91Ó°ÊÓ

Let's first find the critical point. One has to solve the system\({\bf{x' = 0,y' = 0}}\), so, one has:

\(\begin{array}{c}{\bf{0 = 4 - 4y}} \Rightarrow {\bf{y = 1,}}\\{\bf{0 = - 4x}} \Rightarrow {\bf{x = 0}}\end{array}\)

So, the critical point is\(\left( {{\bf{x, y}}} \right){\bf{ = }}\left( {{\bf{0,1}}} \right)\). Now one can solve the related phase plane equation which is:

\(\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}\frac{{\frac{{{\bf{dy}}}}{{{\bf{dt}}}}}}{{\frac{{{\bf{dx}}}}{{{\bf{dt}}}}}}{\bf{ = }}\frac{{{\bf{ - 4x}}}}{{{\bf{4 - 4y}}}}\\{\bf{ = }}\frac{{{\bf{4x}}}}{{{\bf{4y - 4}}}}\\{\bf{(4y - 4)dy = 4xdx}}\end{array}\)

Integrating the previous equation, one will get:

\(\begin{array}{c}\int {{\bf{(4y - 4)}}} {\bf{dy = }}\int {\bf{4}} {\bf{xdx}}\\{\bf{4}}\frac{{{{\bf{y}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - 4y = 4}}\frac{{{{\bf{x}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ + C}}\\{{\bf{y}}^{\bf{2}}}{\bf{ - 2y = }}{{\bf{x}}^{\bf{2}}}{\bf{ + }}\frac{{\bf{C}}}{{\bf{2}}}\\{{\bf{x}}^{\bf{2}}}{\bf{ - (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = - 1 - }}\frac{{\bf{C}}}{{\bf{2}}}\end{array}\)

Introducing a new constant \({\bf{c = - 1 - }}\frac{{\bf{C}}}{{\bf{2}}}\)one has that the solution is\({{\bf{x}}^{\bf{2}}}{\bf{ - (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\)which in \({\bf{xy - }}\)plane represents hyperbole with a focus on the critical point.

The critical point is the saddle and it is unstable. In the figure below are some trajectories with their flow arrows, the critical point, and one particular solution for \({\bf{c = 1}}\) (red line).