00:01
Hi, today we are solving the question in which dx by dt is equal to x into 1 minus x by 4 minus y and dy by dt is equal to y into 1 minus y by 5 minus x.
00:17
So taking x and y is greater than equal is greater than zero.
00:24
So first of all we need to write an equation for the non -zero vertical x -neckline of this system.
00:32
So from here solving it we get it as taking the derivatives equals to zero.
00:44
So for the vertical x -neckline setting dx by dt is equals to zero.
00:52
So from here we get x into 1 minus x by 4 minus y is equals to zero.
00:59
Solving it we get it as 1 minus x by 4 minus y is equals to zero.
01:09
On solving it further we get it as minus x by 4 minus y is equals to minus 1.
01:19
So from here we can take it as x by 4 plus y is equals to 1.
01:26
So we can say that x plus 4y is equals to 4.
01:38
Similarly we can write the equation for y -neckline.
01:43
So y -neckline can be written as dy by dt is equals to zero.
01:50
That means solving it similarly so we get 1 by y minus 5 minus x is equals to zero.
01:58
So from here minus y minus 5 minus x is equals to minus 1.
02:06
So we get x plus y by 5 is equals to 1.
02:10
Solving it we get it as 5x plus y is equals to 5.
02:16
So therefore the required equation will be for first it will x plus 4y is equals to 4 and for y it will be 5x plus y is equals to 5.
02:34
Now we need to find the equilibrium points of the system.
02:41
So equilibrium point is when we set both derivative equal to zero and solve for x and y.
02:51
So if we take x 1 minus x by 4 minus y is equals to zero and x y into 1 minus y by 5 minus x is equals to zero.
03:15
So we get these two equations...