Chapter 3, Problem 1RE

To determine

To classify: The equilibrium solution.

Answer to Problem 1RE

The equilibrium solution is at $y=\frac{1}{2}$, and hence the solution is unstable at that point.

Explanation of Solution

Given:

The given differential equation is $\frac{dy}{dt}=y\left(1-2y\right)$.

Approach:

The differential equation of population model is, $\frac{dy}{dt}=-r\left(1-\frac{1}{A}y\right)\left(1-\frac{1}{B}y\right)y$.

If $f^{\prime }\left(y\right)>0$, then the system is unstable at $y$.

If $f^{\prime }\left(y\right)<0$, then the system is stable at $y$.

Calculation:

The given differential equation is $\frac{dy}{dt}=y\left(1-2y\right)$.

Note that the equilibrium solutions are $y=0,y=\frac{1}{2}$.

Consider the function, $f\left(y\right)=y-2y^2$.

Differentiate the above function as shown below.

$f^{\prime }\left(y\right)=1-4y$

Substitute $y=0$ in above equation.

$\begin{array}{c}{f}^{\prime }\left(0\right)=1-4\left(0\right)\\ =1\end{array}$

Substitute $y=\frac{1}{2}$ in above equation.

$\begin{array}{c}{f}^{\prime }\left(\frac{1}{2}\right)=1-4\left(\frac{1}{2}\right)\\ =-1\end{array}$

Note that at the point $y=\frac{1}{2}$, $f^{\prime }\left(\frac{1}{2}\right)<0$ and at the point $y=0$ $f^{\prime }\left(0\right)>0$.

Therefore, the equilibrium solution is at $y=\frac{1}{2}$ and the function is unstable at that point.