1) Set l = 2 - 3i in the algebraic system to get:
(1 + 3i) A + 2B = 0
-5A + - (1 + 3i) B = 0
A simple, nontrivial solution of this system (using the method to solve for A, B we did before) yields: A = 2 and B = -1 - 3i. Using these one obtains the complex solutions:
y = (-1 +3i) exp (2 - 3i)t
Which can alternatively be written :
x = exp (2t) [(2 cost 3t) - i(2 sin 3t)]
y = exp(2t)[(-cos 3t – 3 sin 3t) - i (3 cos 3t – sin 3t)]
We can write the general solution of the system as:
x = 2exp (2t) (c1 cost 3t - c2 sin 3t)
y = exp(2t)[c1(-cos 3t – 3 sin 3t) - c2 (3 cos 3t – sin 3t)]
Solve, giving the complex and general solutions:
dx/dt = 5x – 2y
dy/dt = 4x – y
Let: x = A exp(lt) and y = B exp(lt)
Substitute the preceding into the system to get:
l A exp(lt) = 5 A exp(lt) - 2B exp(lt)
lB exp(lt) = 4A exp(lt) - B exp(lt)
And the algebraic system is:
(5 - l) A - 2B = 0
4A - (1 - l) B = 0
Yielding the standard determinant form:
[(5 - l) …..- 2]
[4 ……..-(1- l)]
Expand to obtain the characteristic equation:
l2 – 4l + 3 = 0
Or: (l -3) (l - 1) = 0
The roots are: l1 = 3 and l2 = 1. The roots are REAL so there are NO complex solutions!
In the first instance, we substitute the first eigenvalue, l1= 3, into the matrix, whence:
(A - lI)D = 0 =
[2.....-2] [A]
[4 ...2 ] [B]
[4 ...2 ] [B]
Then: 2 A – 2B= 0 so A = 1, B = 1
The first eigenvector is then: K1 =
[1]
[1]
[1]
So: X1 = K1 exp (t)
The second eigenvalue was l2 = 1 so we repeat the process again to obtain the equation to be solved:(A - l2 I)D =
[4.....-2] [A]
[4 ...-.0] [B]
Then: 4A - 2B = 0 or A = 1, B = 2
Then: K2 =
[4 ...-.0] [B]
Then: 4A - 2B = 0 or A = 1, B = 2
Then: K2 =
[1]
[2]
So: X2 = K2 exp (3t)
[2]
So: X2 = K2 exp (3t)
Yielding the general solutions:
x = c1 exp (t) + c2 exp (3t)
y = 2c1 exp (t) + c2 exp (3t)
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