing which we'll do using the
first derivative test.
Remember that the first derivative test allows us to take test values in between each of our critical
points.
Plug them into the first derivative and depending on whether the result is positive or negative that'll
tell us whether the differential equation is increasing or decreasing.
So in order to test the interval to the left of our first critical point negative one will take a test
value of negative 2 because negative 2 is inside of this interval here it's to the left of negative
1.
So we'll take the test value negative 2 and we'll plug it into our first derivative.
So what we want to do is say why Prime of negative 2 is going to be equal to.
And then we'll plug that in for y.
So we'll get negative 2 minus negative 2.
Quantity cubed.
When we simplify this will get negative 2 minus negative 2 cubed is a negative 8.
So we'll get minus a negative 8 are negative signs will cancel and we end up with the negative 2 plus
8 which is going to be equal to a positive six which is positive or greater than zero.
So what we can say then is that the function because we got a positive value the function is increasing
to the left of negative 1.
Next we need to test the interval between negative 1 and 0 between 0 and 1 and greater than 1.
So for example we could use the test values negative 1 1/2 because that would be between negative 1
and 0 positive 1 1/2 because that would be between 0 and 1 and positive 2 because that would be to the
right of 1.
So we would plug negative 1 have positive 1 1/2 and 2 into our first derivative just like we did with
negative 2.
And what we would see is that our results are going to be negative positive and negative.
What that tells us is that our function is increasing to the left of negative one it is decreasing between
negative 1 and 0 increasing between 0 and 1 and decreasing again to the right of one.
So we'll come back to that in a second.
Now we need to test for concavity and the way that we test for concavity.
We again take test values on each interval.
So we're going to have to take a test value to the left of negative one between negative 1 and negative
root 3 over 3.
Remember negative route 3 over 3 is about negative five point eight.
So we could for example take negative three fourths because that would fall within this interval and
therefore represent the function's behavior on this interval.
So maybe we would take negative two negative three fourths negative one fourth positive one fourth positive
three fourths and positive two and we would plug each of those values into the second derivative using
the second derivative test in order to determine the functions concavity on each of these intervals.
So let's for example take the test value of negative 2 and plug it into the second derivative.
So why double prime of negative 2 is going to be equal to plugging it into this equation here.
We're going to get three times negative to the fifth minus four times in negative 2 cubed Plus a negative
2 or just minus 2.
When we simplify here negative 2 to the fit is going to be negative 32 negative 32 times three is a
negative 96.
Here we have negative 2 cubed which is a negative 8 negative 8 times and negative 4 is a positive 32.
And then we have our minus 2.
We can see that this is going to give us negative 66 which is going to be a negative value or less than
zero.
So what we can say then is that the function is concave down on the interval.
Everything left of negative one.
So because we've got a negative answer we put a negative sign here we can say that this is concave down.
Now if we were to test all of those other test values we mentioned the negative three fourths negative
one fourth positive one fourth positive three fourth and positive two in the second derivative our results
would be positive negative positive negative positive.
But keep in mind that we won't always have this perfect alternation so you should always check every
interval for both increasing decreasing in the first derivative and for concavity in the second derivative
to verify the behavior of the function on each interval.
So once you have your positive and negative sign set up for each interval for both increasing decreasing
and concavity.
Well we want to do is move over to a graph to sketch the equilibrium solutions.
So we just have an x y coordinate system.
But of course our differential equation is for the function y in terms of t.
So we've changed the x axis to a t axis so the t axis and the y axis.
And what I've done ahead of time is sketched the critical points.
Remember we said the critical points were negative 1 0 and 1.
So the critical point zero here along the t axis we have negative 1 and we have positive one.
So those are critical points.
And then I also sketched the inflection points.
We have the same values as the critical points.
Negative 1 is zero and positive one.
But then we also have the negative route 3 over threes we'll say negative route 3 over 3.
And we have positive route 3 over 3.
It's customary to see the critical points sketched with a solid line and the inflection points sketched
with a dotted line.
The reason is because the graph is never going to cross one of these critical point lines but it will
cross one of the inflection point lines.
So we like to distinguish between inflection points and critical points so that we can show the graph
crossing an inflection point line but not crossing a critical point line.
So then we just take the data that we've collected about the behavior in each interval and we sketch
a simple curve on each of these intervals.
So here's how you'd match up your number line to your graph.
We have let's say interval number 1 which is everything to the left of negative 1.
We have interval number two which for increasing decreasing behavior would be between negative 1 and
0 but for concavity behavior it's going to be between negative 1 and negative 3.
So because the interval for concavity is more restricted we're just going to define this narrower interval
as interval number two.
Then we have interval number three number four number five and number six here and we would correspond
those to the graph as everything below negative 1 is interval number 1.
This is going to be interval number two number three number four number five.
And then above one is sex.
So now we're just going to sketch the behavior.
So for interval number one we know that the function is increasing and concave down because we have
a positive sign for increasing decreasing.
That tells us increasing because we have a negative sign for concavity that tells us concave down so
increasing and concave down.
So what we'll do is we'll just sketch any point inside of interval number one which is below negative
1 sketch any point and we want to sketch a curve that is increasing but concave down.
And here's what that's going to look like.
This curve is increasing but it is concave down.
And we sketch it so that it does not cross the critical point line.
Then for interval number two we have decreasing but concave up.
So for interval number two we just sketch any point inside of that interval.
It's decreasing and concave up so decreasing and concave up without crossing this critical point line
would look like this for interval number three we have decreasing and concave down.
So interval number three will pick any point we have decreasing and concave down so decreasing and concave
down looks like this.
But then we're crossing an inflection point line and when we cross an inflection point line that means
we're going to change from concave down to concave up.
So the graph starts to then look like this but we don't cross this critical point line then for interval
number four we have increasing and concave up.
So interval number four we have increasing and concave up so it's going to look like this.
But then we cross this inflection point line which means we have to switch to concave down for interval
number five.
We have increasing but concave down.
So for interval number five increasing and concave down looks like this.
And we make sure that we don't cross that critical point line.
And then for interval number six finally we have decreasing and concave up so decreasing and concave
up looks like this and once we've got the behavior of the function sketched over our graph here now
we can look at each critical point and draw a conclusion about its stability.
So if the graph is moving toward the critical point on both sides then we call it a stable equilibrium
solution or a stable solution.
So if we look at the critical point here for y equals one we know that the graph is moving toward this
point on both sides of it.
So we call this here a stable solution the equilibrium solution negative one is the same way we see
the graph moving toward it on both sides.
So this solution is also stable.
In the case of y equals zero we can see the graph moving away from this line on both sides of it.
And that kind of equilibrium solution we call an unstable equilibrium solution.
Keep in mind that it is possible to have the graph move toward the solution on one side and away from
the solution on the other side.
In that case we call the solution a partially stable equilibrium solution.
But remember that that's also pretty rare to see in calculus one or two.
If you think you have a partially stable solution you might want to go back and double check your work
to make sure that you're increasing and decreasing and concavity behavior was evaluated correctly because
chances are all of your solutions are going to be either stable or unstable.
A partially stable solution is much much more rare so make sure to double check that if you have it.
But that's how you find the equilibrium solutions to a differential equation and determine their stability.