dx1/ dt = x1 + x2
dx2/dt = 4x1 + x2
Note the first step, again, is to form a matrix from the coefficients, which we see are (1, 1) for the top, and (4, 1) for the bottom. Thus: A =
(1 .....1)
(4......1)
Then, it must be true from the properties of determinants that:
(A - l) D =
[(1 - l).......1] [k1]
[4 ..... .(1 -l)] [k2]
Again, we allow l to be subtracted from the first element in the upper left, and from the last element in the lower right). Cross-multiplying and using matrix properties we obtain the characteristic equation:
l2 - 2l - 3 = 0
where l1 = 3 and l2 = -1 We need to first find a vector that solves the equation: (A - l1 I)D = 0
In the first instance, we substitute the first eigenvalue, l= 3, into the matrix, whence:
(A - 3I) D = 0 =
[-2.....1] [k1]
[4 ...-2] [k2]
Therefore:
-2k1 + k2 = 0, and 4k1 – 2k2 = 0, So k1 = ½ and k2 = 1
Then our first eigenvector is: K1 =
[½]
[1]
Therefore, the first linearly independent solution for the system is:
X1 = K1 exp (l1 t) = K1 exp (3t)
The second eigenvalue was l2 = -1 so we repeat the process again to obtain the equation to be solved:
(A - l2 I)D = (A - (-1)I) D = (A + I)D
Then, (A + I) D = 0 =
[2.....1] [k1]
[4 ….2] [k2]
Or: 2k1 + k2 = 0 and 4 k1 + 2k2 = 0
So that: k1 = 1, then k2 = - 2
The second eigenvector is then: K2 =
[1]
[-2]
So another linearly independent solution is:
X2 = K2 exp (-t)
Then add the two solutions, to obtain:
X = X1 + X2 = K1 exp (3t) + K2 exp (-t)
With, of course, the 2 column vectors (as computed above) substituted in for K1, K2. This serves as a general approach for solving all such systems.
---------
Problems for the Math Maven:
1) Obtain the general solution(s) for the system:
dx/dt = 2x + y
dy/dt = 2x + 3y
2) Obtain the general solution(s) for the system:
dx1/ dt = 3x1 - 4x2
dx2/dt = x1 - x2
No comments:
Post a Comment