Question

In Problems 13 and 14 the given figure represents the graph of

and, respectively. By hand, sketch a direction field over an appropriate grid for(Problem 13) and then forProblem 14).

FIGURE 2.1.16 Graph for Problem 13

Short answer

In the intervals\( - \infty < y < - \sqrt 3 - 1\)and\( - 1 < y < \sqrt 3 - 1\)we have\(y' > 0\), hence solution increases.

In the intervals \(\sqrt 3 - 1 < y < - 1,\) and \(\sqrt 3 - 1 < y < \infty \) we have\(y' < 0\), hence solution decreases.

Step-by-step solution

Direction field.

If we systematically evaluate \(f\) over a rectangular grid of points in the \(xy\)-plane and draw a line element at each point \((x,y)\)of the grid with a slope \(f(x,y)\)then the collection of all these line elements is called a direction field or a slope field of the differential equation \(\frac{{dy}}{{dx}} = f(x,y)\).

Sketch the graph.

From the graph, observe that graph\(f\) intersects \(y\) axis at\(y = - \sqrt 3 - 1, - 1,\sqrt 3 - 1\).

So the critical points of the differential equation \(\frac{{dy}}{{dx}} = f(y)\) are\(y = - \sqrt 3 - 1, - 1,\sqrt 3 - 1\),

Put the critical points on the vertical line and then divide the line into five intervals as shown:

\(\begin{aligned}{l} - \infty < y < - \sqrt 3 - 1,\sqrt 3 - 1 < y < - 1,\\ - 1 < y < \sqrt 3 - 1,\sqrt 3 - 1 < y < \infty \end{aligned}\)

Sketch of the direction field for \(\frac{{dy}}{{dx}} = f(y)\) is shown below:

Therefore, in the intervals \( - \infty < y < - \sqrt 3 - 1\) and \( - 1 < y < \sqrt 3 - 1\) we have\(y' > 0\), hence solution increases.

In the intervals \(\sqrt 3 - 1 < y < - 1,\) and \(\sqrt 3 - 1 < y < \infty \) we have\(y' < 0\), hence solution decreases.