🎓 University Year 1 · Lesson 14 of 15
The Slope-Field Tracer: Seeing a Differential Equation
A first-order differential equation hands you a slope at every point of the plane; drawing those slopes turns an equation you may not be able to solve into a picture you can follow.
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The big idea: A first-order differential equation dy/dx = f(x, y) is really a rule that assigns a direction to every point in the plane. Draw a short segment with that slope at a grid of points and you get a slope field — a map of tiny arrows. A solution is any curve that stays tangent to the field as it threads through, and picking a starting point selects one particular solution. This lets you understand the qualitative behaviour of an equation — where solutions rise, fall, or settle — even when no tidy formula exists.
An equation about slopes
A first-order differential equation in the form dy/dx = f(x, y) does not tell you the value of y directly. It tells you the slope a solution must have if it passes through the point (x, y). Plug in a point and you get a number: the steepness a solution curve has right there.
So the equation quietly assigns a little slope to every point in the plane. Making that assignment visible is the whole idea of a slope field.
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A slope field is a map of directions
At a grid of points, draw a short line segment whose slope is f(x, y). The resulting forest of tiny segments is the slope field (or direction field). It is a visual portrait of the differential equation: no solving required, just evaluate the slope everywhere. A solution curve is any curve that flows along the field, tangent to the segment at every point it crosses.
\[ \frac{dy}{dx} = f(x, y) \]
The differential equation as a slope rule: at each point (x, y) it prescribes the tangent slope of any solution through that point.
🎮 Slope-Field Tracer LIVE
A differential equation assigns a slope to every point; trace a solution through the field.
Tracing a solution
To find a solution by eye, drop a pencil at any starting point and move so that you are always heading in the direction the nearby segments point. The curve you trace bends to stay tangent to the field. Start somewhere else and you trace a different curve. The whole family of these curves is the general solution; a single initial condition y(x0) = y0 pins down exactly one of them.
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Isoclines: lines of equal slope
A quick way to sketch a field by hand is to find the isoclines — the curves where f(x, y) equals a fixed constant. Along one isocline every segment has the same slope, so you can rule them in all at once. For dy/dx = x, the isoclines are vertical lines; for dy/dx = y, they are horizontal lines.
📝 Worked example: Describe the slope field of dy/dx = x and find the solution through (0, 1).
- The slope depends only on x: it is 0 along the y-axis, positive to the right, negative to the left — segments fan from downhill on the left to uphill on the right.
- Solve by integrating: \( y = \int x\,dx = \tfrac{1}{2}x^2 + C \). The solutions are upward parabolas.
- Apply the initial condition: y(0) = 1 gives \( \tfrac{1}{2}(0)^2 + C = 1 \), so C = 1.
✓ The particular solution is <strong><em>y</em> = ½<em>x</em>² + 1</strong>, the parabola through (0, 1) that runs tangent to the field.
📝 Worked example: For dy/dx = y, where is the slope zero, and what solution passes through (0, 2)?
- The slope equals y, so it is zero exactly on the line y = 0 (the x-axis) — that horizontal line is itself a solution.
- Above the axis slopes are positive and grow with height; below it they are negative — solutions race away from y = 0.
- Solve: this is exponential growth, \( y = C e^{x} \). Apply y(0) = 2: \( C e^{0} = C = 2 \).
✓ The solution through (0, 2) is <strong><em>y</em> = 2<em>e</em><sup><em>x</em></sup></strong>; the line <em>y</em> = 0 is an <em>unstable</em> equilibrium the other solutions flee.
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Equilibria are flat highways
Where the slope is zero along a whole horizontal line y = c, that constant is an equilibrium solution — a solution that never changes. If nearby curves flow toward it the equilibrium is stable (an attractor); if they flow away it is unstable. You can read stability straight off the field without solving anything.
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Solutions can't cross
When f(x, y) is well behaved, exactly one solution passes through each point, so distinct solution curves never cross — if they did, that crossing point would need two different slopes at once. Keep this in mind when tracing: your curve should slot neatly between its neighbours, not intersect them.
Check your understanding
1. What does dy/dx = f(x, y) assign to each point of the plane?
The equation gives the tangent slope of a solution at each point (x, y) — that is exactly what the slope field draws.
2. A solution curve of the equation must…
A solution flows along the field, matching the prescribed slope at every point, so it is tangent to the segments.
3. For dy/dx = y, along which line is the slope zero?
The slope equals y, so it is zero where y = 0. That horizontal line is itself an equilibrium solution.
4. What does an initial condition like y(0) = 1 do?
The general solution is a family of curves; an initial condition picks the single curve passing through the given point.
5. Why can two distinct solution curves never cross (for a well-behaved f)?
Uniqueness means each point has one prescribed slope, so only one solution can pass through it; crossing would demand two slopes at that point.
✅ Key takeaways
- A first-order equation dy/dx = f(x, y) assigns a slope to every point of the plane.
- Drawing those slopes as short segments produces the slope (direction) field — a picture of the equation.
- A solution is a curve tangent to the field everywhere; an initial condition selects one particular solution from the family.
- Isoclines (curves of constant slope) speed up hand sketches, and horizontal lines of zero slope are equilibrium solutions.
- Equilibria that attract nearby curves are stable and those that repel them are unstable, and distinct solutions never cross when f is well behaved.