Author Topic: TT2--P4D  (Read 2587 times)

Victor Ivrii

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TT2--P4D
« on: March 21, 2018, 03:02:25 PM »
Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
-3 & -2\\
2 &-3\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

Jared Jubas-Malz

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Re: TT2--P4D
« Reply #1 on: March 22, 2018, 05:04:05 PM »
Setting $\begin{pmatrix}-3&-2\\2&-3\end{pmatrix} = A$, the eigenvalues can be found by taking $det(A-\lambda I)$:
$$\begin{align}det(A-\lambda I) = det\begin{pmatrix}-3-\lambda&-2\\2&-3-\lambda\end{pmatrix}=\lambda^2+6\lambda+13=0\end{align}$$
The eigenvalues would be $\lambda_{1}=-3+2i \quad and \quad \lambda_{2}=-3-2i$. Using $\lambda_{1}=-3+2i\quad$gives:
$$\begin{align}\begin{pmatrix}-2i&-2\\2&-2i\end{pmatrix}\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \implies ix_{1}=-x_{2}\end{align}$$
Therefore, the corresponding eigenvector would be: $$\textbf{v}=\begin{pmatrix}i\\1\end{pmatrix}$$
We want to find the real/imaginary solutions of $e^{(-3+2i)t}\begin{pmatrix}i\\1\end{pmatrix}$. Separating the exponential then applying Euler's formula:
$$\begin{align}e^{-3t}\times e^{2it}\begin{pmatrix}i\\1\end{pmatrix}=e^{-3t}(cos(2t)+isin(2t))\begin{pmatrix}i\\1\end{pmatrix}=e^{-3t}\begin{pmatrix}icos(2t)&-sin(2t)\\cos(2t)&isin(2t)\end{pmatrix}\end{align}$$
Separating the real and imaginary parts of (3):
$$\begin{align}e^{-3t}\begin{pmatrix}-sin(2t)\\cos(2t)\end{pmatrix}+ie^{-3t}\begin{pmatrix}cos(2t)\\sin(2t)\end{pmatrix}\end{align}$$
Therefore the general real solution will be:
$$\begin{align}\textbf{x}(t)=c_{1}e^{-3t}\begin{pmatrix}-sin(2t)\\cos(2t)\end{pmatrix}+c_{2}ie^{-3t}\begin{pmatrix}cos(2t)\\sin(2t)\end{pmatrix}\end{align}$$
« Last Edit: March 22, 2018, 05:05:55 PM by Jared Jubas-Malz »

Victor Ivrii

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Re: TT2--P4D
« Reply #2 on: March 24, 2018, 10:54:20 AM »
OK

But remember, that sin, cos, ... ln in LaTeX should be escaped as \sin , \cos, ..., \ln 

Then they will be upright and properly spaced from its argument

Please post which program you used to plot