1306
Technical Questions / testing svg
« on: April 19, 2014, 07:55:36 PM »
Testing svg
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1308
Quiz 5 / Re: Problem 2, night sections
« on: November 22, 2013, 09:03:23 AM »I used mathtype in word, is there anything wrong with this s.w?
You definitely used it incorrectly--to type everything, not only math snippets. However I have seen a correct usage of it--and the source is ugly, contains tons of unnecessary elements, and it is very difficult to edit it by hand later or reuse it.
1309
Quiz 5 / Re: Problem 2, night sections
« on: November 21, 2013, 05:35:54 PM »
Again, what crapware produce this code?
Code: [Select]
\begin{array}{l}\det (A - rI) = \left( {\begin{array}{*{20}{c}}{2 - r}&{ - 5}\\1&{ - 2 - r}\end{array}} \right) = {r^2} + 1 = 0\\r = \pm i\\{\rm{when }} r = i\\(2 - i){\xi _1} = 5{\xi _2}{\rm{ and }}{\xi ^1} = \left( {\begin{array}{*{20}{c}}5\\{2 - i}\end{array}} \right)\\{x^1} = \left( {\begin{array}{*{20}{c}}5\\{2 - i}\end{array}} \right){e^{it}} = \left( {\begin{array}{*{20}{c}}5\\{2 - i}\end{array}} \right)(\cos t + i\sin t)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {\begin{array}{*{20}{c}}{5\cos t}\\{2\cos t + \sin t}\end{array}} \right) + i\left( {\begin{array}{*{20}{c}}{5\sin t}\\{2\sin t - \cos t}\end{array}} \right)\\x = {c_1}\left( {\begin{array}{*{20}{c}}{5\cos t}\\{2\cos t + \sin t}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{5\sin t}\\{2\sin t - \cos t}\end{array}} \right)\end{array}
1310
Quiz 5 / Re: Problem 1, night sections
« on: November 21, 2013, 05:33:53 PM »
What a brain dead s.w wrote this code?
$$
\det (A - rI) = \left|\begin{matrix}3 - r &-2\\2& - 2 - r\end{matrix}\right| = r^2-r - 2 = 0\implies r_1=2, r_2=-1
$$
$r = 2\implies \xi_1=2\xi_2$ and we take $\mathbf{\xi}^1= \begin{pmatrix}2\\1\end{pmatrix}$,
$r=-1\implies 2\xi_1=\xi_2$ and we take $\mathbf{\xi}^1= \begin{pmatrix}1\\2\end{pmatrix}$.
Finally
$$
\mathbf{x}= C_1e^{2t}\mathbf{\xi}^1= \begin{pmatrix}2\\1\end{pmatrix}+ C_2e^{-t}\begin{pmatrix}1\\2\end{pmatrix}
$$
As $t\to +\infty$ $|\mathbf{x}(t)|\to \infty $ unless $C_1=0$; then $|\mathbf{x}(t)|\to 0$,
As $t\to -\infty$ $|\mathbf{x}(t)|\to \infty $ unless $C_2=0$; then $|\mathbf{x}(t)|\to 0$.
It is a saddle
$$
\det (A - rI) = \left|\begin{matrix}3 - r &-2\\2& - 2 - r\end{matrix}\right| = r^2-r - 2 = 0\implies r_1=2, r_2=-1
$$
$r = 2\implies \xi_1=2\xi_2$ and we take $\mathbf{\xi}^1= \begin{pmatrix}2\\1\end{pmatrix}$,
$r=-1\implies 2\xi_1=\xi_2$ and we take $\mathbf{\xi}^1= \begin{pmatrix}1\\2\end{pmatrix}$.
Finally
$$
\mathbf{x}= C_1e^{2t}\mathbf{\xi}^1= \begin{pmatrix}2\\1\end{pmatrix}+ C_2e^{-t}\begin{pmatrix}1\\2\end{pmatrix}
$$
As $t\to +\infty$ $|\mathbf{x}(t)|\to \infty $ unless $C_1=0$; then $|\mathbf{x}(t)|\to 0$,
As $t\to -\infty$ $|\mathbf{x}(t)|\to \infty $ unless $C_2=0$; then $|\mathbf{x}(t)|\to 0$.
