Question

n Problems 29 and 30 consider the autonomous differential equationdy/dx = f(y), where the graph of f is given. Use the graph to locatethe critical points of each differential equation. Sketch a phase portrait of each differential equation. By hand, sketch typical solutioncurves in the sub-regions in the xy-plane determined by the graphs ofthe equilibrium solutions.

FIGURE 2.1.19 Graph for Problem 30

Short answer

• The critical points are approximately$0.5,1.6\text{\hspace{0.17em}and\hspace{0.17em}}-2.1$
• The graph is drawn below.

Step-by-step solution

Step 1:Critical points

The zeros of the function f in$dy/dx=f\left(y\right)$are of special importance. Wesay that a real number c is a critical point of the autonomous differential equation $dy/dx=f\left(y\right)$if it is a zero of f—that is,$f\left(c\right)=0$ . A critical point is also called an equilibrium point or stationary point.

Step 2:Critical points and phase portrait

Critical points are the solutions of $f\left(y\right)=0$.

From the graph it is clear that the solutions are approximately$0.5,1.6\text{\hspace{0.17em}and\hspace{0.17em}}-2.1$

We have critical points.$y=-2.1,y=0.5,y=1.6$

From the graph we can also see that before -2.1, f is positive, between -2.1 and 0.5 the graph is negative between 0.5 and 1.6 the graph is positive and after 1.6 the graph is negative, from which we can see that arrows point up, down, up, down.

The phase portrait is shown below:

Since both arrows are pointing away from 0.5, it is an unstable critical point. Since both arrows are pointing towards -2.1 and 1.6, they are stable critical points.

Solution curve

Some typical solution curves look like this,

From the graph we can guess that the critical points are approximately$0.5,1.6\text{\hspace{0.17em}and\hspace{0.17em}}-2.1$

Hence, the critical points are approximately$0.5,1.6\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}-2.1$