Q16RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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.

\(\begin{array}{c}{\bf{x}}'{\bf{ =  - x - }}2{\bf{y}}\\{\bf{y}}'{\bf{ = x}} + {\bf{y}}\end{array}\)

Step-by-Step Solution

Answer

The critical point \(\left( {{\bf{0,0}}} \right)\) is an unstable saddle point.

Step 1: 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\(3y{\bf{ = 0}}\).

Step 2: 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.