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:

l

^{2}– 4l + 3 = 0
Or: (l -3) (l - 1) = 0

The roots are: l1 = 3 and l2 = 1. The

**so there are NO complex solutions!**__roots are REAL__
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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