0:00
Hi.
00:02
Now we have to determine the equilibrium solution and classify them.
00:07
For the differential equation, y -dash is equal to 2 minus y into y -minus 4 whole square, y -minus 5.
00:18
So let us take this as equal to f of y.
00:21
For equilibrium solution, we put y -dash is equal to 0.
00:26
So this implies 2 - minus -y, 4 whole square y minus 5 equal to 0 so from here we have the equilibrium solution comes out to be equal to y is equal to 2 4 and 5 next we check for the stability so for y is equal to 2 we check the of f y on left of y is equal to two and on right of y is equal to two so f y is less than zero on left of y is equal to two and f of y is greater than zero on right of y is equal to two like we can check for the signs for on left of y is equal to two now my f of y now this term will be positive 2 minus y, this is whole square.
01:30
This is in any case will be positive and this will be negative.
01:35
So overall sign of fy is less than 0.
01:38
And on right of y is equal to 2, so this term becomes negative.
01:43
This is also negative...