Follow a solution from its initial point

Tracing a path

The field shows \(\dfrac{dy}{dx} = x - y\). The condition \(y(0) = 1\) means that the solution passes through the orange point \((0, 1)\).

Press "Trace to the right". The path leaves the start along the segment there, where the slope is \(0 - 1 = -1\), so the path falls.

As the path moves to the right, it turns to stay parallel to each segment it meets. At about \(x = 0.7\) it meets a flat segment, and after that the segments it meets rise, so the path rises.

Press "Trace to the left". To the left of the start the segments have negative slopes, so the path rises as it moves left.

One path through each point

The figure does not name the equation of this field. Drag the orange point, and a new path appears through it, both to the right and to the left.

At every point, the path has the slope of the segment there, so it stays parallel to each segment it meets.

Each point has one segment, so only one path passes through each point. Move the slider to show other paths: none of them crosses another.

Notation

1. An initial condition gives the starting point. It means that the solution passes through the point \((x_0, y_0)\).

   \[\cy{y(x_0) = y_0}\]

2. Draw a short piece of the path along the segment at the start.

3. At the end of that piece, turn the path to match the segment there. Repeat this to the right and to the left of the start.

4. The path has the same slope as the field at every point where it passes. The solution is the function \(y(x)\) whose graph is this path.

   \[y'(x) = f\big(x, y(x)\big), \quad \cy{y(x_0) = y_0}\]

   This needs the field to give a slope at every point that the path reaches.

When it fails

Press the button to change the field to \(\dfrac{dy}{dx} = -\dfrac{x}{y}\) and to start at \((0, 2)\). This equation divides by \(y\), so it gives no slope where \(y = 0\).

The path through \((0, 2)\) is the upper half of a circle. Near \((2, 0)\) and \((-2, 0)\) the segments turn vertical, and at those points \(y = 0\), so the path stops at the red rings. The solution exists only for \(x\) between \(-2\) and 2.

Example 1: A rising path

A solution of \(\dfrac{dy}{dx} = x - y\) has \(y(2) = 0\). Describe its path for \(x \ge 2\).

1. At the start \((2, 0)\), the slope is 2, so the path rises steeply.

   \[2 - 0 = 2\]

2. Along a short piece, \(y\) rises faster than \(x\), because the slope is more than 1, so \(x - y\) becomes smaller. At about \((2.5, 0.9)\) the slope is 1.6, so the path becomes less steep.

   \[2.5 - 0.9 = 1.6\]

3. The path continues to rise, and its slope continues to decrease. At \(x = 3\) the slope is about 1.4.

Example 2: A path that bends

A solution of \(\dfrac{dy}{dx} = x - y\) has \(y(-2) = 2\). Describe its path for \(x \ge -2\).

1. At the start \((-2, 2)\), the slope is \(-4\), so the path falls steeply.

   \[-2 - 2 = -4\]

2. A tempting wrong move: Extend the first segment as a straight line.

   \[y = 2 - 4(x + 2)\]

   At \(x = -1\) the line is at \((-1, -2)\), where the segment has slope \(-1 - (-2) = 1\) and rises. The line falls with slope \(-4\), so it does not follow the field.

3. Draw short pieces instead. After each piece, turn the path to match the new segment. Near \(x = -1\) the path is at about \(y = -0.2\), where the slope is only \(-0.8\).

4. At about \(x = -0.4\) the path meets a flat segment, so it is flat there, and after that it rises.

Practice

A solution of \(\dfrac{dy}{dx} = x - y\) has \(y(1) = 3\). Describe its path for \(x \ge 1\).

With the start at \((1, 3)\), follow the path to the right. It is flat where it meets a flat segment. On which line are the segments flat?

Questions

The figure shows the slope field of \(\dfrac{dy}{dx} = 1 - \dfrac{x}{2}\) and the orange start \((0, 0)\). Questions 1 and 2 use the solution through the start.

1. Where is the highest point of the path through the origin?

   • At \(x = 2\)
   • At \(x = 0\)
   • At \(x = 4\)
   Answer

   At \(x = 2\)

2. At which \(x\) does the path through the origin return to the x-axis?

   • \(x = 4\)
   • \(x = 2\)
   • \(x = 3\)
   • \(x = 6\)
   Answer

   \(x = 4\)

3. A solution of \(\dfrac{dy}{dx} = 1 - \dfrac{x}{2}\) has \(y(0) = 3\). What is \(y(2)\)?

   • 4
   • 5
   • 1
   Answer

   4

Summary

You start at the initial point and draw a path with the slope of each segment it meets, to the right and to the left.

This works because the field gives one slope at each point, so one path passes through the start and stays parallel to the segments.

You use it when a problem gives a slope field and an initial condition, and asks for the shape or the value of the solution.

This is one of five skills in topic 7.4 of Which & Why: AP Calculus.