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, unstable, or semistable (see Problem 7 ). $$ d y / d t=y^{2}\left(y^{2}-1\right), \quad-\infty

Short answer

Short Answer: The critical points for the given differential equation, \(y^2(y^2-1)\), are 0 and \(\pm 1\). The equilibrium point 0 is unstable, while the equilibrium points 1 and -1 are asymptotically stable.

Step-by-step solution

Sketch the graph of \(f(y)\) versus \(y\) for the given differential equation

Create a graph for the given differential equation, \(f(y) = \frac{dy}{dt} = y^2 (y^2 - 1)\). This function is a polynomial of degree 4, specifically a quartic function, which means it will have up to 4 roots. You can plot it using any graphing calculator or software.

Determine the critical (equilibrium) points

Critical or equilibrium points are the points where \(\frac{dy}{dt}=0\). In this case, 0 = \(y^2 (y^2 - 1)\). We can find these points by setting each factor equal to zero: (1) \(y^2 = 0\), which implies \(y = 0\). (2) \(y^2 - 1 = 0\), which implies \(y^2 = 1\). Hence, \(y = \pm 1\). So, the critical points are 0 and \(\pm 1\).

Classify each critical point as asymptotically stable, unstable, or semi-stable

To determine the stability of each critical point, we need to analyze the behavior of \(f(y)\) for all critical points in \(y\). (a) For \(y = 0\), 1. If \(y < 0\), then \(f(y) > 0\) (right side of the differential equation is positive), so \(\frac{dy}{dt} > 0\), implying that \(y\) is increasing over time. 2. If \(0 < y < 1\), then \(f(y) > 0\) (right side of the differential equation is positive), so \(\frac{dy}{dt} > 0\), implying that \(y\) is increasing over time. Since in the region of \(y=0\), \(\frac{dy}{dt} > 0\) always holds, \(y=0\) is an unstable equilibrium point. (b) For \(y = 1\), 1. If \(0 < y < 1\), then \(f(y) > 0\) (right side of the differential equation is positive), so \(\frac{dy}{dt} > 0\), implying that \(y\) is increasing over time. 2. If \(y > 1\), then \(f(y) < 0\) (right side of the differential equation is negative), so \(\frac{dy}{dt} < 0\), implying that \(y\) is decreasing over time. As the behavior in the surrounding regions has opposite trends, \(y=1\) is an asymptotically stable equilibrium point. (c) For \(y = -1\), 1. If \(y < -1\), then \(f(y) < 0\) (right side of the differential equation is negative), so \(\frac{dy}{dt} < 0\), implying that \(y\) is decreasing over time. 2. If \(-1 < y < 0\), then \(f(y) < 0\) (right side of the differential equation is negative), so \(\frac{dy}{dt} < 0\), implying that \(y\) is decreasing over time. As before, the behavior in the surrounding regions is opposite, so \(y=-1\) is an asymptotically stable equilibrium point. In summary, the critical points 1 and -1 are asymptotically stable, while the critical point 0 is unstable.