00:01
We are given with the differential equation dy over dt is equals to 18y minus 3y square.
00:13
We have to solve this using the using phase line analysis.
00:18
We have to solve the behavior of y function with respect to t.
00:22
So just for this find the critical point that mean this point at which this derivative is 0.
00:29
So from here 3y taking common we get 6 minus 1.
00:34
It is equals to 0 when y is 0 and y is equals to 6.
00:39
That mean the whole line is divided into three intervals.
00:43
First interval is minus infinity to 0.
00:47
Next interval is 0 to 6 and next is 6 to infinity.
00:51
The function y will change its nature between these intervals.
00:56
So just draw the line to check the phase line analysis.
01:03
It is 0 point, it is 6 and it is less than 0 point and these are greater than 6 points.
01:11
There is t value.
01:13
T value increasing here and this is for y.
01:16
So first interval if our y value is between minus infinity to 0 then our derivative value dy by dt value.
01:28
The 6 is also critical.
01:30
So derivative value when minus infinity to 0 is this.
01:37
So its value is less than 0.
01:39
So it is less than 0.
01:41
That mean it will decrease when it is in this interval.
01:45
It decreases, continuously decreases.
01:50
But at 0 it will be constant 0.
01:59
Now if we have any point, if we take a point from 0 to 6 then derivative value, put values here.
02:09
Y is from 0 to 6 we get derivative value.
02:13
Derivative is greater than 0.
02:16
That mean it increases from 0 to 6.
02:19
So it increases.
02:26
At 6 it is 0.
02:28
So that mean it will not change its nature at 6.
02:33
Because derivative at 6 is 0.
02:35
So it will not change behavior at 6.
02:38
The next interval is left 6 to infinity.
02:41
If we choose a point from 6 to infinity this value, the derivative value will be less than 0...