4
$\begingroup$

I'm trying to solve $Y' = AY$ where $A= \left[ \begin{array}{ c c } -2 & 6 \\ -3 & 4 \end{array} \right]$

I have found the eigenvalue $1 \pm 3i$ with eigenvector for $1+3i: $

$v = \left[ \begin{array}{ c c } 1-i \\ 1 \end{array} \right]$

Which seems to be correct by testing $Av = (1+3i)v$ but when I try to write it as a real solution I don't seem to get the right answer.

$$\left[ \begin{array}{ c c } 1-i \\ 1 \end{array} \right](\cos 3t + i\sin 3t) = \left[ \begin{array}{ c c } \cos 3t + \sin 3t - i\cos 3t + i\sin 3t \\ \cos 3t + i\sin 3t \end{array} \right]$$

$$v_1 = \left[ \begin{array}{ c c } \cos 3t + \sin 3t \\ \cos 3t \end{array} \right],\ v_2 = \left[ \begin{array}{ c c } \cos 3t + \sin 3t \\ \sin 3t \end{array} \right]$$

If I now verify by $v_1' = Av_1$

$$\left[ \begin{array}{ c c } 3(\cos 3t - \sin 3t) \\ -3\sin 3t \end{array} \right] \neq \left[ \begin{array}{ c c } 4\cos 3t - 2\sin 3t \\ \cos 3t - 3\sin 3t \end{array} \right]$$

$\endgroup$
4
$\begingroup$

You are still missing an exponential term, we have:

$e^{\lambda t}v_1 = e^{(1+3i)t}\begin{bmatrix}1-i\\1\end{bmatrix} = e^te^{3it}\begin{bmatrix}1-i\\1\end{bmatrix} = e^t(\cos 3t + i \sin 3t)\begin{bmatrix}1-i\\1\end{bmatrix} = \begin{bmatrix}e^t(\sin 3t + \cos 3t+i (\sin 3 t-\cos 3 t))\\ e^t(\cos 3t + i \sin 3t) \end{bmatrix} $

So, our solution can be written as (because we know that the real and imaginary parts are both independent solutions):

$$Y(t) = c_1 e^t\begin{bmatrix}\sin 3t + \cos 3t\\ \cos 3t \end{bmatrix}+ c_2e^t\begin{bmatrix}\sin 3t - \cos 3t\\ \sin 3t \end{bmatrix}$$

$\endgroup$
  • $\begingroup$ @BabakS.: It takes me 5x longer to post answers than it does to find them! :-) Thank you my friend! $\endgroup$ – Amzoti Jun 8 '13 at 7:19
  • $\begingroup$ @Amzoti: Isee. You know, sometimes, I learn some noble points of your solutions here. Believe me or not. $\endgroup$ – mrs Jun 8 '13 at 7:21
  • $\begingroup$ @BabakS.: You are too kind my friend! Have a great day! $\endgroup$ – Amzoti Jun 8 '13 at 13:16
1
$\begingroup$

The last term in the first component in the vector, you have no $i$. Here is the correction

$$ \left[ \begin{array}{ c c } 1-i \\ 1 \end{array} \right](\cos 3t + i \sin3t) = \left[ \begin{array}{ c c } \cos 3t + \sin 3t - i\cos3t + i\sin 3t \\ \cos 3t + i \sin 3t \end{array} \right] .$$

Added: Here are your eigenvalues and eigenvectors.

$$ \lambda_1=1+3\,i,\, v_1= (1-i,1) $$

$$ \lambda_1=1-3\,i,\, v_1= (1+i,1) $$

Note that, to find the general solution you need to consider the two eigenvectors. It seems you are considering only one. The general solution has the form

$$ Y(t) = c_1v_1 e^{\lambda_1t} + c_2 v_2 e^{\lambda_2 t}. $$

Have you given initial conditions? If yes, then you need to use them to find the constants $c_1$ and $c_2$.

$\endgroup$
  • 1
    $\begingroup$ I'm not sure if i follow $(1-i)(cos 3t + i sin 3t) = cos 3t + i sin 3t - i cos 3t -i^2 sin 3t = cos 3t + sin 3t + i(sin 3t - cos 3t)$ $\endgroup$ – shardy Jun 8 '13 at 0:12
  • $\begingroup$ @shardy: $(1-i)( a+ib )=( a+ib -i(a+ib) ) = a+ib-ia-i^2b=a+ib-ia+b. $ $\endgroup$ – Mhenni Benghorbal Jun 8 '13 at 0:17
  • $\begingroup$ @shardy: You are right. $\endgroup$ – Mhenni Benghorbal Jun 8 '13 at 0:22
  • $\begingroup$ @shardy: Let me check if you got the right eigenvalues. $\endgroup$ – Mhenni Benghorbal Jun 8 '13 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.