Quick questions on Slope fields of first-order differential equations in QCE Specialist Mathematics Unit 4

The field reveals behaviour without an explicit formula. Where $F(x, y) = 0$, segments are horizontal, marking possible turning points or equilibria. Where $F$ is large, the field is steep. Regions where $F > 0$ have solutions increasing left to right; where $F < 0$, decreasing.

If $\frac{dy}{dx} = F(y)$ depends only on $y$, then any value $y = c$ with $F(c) = 0$ gives a constant (equilibrium) solution: a horizontal line that the field segments lie along. Nearby solution curves either approach (stable) or move away from (unstable) this equilibrium.

Given an initial point, start there and draw a smooth curve that follows the direction of the nearby segments, always staying tangent to the field. The curve threads through the segments like a path following arrows. Different initial points give different members of the solution family.

For a well-behaved equation, distinct solution curves never cross, because the slope at any point is uniquely determined by $F(x, y)$. Two curves crossing would require two different tangents at one point, which is impossible.

A slope field is the qualitative companion to the algebraic solution found by separation of variables. The field tells you the shape, the asymptotes and the long-term behaviour at a glance, while the explicit solution gives the exact formula. A good check on an algebraic solution is to confirm it matches the field: its gradient at a few points should agree with $F(x, y)$, and its limiting behaviour should match the equilibria. When an equation cannot be solved in closed form, the slope field is the only practical way to understand the solutions, which is why the syllabus pairs it with numerical methods.

An isocline is the set of points where the slope takes a fixed value $m$, that is the curve $F(x, y) = m$. Along an isocline every field segment has the same gradient, so plotting a few isoclines is an efficient way to construct a field by hand: draw the curve $F(x, y) = 0$ for horizontal segments, then $F(x, y) = 1$, $F(x, y) = -1$ and so on, and fill in parallel segments along each. Recognising the isocline of zero slope is especially useful because it locates every turning point of every solution curve.