00:01
So here we let our system be dx, dt, is going to be negative x plus ky, and then dy, dt is going to be x minus 4y, where k is greater than zero.
00:17
So we have our coefficient matrix, a being the matrix negative 1, k, 1, negative 4.
00:26
And then we have the trace of a is going to be negative 5.
00:32
And the determinant of a is going to be, well, negative 1 times negative 4 minus k.
00:40
So the determinant is going to be 4 minus k.
00:43
And then for our eigenvalues, we take lambda squared minus the trace of a lambda plus determinant of a equals 0.
00:51
And then giving us that our eigenvalues, are going to be negative 5 plus or minus the square root of 9 plus 4k all over 2.
01:08
And therefore, well, since we have that 9 plus 4k is greater than 0, for all k greater than 0, the eigenvalues are going to be real and are going to be distinct.
01:18
And then for the classification by k, if we have k between 0 and 4, the determinant of a equals 4 minus k, greater than 0, and the termination.
01:27
Trace of a is less than 0, so both eigenvalues are negative, implying a stable node.
01:33
And if k equals 4, the determinant of a equals 0 implies the eigenvalues of 0 and negative 5.
01:38
And this is going to be a one -dimensional nal space...