It is a saddle
1311
Quiz 5 / Re: Problem 2, night sections
« on: November 21, 2013, 05:13:08 AM »
Difficult to read.
1312
Quiz 5 / Re: Problem 1, night sections
« on: November 21, 2013, 05:12:03 AM »
One needs to distinguish cases of $t\to +\infty$ and $t\to -\infty$.
1313
Quiz 5 / Problem 2, night sections
« on: November 20, 2013, 08:40:10 PM »
Express the general solution of the given system of equations in terms of real-valued functions.
\begin{equation*}
\mathbf{x}'=\begin{pmatrix}2 &-5\\1 &-2\end{pmatrix}\mathbf{x}.
\end{equation*}
\begin{equation*}
\mathbf{x}'=\begin{pmatrix}2 &-5\\1 &-2\end{pmatrix}\mathbf{x}.
\end{equation*}
1314
Quiz 5 / Problem 1, night sections
« on: November 20, 2013, 08:39:11 PM »
Find the general solution of the given system of equations and describe the behaviour of
the solution as $t\to \infty$:
\begin{equation*}
\mathbf{x}'=\begin{pmatrix}3 &-2\\2 &-2\end{pmatrix}\mathbf{x}.
\end{equation*}
the solution as $t\to \infty$:
\begin{equation*}
\mathbf{x}'=\begin{pmatrix}3 &-2\\2 &-2\end{pmatrix}\mathbf{x}.
\end{equation*}
1315
Quiz 4 / Problem 2 Night Sections
« on: November 13, 2013, 08:33:40 PM »
7.4 p. 395 \#6
Consider the vectors $\mathbf{x}^{(1)}(t) = \begin{pmatrix}t\\1\end{pmatrix}$ and $\mathbf{x}^{(2)}(t) = \begin{pmatrix}t^2\\2t\end{pmatrix}$.
(a) Compute the Wronskian of $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$.
(b) In what intervals are $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$ linearly independent?
(c) What conclusion can be drawn about the coefficients in the system of homogeneous differential equations satisfied by $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$?
Consider the vectors $\mathbf{x}^{(1)}(t) = \begin{pmatrix}t\\1\end{pmatrix}$ and $\mathbf{x}^{(2)}(t) = \begin{pmatrix}t^2\\2t\end{pmatrix}$.
(a) Compute the Wronskian of $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$.
(b) In what intervals are $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$ linearly independent?
(c) What conclusion can be drawn about the coefficients in the system of homogeneous differential equations satisfied by $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$?
1316
Quiz 4 / Problem 1 Night Sections
« on: November 13, 2013, 08:31:56 PM »
5.2 p. 264 \#21 (a)
\begin{equation*}
y''- 2xy'+\lambda y = 0,\qquad -\infty < x < \infty,
\end{equation*}
where $\lambda$ is a constant, is known as the Hermite equation. It is an important equation in mathematical physics.
Find the first four terms in each of two solutions about $x = 0$ and show that they form a fundamental set of solutions.
\begin{equation*}
y''- 2xy'+\lambda y = 0,\qquad -\infty < x < \infty,
\end{equation*}
where $\lambda$ is a constant, is known as the Hermite equation. It is an important equation in mathematical physics.
Find the first four terms in each of two solutions about $x = 0$ and show that they form a fundamental set of solutions.
1318
Quiz 3 / Problem 2 (night sections)
« on: November 06, 2013, 08:12:36 PM »
Find the general solution of the given differential equation. Leave your answer in terms of one or more integrals.
\begin{equation*}
y'''-y'' + y'-y = \sec (t), \qquad -\frac{\pi}{2} < t < \frac{\pi}{2}.
\end{equation*}
\begin{equation*}
y'''-y'' + y'-y = \sec (t), \qquad -\frac{\pi}{2} < t < \frac{\pi}{2}.
\end{equation*}
1319
Quiz 3 / Problem 1 (night sections)
« on: November 06, 2013, 08:11:53 PM »
Find the general solution of the given differential equation.
\begin{equation*}
y'''-y''-y'+ y = 0.
\end{equation*}
\begin{equation*}
y'''-y''-y'+ y = 0.
\end{equation*}
1320
Quiz 2 / Re: Problem 1, Night sections
« on: November 01, 2013, 04:28:08 PM »Question1
What is the reason to post inferior (scanned) solution after a better -- typed has been posted?