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.

At $(0, 0)$: $\frac{dy}{dx} = 0 - 0 = 0$ (horizontal). At $(1, 0)$: $1 - 0 = 1$. At $(0, 1)$: $0 - 1 = -1$.

Setting $x - y = 0$ gives $y = x$: along this line every segment is horizontal. Above the line ($y > x$) slopes are negative; below it ($y < x$) slopes are positive, so curves are pushed toward the line $y = x$ from both sides.

Try $y = x - 1$: then $\frac{dy}{dx} = 1$ and $x - y = x - (x - 1) = 1$, so $y = x - 1$ satisfies the equation exactly. This straight line is one solution, and the field segments align with it.

Starting at the origin with slope $0$, the curve initially is flat, then bends upward as it enters the region $y < x$ where slopes are positive, approaching the line $y = x - 1$ from above as $x$ increases.

At $(0, 0)$ the computed slope is $0$, matching a horizontal start, confirming the sketch begins correctly.