$$x_1^'=x_1+2x_2$$$$x_2^'=3x_1+2x_2$$$$\begin{pmatrix}
x_1\
x_2
\end{pmatrix}'=\begin{pmatrix}
1 & 2\
3 & 2
\end{pmatrix}\begin{pmatrix}
x_1\
x_2
\end{pmatrix}$$$$\begin{vmatrix}
1-\lambda & 2\
3 & 2-\lambda
\end{vmatrix}=(\lambda-4)(\lambda+1)$$$$x=c_1\begin{pmatrix}
2\
3
\end{pmatrix}e^{4x}+c_2\begin{pmatrix}
1\
-1
\end{pmatrix}e^{-x}$$seems legit?

Question

$$x_1^'=x_1+2x_2$$$$x_2^'=3x_1+2x_2$$$$\begin{pmatrix}
x_1\ 
x_2
\end{pmatrix}'=\begin{pmatrix}
1 & 2\ 
3 & 2
\end{pmatrix}\begin{pmatrix}
x_1\ 
x_2
\end{pmatrix}$$$$\begin{vmatrix}
1-\lambda & 2\ 
3 & 2-\lambda
\end{vmatrix}=(\lambda-4)(\lambda+1)$$$$x=c_1\begin{pmatrix}
2\ 
3
\end{pmatrix}e^{4x}+c_2\begin{pmatrix}
1\ 
-1
\end{pmatrix}e^{-x}$$seems legit?

jamesj · Answer

Yes, that looks right.

anonymous · Answer

was feelin a little weird with 4's eigenvector thanks

jamesj · Answer

A good trick for the characteristic equation of a 2x2 matrix A is that it is given by
$$ \lambda^2 - tr(A)\lambda + det(A) = 0 $$
That's what I used here to verify your eigenvalues quickly.

anonymous · Answer

what exactly is the tr() function?

jamesj · Answer

trace of A = sum of the diagonal elements of A

anonymous · Answer

oh i see thats very nice thank u again

jamesj · Answer

Sure.  It's a useful trick.  I can't tell you how many times before I learnt it I would make mistakes and neurotically have to check the characteristic equation again and again to make sure I had it right.

jamesj · Answer

Anyway, looks like you have it under control.  And kudos to you for posting a question with solution, vs. just the question.