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Express the general solution of x'=(3, 4, -2, -1)x in terms of real-valued functions. (this is 2x2 matrix, 3 and 4 on the left, -2 and -1 on the right.)

Sun,

As you will have noticed, the characteristic polynomial,
λ²-2λ+5=0,
will give you two complex conjugate eigenvalues, λ=1±2i, so the corresponding eigenvectors,
x1,2=[1±i,2],
are also complex.
You could write the general solution as the complex-valued vector function

x=c1[1+i,2]e(1+2i)t+c2[1-i,2]e(1-2i)t

and leave it at that, but I suspect you are supposed to give the final answer in real form. For this, you would break the exponentials into a real and complex part,

e(1±2i)t=et e±2it,

and use Euler's identity,

e±iθ=cosθ ± i sinθ

on the complex exponential to get

e(1±2i)t=et (cos (2t) ± i sin (2t) ),

x=c1[1+i,2]et (cos (2t) +i sin (2t) ) + c2[1-i,2] et (cos (2t) - i sin (2t) ).

Finally, you can collect the real and imaginary parts of this expression and redefine the c's so that everything is real (no more i's). I'm not sure if you are supposed to go this far, so I leave this part to you. :)

x=c1et(cos(2t), cos(2t)+sin(2t))+c2et(sin(2t), -cos(2t)+sin(2t)) is the answer. Where cos(2t) and sin(2t) are at the top, and cos(2t)+sin(2t) and -cos(2t)+sin(2t) are at the bottom.
There is a theorem for a 2x2 system, which should be in your textbook, that gets you from the complex eigenvalues and eigenvectors directly to the real solution, without all the steps I did.
It says that if λ=α±iβ are the complex conjugate eigenvalues and a±ib the corresponding eigenvectors, the general real solution is
x=c1(eαtcos(βt) a - eαt sin(βt) b) + c2 (eαtsin(βt) a + eαt cos(βt) b).

See if you can find this theorem, so you can refer to it.

In our case α=1, β=2, a=[1,2], and b=[1,0].

Therefore,
x=c1(et cos(2t) [1,2]  - et sin(2t) [1,0]) + c2 (et sin(2t) [1,2] + et cos(2t) [1,0]).

This looks to me like your answer with c1 and c2 switched, and top and bottom switched.
I got 2a+2ai-2b=0, 4a-2b+2bi=0 for the first eigenvalue, but how do I solve for a and b?
Eigenvectors are unique only up to a constant, so you can choose a to be anything except 0, which will then determine b. Or, vice versa, choose b to be anything except 0 and determine a. It's not obvious to see, but your two equations are identical.

I chose b to be 2 in your second equation, so I got a=1-i and went from there.

To get the final answer you listed, you should choose instead a=1 in your first equation, so then b=1+i and the eigenvector is [1,1+i]. The other one will be [1,1-i].

I still don't know how to get the answer from the theorem that you've provided.
I will work it out for you tomorrow, Sun. Meanwhile, make sure you can find this or a similar theorem in your book or your notes.
Never mind. I solved it and it made sense. Thank you so much for your time and effort.