00:01
For this problem, we want to show that this function, y of t, satisfies the logistic equation, y prime of y equals k times 1 minus y over m.
00:13
So the goal here is to arrive at this form, y prime over y equals k times 1 minus y over m.
00:25
So first you want to get the derivative of y.
00:28
So y prime of t, that's one half times m.
00:35
This times, remember that m and k here, they're both as well as a, they're all constants.
00:45
So we treat one half m as a coefficient, and then you multiply this by the derivative of the inside of the bracket that's zero plus.
00:55
Derivative of hyperbolic tangent, that's hyperbolic secant, but you have to square it, and then of k times t minus a over 2, and then you multiply this by the derivative of the inside of the hyperbolic tangent, so that's k over 2, which means that we have 1 1 half times m times k over 2 times the square of the hyperbolic secant of k times t minus a over 2.
01:36
So our y prime over y, that's equal to 1 half times m times k over 2 times the square of the hyperbolic secant of k times t minus a over 2.
01:53
Divided by 1 half times m times 1 plus hyperbolic tangent of k times t minus a all over 2...