- #1
Benny
- 584
- 0
Hi, I'm unsure about how to do the following question.
I am given the following system for which I first need to find the general solution.
[tex]
\left[ {\begin{array}{*{20}c}
{\mathop x\limits^ \bullet } \\
{\mathop y\limits^ \bullet } \\
\end{array}} \right] = \left[ {\begin{array}{*{20}c}
3 & { - 2} \\
6 & { - 5} \\
\end{array}} \right]\left[ {\begin{array}{*{20}c}
x \\
y \\
\end{array}} \right]
[/tex]
I found the general solution to be:
[tex]
\left[ {\begin{array}{*{20}c}
x \\
y \\
\end{array}} \right] = c_1 \left[ {\begin{array}{*{20}c}
1 \\
1 \\
\end{array}} \right]e^t + c_2 \left[ {\begin{array}{*{20}c}
1 \\
3 \\
\end{array}} \right]e^{ - 3t}
[/tex]
I then sketched the phase portrait which in this case is a saddle. The question asks me to use this phase portrait to sketch x(t) and y(t) on the same graph for the case of x = 1, y = 2.9 when t = 0.
I can't think of a way to do this. The phase portrait I've sketched is for the general solution. I've tried a few things with the point (x,y) = (1,2.9) including shifting the 'centre' of the saddle to that point but none of the things I've tried have any reasoning behind them. They're just random things I've tried which haven't lead me anywhere. Can someone please help me out? Thanks.
I am given the following system for which I first need to find the general solution.
[tex]
\left[ {\begin{array}{*{20}c}
{\mathop x\limits^ \bullet } \\
{\mathop y\limits^ \bullet } \\
\end{array}} \right] = \left[ {\begin{array}{*{20}c}
3 & { - 2} \\
6 & { - 5} \\
\end{array}} \right]\left[ {\begin{array}{*{20}c}
x \\
y \\
\end{array}} \right]
[/tex]
I found the general solution to be:
[tex]
\left[ {\begin{array}{*{20}c}
x \\
y \\
\end{array}} \right] = c_1 \left[ {\begin{array}{*{20}c}
1 \\
1 \\
\end{array}} \right]e^t + c_2 \left[ {\begin{array}{*{20}c}
1 \\
3 \\
\end{array}} \right]e^{ - 3t}
[/tex]
I then sketched the phase portrait which in this case is a saddle. The question asks me to use this phase portrait to sketch x(t) and y(t) on the same graph for the case of x = 1, y = 2.9 when t = 0.
I can't think of a way to do this. The phase portrait I've sketched is for the general solution. I've tried a few things with the point (x,y) = (1,2.9) including shifting the 'centre' of the saddle to that point but none of the things I've tried have any reasoning behind them. They're just random things I've tried which haven't lead me anywhere. Can someone please help me out? Thanks.