00:01
All right, so we're given this differential equation here.
00:04
And i've already started some work here because there's a lot of algebra here.
00:07
So i kind of jumped ahead here for us.
00:10
But here's our differential equation.
00:11
And looking at the equation, we have negative y times y minus 1, times 2 plus 5.
00:16
What are the y values that are significant? the equilibrium points, 0, 1, negative 5.
00:25
There we go.
00:26
And we're going to see that this is actually stable here.
00:29
Y5 is stable.
00:30
We'll show y in a second.
00:32
Well, a few minutes.
00:33
And then here, that's for part a.
00:36
And then for part b, we're given this initial value, where y of negative 5 is 2.
00:40
And so we want to find out what happens to y of x as the limit is x becomes infinite here.
00:48
So it's a separable difference equation, so we need to find our initial function.
00:52
So we can do that by separating the variables.
00:56
Y is on one side, x on the other.
00:58
But we can't integrate this term on the left without a little bit of work.
01:01
So we do separate into linear terms or linear terms of linear denominators.
01:09
Here's this work.
01:11
There's lots of different ways to do this.
01:13
This is where i did it.
01:16
You rewrite it like this, this is what you want.
01:20
A over a term plus b over term plus c over term, multiplied by the whole denominator that you have.
01:25
And then you cancel a bunch of stuff out.
01:27
Here's a quadratic terms, linear terms, and then constant terms.
01:32
And just look at a plus b plus c for the quadratic is zero because there's no quadratics.
01:37
4a plus 5b my c is zero because there's no linear terms, but there's a constant term, negative 1.
01:44
So negative 5, a is 1.
01:46
A is negative a fifth.
01:48
Use that into both of these equations and then add them together because the negative c and positive c would cancel out, leaving b is a sixth.
01:58
Use that that was the sixth to figure out what c is in this quadratic equation here and we get it 30th this gives us our function all right which we can then integrate so i'm going to get rid of some of this stuff here since we don't need it anymore so let's go do this so we've got d y over or excuse me no not d y over um we'll be right here this is going to be equal to one negative 1 over 5 y plus 1 over 6 y minus 1 plus 1 over 30 times y plus 5 d y is negative d x great now we can integrate as we wish because this will be easy to integrate so i'm going to put this next part down here so i get negative 1 5th l and y and plus 1 6th l .m .1 .1 .1 .1 plus 1 .1 plus 1 .5th l .m.
03:06
1 .1 plus 1 .5 30th, ln, y plus 5 equals negative x plus c.
03:25
I got to remember that.
03:26
I'm going to simplify this just to make it a little easier to work with.
03:30
So it's natural log of y to the negative 1 5th.
03:37
And then we have, no, sorry, y minus 1 to the 1 6th.
03:48
And then we have y plus 5 to the 1 30th is negative x plus c great so now we can answer our par b here when x is negative 5 we have y is 2 so i'm going to do this here i'm going to leave this alone because we're going to come back to in a moment this is again for part b so we've got natural log of 2 to negative 1 5th uh this is 2 2 minus 1 is just 1, so we'll kind of ignore that.
04:22
And then we have 2 plus 5, 7 to the 1 30th, and that equals 5 plus c.
04:34
And we're at c then.
04:37
When we do this whole calculation, we end up with c equals negative 5 .0738.
04:47
So then our equation is then, let's see, let's see, let's go back here.
04:57
Natural log of y than negative one fifth, one sixth of y plus five, one thirtieth equals negative x minus 5.
05:22
0 .0738.
05:24
Okay, great.
05:26
We're getting there.
05:27
So now we're going to let x become infinite, so this doesn't really help us.
05:31
We'll use the property of natural logs in e, take e to both sides.
05:36
Then we get y to the negative 1 5th, 1 minus 1, 6th, 1 plus 5 to the 1 30th, z to the negative x minus 5 .0738.
05:55
All right, now we're going to let x become infinite.
05:59
So we're at the limit of y of x, as x becomes infinite...