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

Syllabus dot point

“Construct and interpret slope (direction) fields of a first-order differential equation, sketch solution curves through given points, and relate the qualitative behaviour of solutions to the field”

QCESpecialist MathematicsUnit 4: Further calculus, and…8 min read

Quick answer

A slope field for dydx=F(x,y)\frac{dy}{dx} = F(x, y) draws a short segment of gradient F(x,y)F(x, y) at each grid point, and solution curves stay tangent to these segments everywhere. Horizontal segments occur where F=0F = 0, marking equilibria; if FF depends only on yy, each root of F(y)=0F(y) = 0 gives a constant solution. Sketch a solution by starting at the initial point and following the field direction. Distinct solution curves never cross.

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  1. What this dot point is asking
  2. The answer
  3. Exam-style questions

What this dot point is asking

QCAA wants you to understand a first-order differential equation dydx=F(x,y)\frac{dy}{dx} = F(x, y) geometrically, before solving it. A slope field draws a short line segment at each grid point with gradient given by FF, and the solution curves follow these segments. The assessable skill is reading and constructing the field and sketching solution curves that stay tangent to it.

The answer

What a slope field shows

A first-order differential equation dydx=F(x,y)\frac{dy}{dx} = F(x, y) assigns a slope to every point (x,y)(x, y) in the plane. The slope field (direction field) is a grid of short segments, each drawn with the gradient F(x,y)F(x, y) at its location. A solution curve is any curve that is tangent to these segments everywhere it passes.

Constructing the field

At each grid point, evaluate F(x,y)F(x, y) to get the slope, then draw a short segment with that gradient. For example, if dydx=x\frac{dy}{dx} = x, the slope at (2,1)(2, 1) is 22, so the segment there rises steeply; along the yy-axis (x=0x = 0) all segments are horizontal.

Reading qualitative behaviour

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

Equilibrium solutions

If dydx=F(y)\frac{dy}{dx} = F(y) depends only on yy, then any value y=cy = c with F(c)=0F(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.

Sketching a solution through a point

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.

Solutions do not cross

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

Stable and unstable equilibria

When dydx\dfrac{dy}{dx} depends only on yy, the equilibria are the roots of F(y)=0F(y) = 0, and their stability is read from the sign of FF on either side. If F>0F > 0 just below an equilibrium and F<0F < 0 just above it, solutions are drawn toward the equilibrium from both sides, making it stable (an attractor). If the signs are reversed, solutions move away and the equilibrium is unstable (a repeller). This sign analysis matches the slope-field picture: stable equilibria have segments pointing inward, unstable ones have segments pointing outward, and it connects directly to the logistic model, whose carrying capacity is a stable equilibrium.

Linking the field to the explicit solution

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)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.

Isoclines

An isocline is the set of points where the slope takes a fixed value mm, that is the curve F(x,y)=mF(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)=0F(x, y) = 0 for horizontal segments, then F(x,y)=1F(x, y) = 1, F(x,y)=−1F(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.

Worked example

For dydx=x−y\dfrac{dy}{dx} = x - y, describe the slope field and sketch the solution through (0,0)(0, 0).

To understand the field, we find where segments are horizontal, then identify key solution curves, and finally trace the curve through the given initial point.

Step 1: Evaluate the slope at sample points

At each grid point substitute into F(x,y)=x−yF(x,y) = x - y:

At (0,0)(0, 0): dydx=0−0=0\frac{dy}{dx} = 0 - 0 = 0 (horizontal). At (1,0)(1, 0): 1−0=11 - 0 = 1. At (0,1)(0, 1): 0−1=−10 - 1 = -1. At (2,1)(2, 1): 2−1=12 - 1 = 1.

This pattern already shows that the slope depends on the gap between xx and yy.

Step 2: Find the isocline of zero slope

Setting x−y=0x - y = 0 gives the line y=xy = x. Along this line every segment is horizontal. Above the line (y>xy > x) slopes are negative; below it (y<xy < x) slopes are positive. Curves on both sides are pushed toward the line y=xy = x.

Step 3: Identify a special solution

Try the straight line y=x−1y = x - 1. Then dydx=1\dfrac{dy}{dx} = 1 and x−y=x−(x−1)=1x - y = x - (x - 1) = 1, so the equation is satisfied exactly. This straight line is one particular solution, and the field segments align parallel to it everywhere.

Step 4: Sketch the solution through (0,0)(0, 0)

Starting at the origin, the slope is 00, so the curve begins flat. Below the line y=xy = x the slopes are positive, so the curve bends upward. As xx increases the curve stays below y=xy = x and approaches the straight-line solution y=x−1y = x - 1 from above.

Step 5: Verify the start of the curve

At (0,0)(0, 0) the computed slope is F(0,0)=0−0=0F(0,0) = 0 - 0 = 0, confirming the sketch begins with a horizontal tangent.

Final answer: The slope field has horizontal segments along y=xy = x, positive slopes below and negative slopes above. The solution through (0,0)(0, 0) starts flat, rises, and approaches the line y=x−1y = x - 1.

Common trap

A common error is drawing solution curves that cross one another; for a first-order equation the slope at each point is unique, so distinct solutions cannot intersect. Another slip is sketching a curve that ignores the field, for instance cutting across segments rather than staying tangent to them. Always make the curve follow the local segment direction, and identify lines of zero slope (where F=0F = 0) first, since they anchor the picture.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Paper 2 (complex familiar). A slope field is shown for dydx=−0.5(y−4)x\dfrac{dy}{dx} = -\dfrac{0.5(y - 4)}{x}, x≠0x \neq 0. Point AA is at (−4,−2)(-4, -2). (a) Determine the slope at AA. (b) Use the field to sketch the solution curve through the point where x=−6x = -6, y=3.5y = 3.5.
Show worked answer →

(a) Substitute A=(−4,−2)A = (-4, -2): dydx=−0.5(−2−4)−4=−0.5(−6)−4=−34.\dfrac{dy}{dx} = -\dfrac{0.5(-2 - 4)}{-4} = -\dfrac{0.5(-6)}{-4} = -\dfrac{3}{4}.

(b) From (−6,3.5)(-6, 3.5), the curve sits where y<4y < 4 and x<0x < 0, so the slope is negative and the curve falls as it moves right. As x→0−x \to 0^- the 1x\dfrac{1}{x} factor steepens the slope, so the curve is asymptotic to the yy-axis. The sketch must stay tangent to the segments and not cross x=0x = 0 (excluded).

Markers reward the substituted slope value and a solution curve that follows the field, falls, and is asymptotic near x=0x = 0.

Practice3 marks
Paper 1 (technique). A differential equation is dydx=y(2−y)\dfrac{dy}{dx} = y(2 - y). (a) Determine the equilibrium solutions. (b) State, with reason, whether each is stable or unstable.
Show worked answer →

(a) Equilibria where dydx=0\dfrac{dy}{dx} = 0: y(2−y)=0y(2 - y) = 0, so y=0y = 0 and y=2y = 2.

(b) For 0<y<20 < y < 2, dydx>0\dfrac{dy}{dx} > 0, so solutions rise toward y=2y = 2; for y>2y > 2, dydx<0\dfrac{dy}{dx} < 0, so solutions fall toward y=2y = 2. Hence y=2y = 2 is stable. Near y=0y = 0, solutions with y>0y > 0 move away, so y=0y = 0 is unstable.

Markers reward both equilibria and a sign-based justification of stability.

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