A slope field can be read at three levels of confidence. Some things follow from the arithmetic of the slope: a horizontal solution, the direction of travel on each side of it, and the fact that two solutions cannot cross. Some things the picture only suggests: where a curve is heading in the long run, and whether it leaves the window because it grows large or because it ends. Reading well means keeping the two apart.
Core ideas
Three kinds of reading, from the surest to the least sure.
If f(t, c) = 0 for all t, the constant function y = c has derivative zero and the equation says its slope should be zero, so it is a solution. For y′ = y·(1 − y) the right side vanishes at y = 0 and at y = 1, and both lines are solutions. This is settled by arithmetic, and the lab lists them.
At a point the field has one slope, so two solutions through the same point with different slopes are impossible: that much the arithmetic settles. A curve that reached a horizontal solution would arrive flat, with the same slope 0, and ruling that out takes one more fact: through a given point there is exactly one solution. This course states it, uses it and does not prove it; it holds for every polynomial right side, so a curve that starts between two horizontal solutions stays between them.
Where a curve is heading as t grows is something you see, and the sign of the slope supports it, but the field does not prove a limit. A curve between y = 0 and y = 1 is rising and cannot reach 1; that it approaches 1 is what the picture suggests and Equilibria, Stability and Phase Lines establishes.
The material
The horizontal solutions, the sign of the slope between them, the no-crossing rule, and two fields where reading further would be a mistake.
Start with y′ = y·(1 − y). The right side is a product, and a product is zero when either factor is. So the slope is 0 exactly where y = 0 or y = 1, at every t: both lines are horizontal solutions, and the lab lists them in the tile named Horizontal solutions.
Between the lines and outside them the sign of the slope is fixed. For y between 0 and 1 both factors y and 1 − y are positive, so the slope is positive and the segments rise. For y above 1 the factor 1 − y is negative and the slope is negative. For y below 0 the factor y is negative and 1 − y is positive, so the slope is negative again.
A solution that starts at y = 1/2 rises, because the slope there is positive. Could it reach y = 1? Not by crossing at an angle: the field gives one slope at each point, and two curves through one point with two slopes cannot both be solutions. But a curve that reached y = 1 would arrive flat, since the slope there is 0 by the equation, and the one-slope argument has nothing to say against that. What rules it out is the uniqueness claim: through a given point there is exactly one solution, and the horizontal line is already it. For a polynomial right side the claim holds, and with it a solution that starts below 1 stays below it for ever, and the same argument keeps it above 0.
Two different solutions of y′ = f(t, y) never cross at an angle, because they would need two slopes at one point. For the equations of this course they never touch either: through each point there is exactly one solution, which is the uniqueness statement of “Blow-Up and the Interval of Existence”.
The first sentence is settled by the arithmetic of the slope, and the lab demonstrates it. The second is a claim this course states, uses and does not prove; it is what makes a horizontal solution a wall and not merely a line of flat segments. For the rest of this course, take both as the working rule.
In the second preset the equation is y′ = t − y and the drawn solution starts at (0, 2). The line y = t − 1 is itself a solution: its derivative is 1, and t − (t − 1) = 1, so the residual is zero, by the test of “What a Differential Equation Is”. The drawn curve bends towards that line. That is what the picture suggests; First-Order Linear Equations is where the claim is established.
Take y′ = y² with the solution through (0, 1), and look at the third preset. The drawn curve climbs, steeper and steeper, and leaves the window before t = 1.
What the field supports: the slope y² is never negative, so the curve never falls, and it grows steeper as y grows. What it does not settle is whether the curve is merely large at the window's edge or has stopped existing there. “Blow-Up and the Interval of Existence” answers that.
Hold that distinction in the examples that follow. The arithmetic of the slope settles the horizontal solutions, the sign on each side and the no-crossing rule. The look of the picture suggests the far future, and the drawn curve is a floating-point drawing that says so in the legend.
Interactive
The slope at every point, before any solving
Method
Settled things first, suggested things last, and the line between them written down.
Solve f(t, c) = 0 for all t. When f has no t, these are the roots of f in y. Mark each line y = c.
Between and beyond the horizontal solutions, decide whether f is positive or negative, by testing one point or by signs of factors. That is the direction of travel in the region.
A solution that starts in a region between two horizontal solutions stays in it: one slope per point forbids a crossing, and the uniqueness claim forbids a touch. State what that bounds.
Draw where the curves seem to head, and write beside it that this is read from the picture and not proved by it. Do not state a limit as a fact.
Read the field of y′ = y·(1 − y) and say what happens to a solution that starts at y = 1/2.
What the arithmetic settled: the two horizontal solutions and the direction of travel in each region. What the uniqueness claim adds: the curve from 1/2 stays strictly between 0 and 1, because meeting either line would put two solutions through one point. What the picture only suggests is that the curve climbs towards 1. Rising and bounded above does not by itself fix where it ends, and the picture shows it levelling off without proving it.
For a rehearsal, take a start at y = 2. The slope there is 2·(1 − 2) = −2, so the curve falls, and by the same argument it stays above 1.
You should be able to find the horizontal solutions of an equation with no t in it, read the sign of the slope in each region, apply the no-crossing rule to bound a solution, and state which of your conclusions come from the arithmetic and which only from the look of a drawn curve.
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