# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on March 21, 2018, 03:02:25 PM

Title: TT2--P4D
Post by: Victor Ivrii 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.
Title: Re: TT2--P4D
Post by: Jared Jubas-Malz 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}
Title: Re: TT2--P4D
Post by: Victor Ivrii 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