00:01
So first of all, we've got to solve the differential equation.
00:03
Dx by d t equals t over x squared.
00:07
And we're going to separate variables.
00:09
So the integral of x squared the x is going to be the integral of t v t.
00:16
And when t is zero, x is going to be x naught.
00:20
And when t goes to t, x is going to be x of t.
00:24
So this is telling us that one third of x of t cubed minus x nought cubed.
00:32
Is equal to the integral of t from zero to t, so that's a half t squared.
00:41
So x of t cubed is going to be x nought cubed plus three halves t squared.
00:49
So x of t is going to be x nought cubed plus three halves t squared to the power of one third.
01:02
Now we're given a population and we're told dp by d t is equal to p times and now beta we're told is 1 and sigma is 10 to the minus 3 and we're also told that p of 0 equals n equals 800.
01:28
Okay, so let's solve this.
01:31
So the integral of 1 over p 1 minus 10 to the minus 3 p is the integral d t.
01:39
And when t is equal to 0, p is equal to n.
01:46
So, and then when t is t, we've got p of t, and this is going to be dp.
01:52
So let's use partial fractions.
01:55
So this side we get t.
01:58
On this side, we're integrating from n to p of t.
02:02
And let's do, so dp, we're going to have something over p plus something over 1 minus tenth and minus 3 p.
02:12
So what we have is that let's say this is a and this is b.
02:16
Then we've got b minus 10 to the minus 3a is the coefficient of p, which is zero.
02:23
So b equals 10 to the minus 3a.
02:27
And we also have a equals 1.
02:32
So that tells us that a is 1, b is 10 to the minus 3.
02:43
And then you can see that this is all going to cancel out when we multiply, when we combine these together.
02:48
So this gives us log p of t over n.
02:53
So integrating 1 over p, we get log of p.
02:56
And then between these two limits, this is what we get.
03:00
And then on this side, we're going to get minus log of 1 minus 10 to minus 3 p.
03:07
So the minus sign comes from this minus sign when we take the derivative.
03:11
The 10 to minus 3 will cancel out.
03:14
And we get 1 minus 10 to the minus 3 p of t divided by 1 minus 10.
03:20
To the minus 3n, and n is 800.
03:24
So 10 to the minus 3n is 0 .8.
03:27
So 1 minus 0 .8 is 0 .2.
03:29
And 1 divided by 0 .2 is 5.
03:32
So let's just put a 5 here...