Problem: We were to obtain the general solution(s) for the system:
dx/dt = 2x + y
dy/dt = 2x + 3
The first step, again, is to form a determinant from the coefficients, which we see are (2, 1) for the top, and (2, 3) for the bottom. Thus:
A =
(2 .....1)
(2......3)
Then, it must be true from the properties of determinants (And recall again that I is the identity matrix) that:
(A - LI) D =
[(2 - L)......1] [d1]
[4 .....(3 -L)] [d2]
For any determinant D such that D =
[a ……b]
[c……..d]
We compute according to (a*d – b*c) in order to find the characteristic equation.
We then obtain for this:
L^2 - 5L +6 = 0, for which:
L1 = -2 and L2 = -3
We need to find a vector that solves the equation:
(A - LI)D = 0
Now, in the first instance, we substitute the first eigenvalue, L= -2, into the matrix for L,
whence:
(A +2 I) D = 0 =
[4.....1] [d1]
[4 ...5] [d2]
Therefore, combining equations in d1, d2:
8 d1 + 6 d2 = 0, or d1 = (-3/4) d2
Let d2 = 4, then d1 = -3/4 (4) = -3
Then our first eigenvector is: D1 =
[-3]
[4]
Therefore, the first linearly independent solution for the system is:
X1 = D1 exp ( L1t) = D1 exp (-2t)
since L1 = -2
The second eigenvalue was L2 = -3 so we repeat the process again to obtain the equation to be solved:
(A - L2 I)D = (A - (-3)I) D = (A + 3I)D
Then, (A + 3 I) D = 0 =
[5.....1] [d1]
[4 ...6] [d2]
Or, combining equations:
- d1 – 5 d2 = 0, so d2 = -d1/5
Now, let: d1 = 5, then d2 = - (5)/5 = -1
The second eigenvector is then: D2 =
[5 ]
[-1]
And another linearly independent solution is:
X2 = D2 exp (-3t)
Where D2 =
[5]
[-1]
The full general solution is:
X = D1 exp (-2t) + D2 exp (-3t)
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