0:00
Hello.
00:01
So here we're given the linear system where we have that dx, dt, is going to be equal to negative py, and we have that dydt is equal to negative qx, where p and q are greater than zero.
00:17
So we eliminate y.
00:19
We get x is equal to negative py implies that y is going to be equal to negative 1 over p times x.
00:27
And then we differentiate x.
00:28
So we have that x equals.
00:30
Negative p y it's negative p times negative q x which is equal to p q x so we have that x minus p q x is equal to zero and we have then a second order equation here so we look at the characteristic qualinomial a characteristic equation r squared minus pq equals zero implies that r is equal to plus or minus the square root of pq and then we have that that x of t is going to be equal to c1 times e to the square root of p q t plus our constant c2 times e to the negative square root of pqt.
01:17
And then we recover y of t using y is equal to negative 1 over px.
01:24
And then get the matrix form from the eigenvalues where our matrix a is going to be the matrix 0, negative p, negative q, zero.
01:36
Giving us that lambda squared minus pq equal zero, implying that lambda is equal to plus or minus the square root of pq...