Curves that obey every segment AB & BC

Given a field and an initial point, the curve is almost determined before you draw anything.

1. Start at the given point. Not at the origin, not at the left edge of the picture. The initial condition is what makes the solution particular.
2. Follow the segments in both directions. A solution exists to the left of the initial point as well as to the right, and AP prompts often want both.
3. Stay tangent. The curve's slope where it passes a lattice point is that segment's slope. If your curve cuts across a segment, one of the two is wrong, and it is not the segment.

The result is smooth, since $\frac{dy}{dx}$ exists everywhere the field is drawn. Corners do not appear in solution curves, and neither do vertical jumps.

Concavity is readable too. If the segments get steeper as the curve moves right, the slope is increasing and the curve is concave up. That is worth noticing because it distinguishes two sketches that reach the same place by different routes.

If $\frac{dy}{dx} = f(x, y)$ is zero along an entire horizontal line $y = c$, then the constant function $y = c$ is a solution: its derivative is zero and so is the right side, everywhere. That is an equilibrium solution.

For $\frac{dy}{dx} = y(2 - y)$ the right side vanishes when $y = 0$ and when $y = 2$, so there are two of them. The whole field organises itself around those two lines. Between them the product is positive, so solutions rise; above $y = 2$ the second factor is negative, so solutions fall; below $y = 0$ the first factor is negative and the second positive, so solutions fall away.

Equilibria come in two kinds and the field shows which is which. An equilibrium that nearby solutions approach from both sides is attracting; here $y = 2$ is. One that nearby solutions move away from is repelling; here $y = 0$ is. That reading is the answer to almost every long-run question in this topic.

Note what an equilibrium is not: it is not the maximum of any solution, and it is not a value any non-constant solution ever attains. A solution starting at $y = 0.4$ climbs toward $2$ and stays below it for all time.

Two distinct solution curves never touch. If they did, the point where they met would have two different solutions through it, and the segment there gives only one slope.

The immediate consequence is that an equilibrium line is a barrier. A solution that starts below $y = 2$ stays below $y = 2$ forever, in both directions, because crossing would mean touching the constant solution $y = 2$. Drawing a curve that sails through an equilibrium is the most common sketching error in this topic, and it is not a matter of accuracy: the curve it produces is not a solution of anything.

Three more things a sketched curve must not do:

1. Cross another sketched solution. Same argument, no equilibrium required.
2. Contradict a segment. A curve falling through a region where every segment rises is reporting the opposite of what the field says.
3. Miss the initial point. A curve of the right shape through the wrong point is a different solution.

Together these say something simple: through each point of the field there is exactly one solution curve, and the field draws its tangent.

Equilibria have an algebraic life too, and it is where they get lost. Solving $\frac{dy}{dx} = y^{2}$ by separating means dividing both sides by $y^{2}$, which is legal only when $y \neq 0$. Carrying on gives $-\frac{1}{y} = x + C$, so $y = -\frac{1}{x + C}$.

Now look at what is missing. The constant function $y = 0$ satisfies the original equation, since both sides are zero. It is not a member of that family for any value of $C$. The division discarded it, silently, at the first step.

The habit that fixes this: before dividing by an expression, write down what happens when that expression is zero. If the resulting constant function solves the equation, it is a solution and it belongs in the answer.

Sometimes the family does recover it. Separating $\frac{dy}{dt} = ky$ gives $y = Ce^{kt}$, and the lost solution $y = 0$ reappears as $C = 0$. Sometimes it does not, as above. Either way the check is the same three seconds, and on a field the lost solution is the horizontal line you were about to draw a curve straight through.