00:01
Consider the given system of differential equation that is d x over dt is equal to y negative 1, dy over dt is equal to 4 x squared negative y square.
00:14
In the first a part we need to find the equilibrium point.
00:19
So first of all, substitute d x over dt is equal to 0 and dy over dt is equal to 0.
00:28
So from here, y negative 1 is equal to 0 and 4 times of x square negative y square is equal to 0.
00:39
From here, y is equal to 1 and from here x square is equal to y square over 4.
00:48
Substitute the value of y here, we can find x square is equal to 1 over 4 from here value of x equal to positive negative 1 over.
00:56
Therefore, our equilibrium point are x comma y is equal to 1 over 2 .1 or x comma y is equal to negative 1 over 2 .1.
01:21
These are our required equilibrium points.
01:31
Now in part b, we need to classify the equilibrium point.
01:36
So first of all, let us suppose the function f of x comma y is equal to y negative and g of x comma y is equal to 4 x square negative y square.
01:54
Now find the jacobr matrix y that is f x x x x x x so differentiate the function f with instructive x it gives 0 here it is 1 here it is at x here it is negative 2y now item point 1 .2 .1 the jacobin matrix j become 0 1 1 4 negative now find its eigenvalue now its eigenvalue calculated by j negative lambda i is equal to so this implies negative lambda 1, 4, negative twice of negative lambda is equal to 0.
02:51
Its determinant will be negative lambda multiplied to negative 2 negative lambda, negative 4 is equal to 0.
02:59
So from here the lambda square plus twice lambda negative 4 is equal to 0.
03:06
So this implies lambda plus 1 the whole square negative 5 is equal to 0.
03:14
So from here, lambda is equal to negative 1, positive negative root 5.
03:21
So this implies the first again value that is negative 1 plus root 5 which is positive value.
03:29
Second argument that is negative 1, negative root 5.
03:32
This is negative value...