See the field before solving

Chapter 2 - Turn a rate rule into a picture

An equation $y'=f(t,y)$ assigns one slope to every point $(t,y)$ . A direction field draws a small segment with that slope at each point. A solution curve does not invent its path; it threads through the assigned segments.

By the end of this chapter, you should be able to:

read increasing, decreasing, and equilibrium behavior from a field;
sketch solution curves through initial conditions;
connect features of $f(t,y)$ to visible features of the field; and
identify what a numerical or symbolic answer must look like before calculating it.

Lab: move the initial condition

Direction-field explorerEvery short mark is a local instruction.

Rate rule

local slopesolution curveinitial condition

Try these moves before reading on:

Choose $y'=-0.65y$ . Place the initial point above, on, and below $y=0$ .
Choose the logistic equation. Find every horizontal row where the marks go flat.
Choose $y'=\sin t-0.35y$ . Notice that equal heights can have different slopes at different times.

Read a field in four passes

1. Find zero-slope sets

Where $f(t,y)=0$ , the field is horizontal. If the equation is autonomous, $y'=f(y)$ , its rate depends only on the state, and a horizontal zero-slope row often gives an equilibrium solution: a state that stays constant.

For $y'=y(4-y)$ , the zero-rate states are $y=0$ and $y=4$ .

2. Mark the sign regions

$y'\gt 0$ : solution curves rise as $t$ increases.
$y'\lt 0$ : solution curves fall as $t$ increases.
large $|y'|$ : solution curves look steep.

For $y'=y(4-y)$ , the rate is positive between 0 and 4 and negative above 4. Positive solutions move toward 4.

3. Ask what stays constant

If $f$ depends only on $y$ , every mark on one horizontal row has the same slope. If $f$ depends only on $t$ , every mark in one vertical column has the same slope. These repeated patterns identify structure before algebra does.

4. Thread, do not connect dots

A solution curve stays tangent to the field and remains smooth where $f$ is smooth. It cannot turn at a point where the assigned slope is strictly positive or cross an equilibrium when uniqueness applies.

Field fingerprint. A direction field looks identical along every horizontal row but changes from one row to the next. What form is most likely?

Worked example: predict without solving

Read $y'=2-y$ , $y(0)=5$ fully worked

Zero rate: $2-y=0$ at $y=2$ . The row $y=2$ is horizontal.
Sign: above $y=2$ , $y'\lt 0$ ; below it, $y'\gt 0$ . Arrows point toward 2 from both sides.
Initial motion: $y'(0)=2-5=-3$ , so the curve starts downward with slope $-3$ .
Long-run prediction: the rate weakens as $y$ nears 2, so the curve falls and flattens toward $y=2$ .
Shape check: differentiating the equation gives $y''=-y'$ . While the solution falls, $y'\lt 0$ , so $y''\gt 0$ . The curve is concave up.

Same structure, less support

For $y'=1-\tfrac12 y$ , $y(0)=-2$ :

the equilibrium is $y=\underline{\hspace{2em}}$ ;
the initial slope is $\underline{\hspace{3em}}$ ;
the solution moves [toward / away from] the equilibrium; and
while it rises, its concavity is [up / down].

Check the four predictions

The equilibrium is $y=2$ ; the initial slope is $2$ ; the solution moves toward 2; and $y''=-\tfrac12 y'\lt 0$ , so it is concave down while rising.

What can go wrong?

Error clinic: treating a slope mark as a point to hit

The small segment at $(1,2)$ reports the slope a solution would have if it passed through $(1,2)$ . It does not require every solution to pass through that point. A field encodes infinitely many possible initial-value problems at once.

Error clinic: drawing vertical solution segments

A classical solution $y(t)$ is a function of $t$ , so its graph cannot double back or become vertical. Very steep is possible; vertical is not a finite derivative.

When may solution curves cross?

If $f(t,y)$ and its dependence on $y$ are sufficiently well behaved near a point, one initial condition selects one local solution. Two distinct solution curves cannot cross there: the crossing point would give the same initial condition two different futures.

This is a conceptual use of uniqueness. You do not need the full theorem yet. Use the practical rule: for the smooth equations in the early chapters, never sketch two solution curves crossing.

Practice

Practice - qualitative
2.1 For $y'=y-3$ , identify the equilibrium, sign regions, and long-run direction for initial values 1, 3, and 5.
Detailed solution
Practice - field fingerprint
2.2 Describe how the fields for $y'=\cos t$ , $y'=\cos y$ , and $y'=\cos(t+y)$ differ in repeated pattern.
Detailed solution
Integrated - derivatives
2.3 A solution satisfies $y'=y(1-y)$ and $y(0)=1/4$ . Predict monotonicity, limiting state, and the sign of $y''$ before and after $y=1/2$ .
Detailed solution
Challenge - nonuniqueness
2.4 Verify that both $y=0$ and $y=(t/2)^2$ for $t\ge0$ solve $y'=\sqrt{y}$ with $y(0)=0$ . What field-sketching rule must you suspend at this nonsmooth point?
Detailed solution
Practice - extension
2.5 For $y'=t-y$ , mark the zero-slope line and sign regions. Is the line $y=t$ itself a solution? For the solution through $(0,0)$ , compute $y'(0)$ and $y''(0)$ , then predict its initial motion for small positive time.
Detailed solution
Challenge - transfer
2.6 For $y'=t^2-y$ , derive $y''$ along a solution. A solution passes through $(1,1)$ , while another passes through $(-1,1)$ . Both have zero slope there. Determine which point is a local minimum and which is a local maximum, and explain why horizontal slope alone does not settle the question.
Detailed solution

Spaced retrieval 1

Without looking back, translate: "A quantity moves toward 10 at a rate equal to twice its signed distance from 10." Check the sign by testing a state above 10.

Reveal after committing

$y'=-2(y-10)=2(10-y)$ . At $y=12$ , the rate is $-4$ , so the state moves down toward 10.

I manipulated each field, made predictions first, and completed the spaced retrieval.

Next: Learn the structural fingerprints that select a solution method.

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Differential Equations as a Second Language