Find the general solution of $\textbf{x}^{'}=\begin{pmatrix} -1&-4\\1&-1\end{pmatrix}\textbf{x}$.

The eigenvalues I found are $-1 \pm 2i $ and I chose $-1-2i$ to be my eigenvector and found it to be $e_1=\begin{pmatrix} 2i\\-1 \end{pmatrix}$ Now plugging this in to the formula we have $$e^{-t}\begin{pmatrix} 2i \\-1 \end{pmatrix}(\cos 2t-i\sin2t)=c_1e^{-t}\begin{pmatrix} 2\sin 2t\\-\cos 2t \end{pmatrix} + c_2e^{-t}\begin{pmatrix} 2 \cos 2t\\ \sin 2t \end{pmatrix}$$ The book however has: $c_1e^{-t}\begin{pmatrix} - 2\sin 2t\\\cos 2t \end{pmatrix} + c_2e^{-t}\begin{pmatrix} 2 \cos 2t\\ \sin 2t \end{pmatrix}$. How did the signs get switched because I tried to do this with $e_2=\begin{pmatrix} -2i\\-1 \end{pmatrix}$ and all that happens is the right side becomes negative which still doesn't explain how the signs are switched.

  • $\begingroup$ Your $c_1$ is the negative of the book's $c_1$. $\endgroup$ – John Habert Mar 27 '14 at 17:33
  • $\begingroup$ but how did that happen is this correct? $\endgroup$ – adam Mar 27 '14 at 17:34
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    $\begingroup$ Since the constants are arbitrary until you use initial conditions to solve for them, it is fine. It is a similar trick as to how you got all the $i$ terms to drop out by being absorbed into $c_2$. $\endgroup$ – John Habert Mar 27 '14 at 17:36
  • $\begingroup$ Thanks I understand that part, but what I dont understand is why they would do something random and factor out a negative 1. I understand that $\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$ because we dont want radicals in the denominator. Now why factor out a $-1$? $\endgroup$ – adam Mar 27 '14 at 17:41

Both answers are correct. To expand on John Habert's comment, since the coefficient $c_1$ is arbitrary, your $c_1$ is the negative of the book's $c_1$. To see what I mean, replace your $c_1$ with $-c_1$: you'll get the book's answer.

  • $\begingroup$ Is there any particular reason why they factor out a $-1$? I just think that can be confusing to do random things like that. $\endgroup$ – adam Mar 27 '14 at 17:37
  • $\begingroup$ There is no good reason for it. Whoever worked out the solution did so in a slightly different way than you did and the final form of the answers was different, though they are the same answer. $\endgroup$ – user134824 Mar 27 '14 at 17:40
  • $\begingroup$ If this is an initial value problem, then it is possible this choice of signs makes finding $c_1,c_2$ a little easier. $\endgroup$ – John Habert Mar 27 '14 at 17:40
  • $\begingroup$ Ahhh okok I got it. $\endgroup$ – adam Mar 27 '14 at 17:42

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