- #1
- 2,076
- 140
Homework Statement
Solve the system:
##x' = 5x - y##
##y' = 4x + y##
Homework Equations
##t## is transpose.
The Attempt at a Solution
I'm a bit rusty with these and I had a small question.
I put the system into the form ##x' = Ax## and proceeded to solve for the eigenvalues. I found that ##\lambda_1 = 3## was the only eigenvalue of multiplicity 2.
I then solved for the eigenspace ##ε_A(\lambda_1) = null(A - \lambda_1I)## and found that the only eigenvector in the span was ##v_1 = [1, 2]^t##.
This yielded my first independent solution ##x_1(t) = e^{3t} [1, 2]^t##.
I need a second independent solution now of the form ##x_2(t) = e^{3t}(tv_1 + u_1)## where ##u_1## is a solution of the system ##(A - \lambda_1I) = v_1##.
Super easy system to solve as it required only two row operations and it left me with:
[2 -1 | 1]
[0 0 | 0]
Hence ##u_1 = span\{ [1/2, 0]^t, [1/2, 1]^t \}## where I've noticed the second vector in the span is simply ##\frac{1}{2}v_1##.
Now my problem. The book lists ##u_1 = [0, -1]^t## as the generalized eigenvector they have used to obtain ##x_2(t)##. I understand how they have obtained this solution, but I'm wondering if my solution is also correct. I would wind up using ##[1/2, 0]^t## as my generalized eigenvector or any multiple of it. Preferably I would use ##[1, 0]^t##.
Then of course ##X(t) = (x_1(t), x_2(t))## and ##X(t)C## is the general solution.