00:01
So we're given the differential equation x ' equals 3, 2, negative 5, negative 3, x.
00:11
Okay, so for a, you want to find the eigenvalues and eigenvectors for the coefficient matrix.
00:18
So in order to do this, we're going to calculate the determinant of, let's call this matrix a, a minus lambda times the identity matrix, 3 minus lambda, 2, negative 5, negative 3, minus lambda, which results in 19 minus 6 lambda plus lambda squared, set it equal to 0.
00:58
We can solve this and find that lambda 1 equals 3 plus i square root of 10, lambda 2 equals 3 minus i square root of 10.
01:13
Now all we need to do is to find the eigenvectors that is associated with it.
01:21
So for the first vector v1, we're gonna to set it equal to lambda 1 times v1 and then you get v1 equals negative i square root of 2 over 5 and 1 and for the second vector we do lambda 2 times v2 we get v2 equals i times square root of 2 over 5 and 1.
02:02
Okay, so the solution to this differential equation can be written as x1 equals e to the 3t c1 times cosine of square root of 10 t plus square root of 2 over 5 e to the 3t e to 3t c2 times sine of square root of 10 times t...