00:01
Hello, so here we let x of t and y of t satisfied, where we have dx d t be equal to 1 .4x minus 1 .2y, and we have that the derivative, the y, d t, to be equal to 0 .8 x minus 1 .4 y.
00:23
With our matrix a being a equals 1 .4, negative 1 .2, 0 .8, negative 1 .4.
00:31
And then part a here, for the type of interaction, well, the term, the plus 0 .8x in the dydt, that means the species y benefits from the presence of x.
00:46
And the term negative 1 .2y in dxdt means the species y is going to hurts x.
00:53
So species y also dies out without x, the negative 1 .4y, while x can grow on its own, the positive 1 .4x.
01:03
So that is going to be the predator prey pattern, where we have x is equal to the prey, and y is the predator.
01:15
This is the predator prey pattern.
01:22
And then for part b, we solve here, dx, dt equals zero, and d .y, dt equals zero.
01:30
So from 1 .4x minus 1 .2y equals 0.
01:38
This implies that y equals 76x.
01:42
And then we have 0 .8x minus 1 .4y equals 0.
01:50
This implies that y is equal to 4 -7s x.
01:54
So then the lines intersect only at xy equals 0 -0...