Question

Involve equations of the form \(d y / d t=f(y)\). In each problem sketch the graph of \(f(y)\) versus \(y,\) determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. $$ d y / d t=-2(\arctan y) /\left(1+y^{2}\right), \quad-\infty

Short answer

Answer: The equilibrium point at \(y = 0\) is asymptotically stable.

Step-by-step solution

Understand the given differential equation

The given differential equation is: $$ \frac{dy}{dt} = -2\frac{\arctan{y}}{1+y^2} $$ This is an autonomous first-order differential equation because there is no explicit time dependence in the equation.

Sketch the graph of \(f(y)\) versus \(y\)

To sketch the graph of \(f(y) = -2\frac{\arctan{y}}{1+y^2}\) versus \(y\), we can consider the behavior of \(f(y)\) at some key points. 1. When \(y = 0\), we have \(f(0) = -2\frac{\arctan{0}}{1+0^2} = 0\). 2. When \(y \rightarrow \infty\), since \(\arctan{y}\) approaches \(\frac{\pi}{2}\), we have \(f(y) \rightarrow -2\frac{\frac{\pi}{2}}{1+y^2} = 0\). 3. When \(y \rightarrow -\infty\), since \(\arctan{y}\) approaches \(-\frac{\pi}{2}\), we have \(f(y) \rightarrow -2\frac{-\frac{\pi}{2}}{1+y^2} = 0\). From these results, we can observe that the function \(f(y)\) is a continuous and bounded function in the range \(-\infty < y < \infty\). It starts at zero for \(y \rightarrow -\infty\), goes to zero at \(y=0\), and finally approaches zero as \(y \rightarrow \infty\). Now we sketch the graph of \(f(y)\) versus y, using the above insights.

Determine the critical (equilibrium) points

Critical points (equilibrium points) occur when \(\frac{dy}{dt} = 0\). This implies the following equation: $$ -2\frac{\arctan{y}}{1+y^2} = 0 $$ Since the only case when the equation can be true is when the numerator is zero (\(\arctan{y} = 0\)), the only equilibrium point is \(y = 0\).

Classify each critical point as asymptotically stable or unstable

Now we will determine if the equilibrium point is asymptotically stable or unstable by examining the behavior of \(f(y)\) in the vicinity of \(y = 0\). For \(y > 0\), we have \(\arctan{y} > 0\), and since \(1+y^2 > 1\), then \(\frac{\arctan{y}}{1+y^2} > 0\). Therefore, \(f(y) = -2\frac{\arctan{y}}{1+y^2} < 0\). This means that if \(y > 0\), then \(\frac{dy}{dt} < 0\), indicating that \(y\) will decrease towards the equilibrium point. For \(y < 0\), we have \(\arctan{y} < 0\), and since \(1+y^2 > 1\), then \(\frac{\arctan{y}}{1+y^2} < 0\). Therefore, \(f(y) = -2\frac{\arctan{y}}{1+y^2} > 0\). This means that if \(y < 0\), then \(\frac{dy}{dt} > 0\), indicating that \(y\) will increase towards the equilibrium point. Since both conditions \(y > 0\) and \(y < 0\) show that the system will converge to the equilibrium point at \(y = 0\), we can conclude that this point is asymptotically stable.