12-267/Homework Assignment 6: Difference between revisions
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This assignment is due in class on Friday November 9. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if your readers have a hard time deciphering your work they will give up and assume it is wrong. |
This assignment is due in class on Friday November 9. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if your readers have a hard time deciphering your work they will give up and assume it is wrong. |
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# <math>v'=\begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix}v + \begin{pmatrix} e^t \\ t \end{pmatrix}</math>. |
# <math>v'=\begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix}v + \begin{pmatrix} e^t \\ t \end{pmatrix}</math>. |
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# <math>v'=\begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix}v + \begin{pmatrix} -\cos t \\ \sin t \end{pmatrix}</math>. |
# <math>v'=\begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix}v + \begin{pmatrix} -\cos t \\ \sin t \end{pmatrix}</math>. |
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'''Task 4.''' Assume <math>t>0</math>. For the following equation, |
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<center><math>tv'=\begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix}v + \begin{pmatrix} 1-t^2 \\ 2t \end{pmatrix}</math></center>, |
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it is given that a solution of the homogeneous version is |
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<center><math>v(t) = c_1\begin{pmatrix}1\\1\end{pmatrix}t + \begin{pmatrix}1\\3\end{pmatrix}t^{-1}</math>.</center> |
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Use "fundamental solutions" to find a solution of the full equation. |
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'''Task 5.''' (Not for grade). Find a quadratic differential equation whose phase portrait is as below. |
'''Task 5.''' (Not for grade). Find a quadratic differential equation whose phase portrait is as below. |
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Revision as of 20:33, 2 November 2012
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This assignment is due in class on Friday November 9. Here and everywhere, neatness counts!! You may be brilliant and you may mean just the right things, but if your readers have a hard time deciphering your work they will give up and assume it is wrong.
Task 1. Draw the phase portraits for the following systems, near [math]\displaystyle{ (x,y)=(0,0) }[/math]:
- [math]\displaystyle{ \begin{cases} \dot{x}=2x+y \\ \dot{y}=-x+4y \end{cases} }[/math].
- [math]\displaystyle{ \begin{cases} \dot{x}=4x-5y \\ \dot{y}=4x-4y \end{cases} }[/math].
- [math]\displaystyle{ \begin{cases} \dot{x}=x-2y \\ \dot{y}=-2x+4y \end{cases} }[/math].
- [math]\displaystyle{ \begin{cases} \dot{x}=-x+y \\ \dot{y}=-5x+3y \end{cases} }[/math].
- [math]\displaystyle{ \begin{cases} \dot{x}=-5x+4y \\ \dot{y}=-8x+7y \end{cases} }[/math].
Task 2. Draw the phase portrait of the system
[math]\displaystyle{ \begin{cases}\dot{x}=17+x-9y+\sin(2-2x-y+xy)\\\dot{y}=7+2x-5y+\cos(x-1)\end{cases} }[/math]
near the point [math]\displaystyle{ (x,y)=(1,2) }[/math].
Task 3. Solve using diagonalization (one solution is enough):
- [math]\displaystyle{ v'=\begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix}v + \begin{pmatrix} e^t \\ t \end{pmatrix} }[/math].
- [math]\displaystyle{ v'=\begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix}v + \begin{pmatrix} -\cos t \\ \sin t \end{pmatrix} }[/math].
Task 4. Assume [math]\displaystyle{ t\gt 0 }[/math]. For the following equation,
,
it is given that a solution of the homogeneous version is
Use "fundamental solutions" to find a solution of the full equation.
Task 5. (Not for grade). Find a quadratic differential equation whose phase portrait is as below.
Hint. "Monkey Saddle".