Read local directions before solving

Try

The equation y′=t−y gives a slope at every point in the time-state plane. You do not need a solution formula to draw a short segment at (0,1): its slope is −1. At (1,0), the slope is 1. Draw those two segments with time increasing to the right. These segments describe possible local directions, not jumps between grid points.

Choose the initial point (0,0) and imagine a smooth curve following the directions. It begins horizontally. Does that mean it remains horizontal forever? Keep this question visible while building a larger field.

Predict

Every point on the line y=t has slope zero. Predict whether the line itself is a solution. Then differentiate y=t. Its derivative is one, whereas the equation demands zero on the line. A nullcline, a set where the slope vanishes, need not be a trajectory. This distinction becomes especially useful when the equation depends explicitly on time.

Experiment

Calculate the field below before copying its slopes onto a grid. Use identical horizontal and vertical scale units, and draw short segments rather than arrows whose length could be mistaken for speed.

State y	Slope at t=0	Slope at t=1	Slope at t=2
−1	1	2	3
0	0	1	2
1	−1	0	1

Starting at (0,0), compare the direction at the starting point with directions just to its right and above it. The curve can leave its horizontal tangent because the field changes with t. Differentiate the equation along a solution: y″=1−y′=1−t+y. At the initial point y″=1, so the curve initially bends upward. This calculation refines the visual prediction without requiring the full solution.

Now reverse the task. The following different supplied field is already drawn; identify its equation from competing candidates. Do not copy t−y from the preceding worked example. First read the inclination at (1,0), then at (0,1), and identify a line of horizontal segments. Record your prediction and two reasons before revealing the explanation. The expandable numerical table supplies the same information as the diagram.

Read the field before selecting an equation

These exact synthetic fields are supplied first. Each segment has the same geometric length; its inclination represents dy/dt, not speed or a step of a numerical solution. Both axes use 85 SVG coordinate units per mathematical unit.

Supplied case Field A: guided identificationField B: fresh transfer case

Equivalent numerical slopes for every plotted point

Each entry is dy/dt, not the solution value y.

State y / time t	-2	-1	0	1	2
2	0	1	2	3	4
1	-1	0	1	2	3
0	-2	-1	0	1	2
-1	-3	-2	-1	0	1
-2	-4	-3	-2	-1	0

Before revealing: identify two diagnostic points and one zero-slope pattern. Explain which candidate each observation rules out. A pattern visible at finitely many points does not uniquely identify an arbitrary unknown function.

Predicted equation y′ = t − yy′ = y − ty′ = t + yy′ = y(1 − y)

Compare prediction with fieldReset

Changing either selection hides the preceding explanation. This temporary activity saves no answer or mastery credit; record your diagnostic points and justification in your lesson notes.

Field A matches y′=t+y among the listed candidates. At (1,0) the slope is 1, excluding y−t and y(1−y). At (0,1) it is also 1, excluding t−y. This is a proof by diagnostic values within the stated candidate set, not a claim that 25 observations determine every possible function of two variables.

For Field A, consider the initial point (0,1). The initial tangent has slope 1. Differentiating the equation along a solution gives y″=1+y′, so the initial second derivative is 2: the solution is increasing and initially concave up. The horizontal-segment diagonal y=−t is not a solution because its own derivative is −1. These are distinct questions: which equation matches the field, what happens along one initial solution, and which curves are nullclines.

Understand

For an autonomous equation y′=f(y), the rate depends on the state but not explicitly on time. A zero of f then produces a constant solution, called an equilibrium. Consider y′=y(3−y). The equilibria are zero and three. The sign chart contains more information than the locations alone.

State interval	Sign of y	Sign of 3−y	Direction
Below 0	Negative	Positive	Decreasing
Between 0 and 3	Positive	Positive	Increasing
Above 3	Positive	Negative	Decreasing

Nearby states move away from zero and toward three, so zero is unstable and three is asymptotically stable. These are local behavior statements as time increases. A physically meaningful population might restrict y to nonnegative values, but the mathematical equation also has negative-state trajectories. State which domain you are discussing.

A field provides qualitative evidence. A finite grid does not prove that solutions cannot cross, that a solution exists forever, or that a numerical curve has the correct error. Such conclusions need assumptions and arguments. In particular, noncrossing follows from appropriate uniqueness conditions, not from the resolution of a drawing.

Apply

Make a phase line for y′=2−y. Mark the equilibrium, test one value on each side, and draw arrows. Compare this autonomous nullcline with the moving nullcline y=t from the earlier example. Explain why one is a constant solution and the other is not.

For the original field, identify two points with the same slope and two points with opposite slopes. Then decide where a solution is increasing, decreasing, or instantaneously horizontal. Treat “horizontal now” and “constant for all time” as separate claims in your written response.

For unfamiliar practice, select Field B in the explorer and close the explanation. Compare points (0,0), (1,0) and (0,1), then inspect whether equal-state slopes remain unchanged when time changes. Choose among its four new candidates, give a diagnostic rejection of each incorrect equation, and identify every visible zero-slope line. Finally predict whether the solution beginning at (0,0) can leave its horizontal tangent and which way it bends.

Check the Field B argument after writing your prediction

The matching equation is y′=t(1−y). Its zero-slope sets are t=0 and y=1. At a fixed state other than 1, reversing the sign of time reverses the slope, so the equation is not autonomous. The constant function y=1 is a solution, but the vertical time slice t=0 is not a graph y(t) defined over a time interval. Along a solution, y″=1−y−ty′; at (0,0), y″=1, so the solution initially bends upward despite its horizontal tangent. An exact check is y(t)=1−exp(−t²/2), whose derivative is t exp(−t²/2). This verifies the equation and initial condition without relying on a plotted numerical trajectory.

The solution is not determined by a horizontal mark alone. For these smooth right-hand sides, local existence and uniqueness can be justified by the theorem in the next lesson. A finite picture by itself supplies neither that theorem nor a guarantee for an arbitrary differential equation.

Continue

The next lesson asks when an initial point selects exactly one trajectory. Keep your slope-field drawing as intuition, but use the existence and uniqueness hypotheses to justify conclusions that the picture alone cannot establish.