Slope Fields
A first-order differential equation \(y' = f(x, y)\) defines a slope field: a grid of short segments whose slopes visualize the whole family of solution curves, so you can sketch a particular solution through any given point without solving anything symbolically.
By the end you'll be able to compute the slope \(f(x, y)\) a field prescribes at any point, and trace the solution curve through a given initial point by following the segments — no formula required.
Predict: drop the point above the line \(y = x\) — does the solution curve rise or fall from there? Drag the orange point (or use the sliders) to check.
This is the slope field of \(y' = y - x\): at every grid point the short segment has slope \(y - x\). The blue curve is the particular solution through your initial point — it stays tangent to every segment it passes. Along the dashed line \(y = x\) the slope is 0: that's an isocline of horizontal segments. Hover the grid for the exact slope at any point.
Initial point (0.00, 1.00) · slope there y′ = y₀ − x₀ = 1.00 — above the line y = x, so the solution rises from the start.
A slope field turns the equation \(y' = f(x, y)\) into a picture: tiny tangent marks everywhere, which every solution curve must run parallel to wherever it goes.
Reading a segment: substitute the point
The equation \(y' = f(x, y)\) doesn't name one function — it describes a whole family: any solution curve must, at every point \((x, y)\) it passes through, have slope \(f(x, y)\). So to build the field you pick a grid of points, substitute each one into \(f\), and draw a short segment with that slope. That's the entire construction — one substitution per grid point.
Before drawing segment by segment, find the structure: where is \(f(x, y) = 0\)? Along those curves the segments are horizontal, and they split the plane into regions. In the demo above, \(y - x = 0\) exactly on the dashed line \(y = x\) — above it \(y' > 0\) and segments tilt up, below it \(y' < 0\) and they tilt down. Sample one test point per region for the sign and rough size of \(f\), then fill in the grid so the segments visually flow in a consistent pattern.
Think of the segments as a trail of signposts. To sketch the particular solution through an initial point, start there and draw a curve that is tangent to the segment at every point it passes — the field steers you. You never need the formula: the sketch already shows where solutions grow, decay, level off toward an equilibrium, or blow up — for every initial condition at once. As Paul's Online Math Notes puts it, the arrows in a direction field are tangents to the actual solutions.
Worked example — build and read a field
Build the slope field of \(y' = 3x + 2y - 4\) at the point \((1, 2)\), then describe the solution through \((0, 1)\).
Segment: substitute the point into the right-hand side: \(y' = 3(1) + 2(2) - 4 = 3\). So through \((1, 2)\) you draw a short segment of slope 3. Repeat over the grid.
Trace: for the initial-value problem \(y(0) = 1\), start at \((0, 1)\) and follow the segments. The traced curve matches the exact solution \(y = -\tfrac{3}{2}x + \tfrac{5}{4} - \tfrac{1}{4}e^{2x}\) — but you sketched its shape, concavity, and long-run behavior without ever finding that formula.
Same field, a different grid point: what slope does \(y' = 3x + 2y - 4\) prescribe at \((-1, 4)\)? Substitute: \(y' = 3(-1) + 2(4) - 4 = \) ____ , so the segment there tilts ____ .
Reveal the answer
\(y' = -3 + 8 - 4 = \) 1 — positive, so the segment tilts upward with slope 1, and any solution passing through \((-1, 4)\) is rising there.
More info — the field as a bridge to solving methods
A slope field is the qualitative counterpart to two things you'll meet next. Symbolic techniques (like separation of variables, the next lesson) find the exact family the field is picturing; numerical methods like Euler's method literally walk along the field step by step — start at the initial point, step in the direction of the local segment, recompute, repeat. If you can read a slope field, you already understand what both are doing. For the general/particular-solution vocabulary behind this, see the OpenStax review under Dive deeper below.
Check your understanding
For the slope field of \(y' = y - x\), what slope does the short segment drawn at the point \((2, 4)\) have?
You want to sketch the particular solution of \(y' = y - x\) through \((0, 1)\) using its slope field. What must you do first?
For the slope field of \(y' = 3x + 2y - 4\), what is the slope of the segment at the point \((2, 0)\)?
In the slope field of \(y' = y - x\), you start a solution curve at \((0, -2)\). What does the curve do just to the right of that point, and why?
Recap
- \(y' = f(x, y)\) describes a family of curves: every solution must have slope \(f(x, y)\) at each point it passes through.
- To build the field, substitute each grid point into \(f(x, y)\) and draw a short segment with that slope — e.g. for \(y' = y - x\), the segment at \((2, 2)\) has slope 0.
- Curves where \(f(x, y) = 0\) are isoclines of horizontal segments; they split the plane into rising (\(y' > 0\)) and falling (\(y' < 0\)) regions.
- To sketch the particular solution through an initial point, trace a curve tangent to the segments — no closed-form solution needed.
Dive deeper
- Direction Fields — Paul's Online Math Notes Construct a direction field and sketch a solution from it.
- OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations Review general vs. particular solutions and initial-value problems.
Sources
- Slope fields (direction fields) for first-order differential equations