00:01
In order to find the equilibrium solution and increasing, decreasing intervals, the equilibrium solution is easiest because we could set this equal to zero or this equal to zero.
00:11
If we set the 1 .2p equal to zero, that would occur when p equals zero.
00:18
So that's one solution.
00:20
Or, alternatively, if we set 1 minus p over 4 ,200 equal to zero, equilibrium is when the slope would be zero or when the derivative d p d t equals zero well then we could just move over that negative term and then set these equal multiply the 4200 over so p is equal to 4 ,200 is another equilibrium solution which makes sense because if we substituted that value back in here it would be 1 minus 4 ,200 divided by 4 ,000 200, which is 1 minus 1, which is 0.
01:08
Ok, to find the other two, we're going to set up an interval with 0 and 4200.
01:15
And then we're going to test values inside and outside of these points here.
01:20
So for instance, let's just test the value negative 1, 1, and 4 ,2001, for instance.
01:28
We're going to plug them back into the whole differential equation here and see if they give us positives or negative.
01:35
If we plug in negative, we're going to get a negative here and then a positive.
01:43
So this would give us negative, derivative...