Q13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11–14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows).

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{(}\mathit{y}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{(}\mathit{y}\mathbf{-}\mathbf{1}\mathbf{)} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{(}\mathit{x}\mathbf{-}\mathit{y}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)} \end{gathered}$$

Step-by-Step Solution

Answer

The solution is $\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathit{c}$.

Step 1: Find phase plane equation

Here the system is:

$$\begin{gathered} \frac{dx}{dt}=(y-x)(y-1) \\ \frac{dy}{dt}=(x-y)(x-1) \end{gathered}$$

And the phase plane equation is:

$$\begin{gathered} \frac{dy}{dx}=\frac{(x-y)(x-1)}{(y-x)(y-1)} \\ \frac{dy}{dx}=\frac{1-x}{y-1} \end{gathered}$$

Step 2: Solve the equation

Here the equation is $\frac{dy}{dx}=\frac{1-x}{y-1}$.

Solving by variable separating. Then,

$$\begin{gathered} \int \left(y-1\right)dy=\int -\left(x-1\right)dx \\ {\left(y-1\right)}^2+{\left(x-1\right)}^2=c \end{gathered}$$

Since the solutions are centered at (1,1). And line y = x.

Step 3: Sketch some trajectories.

Therefore, the solution is $\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathit{c}$.