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Mobile App Chapter 5: Q16RP (page 306)
Sketch some typical trajectories for the given system and by comparing with Figure\({\bf{5}}{\bf{.12}}\), page\({\bf{267}}\), identify the type of critical point at the origin.
\(\begin{array}{c}{\bf{x}}'{\bf{ = - x - }}2{\bf{y}}\\{\bf{y}}'{\bf{ = x}} + {\bf{y}}\end{array}\)
Short Answer
The critical point \(\left( {{\bf{0,0}}} \right)\) is an unstable saddle point.
Step by step solution
Finding the value of \({\bf{x,y}}\)
One can solve the critical point. To do so we need to solve the system\({\bf{x}}'{\bf{ = 0,y}}'{\bf{ = 0}}\), so one has
\(\begin{array}{c}{\bf{0 = - x - }}2{\bf{y}}\\{\bf{0 = x}} + {\bf{y}}\end{array}\)
The first equation gives us that\(x{\bf{ = 2 }}y\), so substituting this into the second equation one gets that\(3y{\bf{ = 0}}\).
Finding the critical point
So, one has that \({\bf{x = 0, y = 0}}\) and the critical point is\(\left( {{\bf{x, y}}} \right){\bf{ = }}\left( {{\bf{0,0}}} \right)\).
Comparing this picture with the Figure \({\bf{5}}{\bf{.12}}\) in the Textbook one can conclude that the critical point \(\left( {{\bf{0,0}}} \right)\) is an unstable saddle point.
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