Given the first order linear homogeneous system of equations

Question

Given the first order linear homogeneous system of equations

x'(t)=-x(t)+2y(t)

y'(t)=-2x(t)+3y(t)

Its coefficient matrix A=[-1 2]

[-2 3]

has two represented real eigenvalueλ₁=λ₂=1 but only one independent eigenvector

v=[2]

[2]

What would be the general solution?

A.) C₁e^t[2]+C₂te^t[2]

[2] [2]

B.)C₁e^t[2]+C₂e^t[1]

[2] [1]

C.)C₁e^t[2]+C₂e^t([-1]+t [2])

[2] [1] [2]

D.)C₁e^t[2]+C₂e^t([-1]+t [2])

[2] [0] [2]

Answer 1

Given what we already know about the eigenvector {2, 2}, let's augment that eigenvector against A-1I:

[ -2 2 2 ]

[ -2 2 2 ]

Putting this system into reduced row-echelon form gets us:

[ 1 -1 -1 ]

[ 0 0 0 ]

This means that x - y = -1, and that y is a free variable. We can re-express this as x = y - 1, and y is a free variable. Here, we can pick different values of y (since y is free to vary), and this should produce a linearly independent eigenvector. Picking y = 0 shows that x = -1 and y = 0, or {-1, 0}.

The first eigenvector {2, 2} is free to take on any scalar multiple, so there should be a c₁{2, 2} term in the answer. The next part of the answer will feature {2, 2}, but multiplied by t. Failure to include that t would produce a result that demonstrates linear dependence. The new eigenvector {-1, 0} does not need to be multiplied by anything else, although it should be multiplied from the outside by an arbitrary c₂.

And of course, our eigenvalues are both 1.

D.) C₁e^t[2]+C₂e^t([-1]+t [2])

[2] [0] [2]