Q12RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 12 and 13, solve the related phase plane equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows) and describe the stability of the critical points (i.e., compare with figure 5.12, page 267).

\({\bf{x' = y - 2,y' = 2 - x}}\)

Step-by-Step Solution

Answer

The critical point is (2, 2) and the solution is\({{\bf{(y - 2)}}^{\bf{2}}}{\bf{ + (x - 2}}{{\bf{)}}^{\bf{2}}}{\bf{ = }}{{\bf{c}}^{\bf{2}}}\).

 

Step 1: Find the critical points

Here\({\bf{x' = y - 2,y' = 2 - x}}\)

 

For the critical points put the system equal to zero.

 

\(\begin{array}{c}{\bf{x' = 0}}\\{\bf{y - 2 = 0}}\\{\bf{y = 2}}\\{\bf{y' = 0}}\\{\bf{2 - x = 0}}\\{\bf{x = 2}}\end{array}\)
 

So, the critical point is (x, y) = (2, 2).

Step 2: Solve for phase equation

The phase plane equation is:

 

\(\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}\frac{{{\bf{2 - x}}}}{{{\bf{y - 2}}}}\\\int {{\bf{(y - 2)dy}}}  = \int {{\bf{(2 - x)dx}}} \\\frac{{{{\bf{y}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - 2y = 2x - }}\frac{{{{\bf{x}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ + c}}\\{{\bf{(y - 2)}}^{\bf{2}}}{\bf{ + (x - 2}}{{\bf{)}}^{\bf{2}}}{\bf{ = 2c - 8}}\\{{\bf{(y - 2)}}^{\bf{2}}}{\bf{ + (x - 2}}{{\bf{)}}^{\bf{2}}}{\bf{ = }}{{\bf{c}}^{\bf{2}}}\end{array}\)

Step 3:Sketch

Therefore, this is the required result.