Direction Fields and Solution Curves

The idea

A direction field places a short segment of slope $f(t,y)$ at each sample point. A solution curve stays tangent to those segments. The field can reveal growth, decay and equilibria without an explicit solution formula. It shows local directions, not equal travel speeds along different segments.

Visual guide

VISUAL GUIDEThe curve stays tangent to the direction field

Solution y = 2e⁻ᵗ

For y′ = −y, slopes repeat along each horizontal level. Above zero they point downward; below zero they point upward. The blue solution from y(0) = 2 follows those directions and approaches the equilibrium without crossing it. Arrow lengths are normalized to emphasize slopes.

Method and assumptions

Evaluate $f$ on a grid, draw segments with those slopes, and trace from the initial point. Isoclines satisfy $f(t,y)=c$ and group equal slopes. For an autonomous equation $y'=f(y)$ , slopes repeat along horizontal lines. Zeros of $f$ give equilibrium levels.

Worked example: exponential decay

For $y'=-y$ , positive solutions slope downward and negative solutions upward. The line $y=0$ is an equilibrium. The solution through $(0,2)$ is $2e^{-t}$ and approaches zero without crossing it.

Worked example: a moving zero-slope line

For $y'=t-y$ , the line $y=t$ has horizontal direction segments. It is not a solution: the function $y=t$ has derivative $1$ , not zero. Solving gives $y=t-1+Ce^{-t}$ , whose slope changes according to its position relative to the isocline.

Interpreting the result

Under uniqueness, distinct solution curves cannot cross at the same $(t,y)$ . A coarse slope field can hide rapid changes; numerical traces need a step-size check as well as a visually plausible curve.

Explore

Exact decay versus Euler steps

Try this. Keep the decay rate at 1 and compare 4, 8 and 16 steps. Then use rate 3 with one step: the numerical result fails to follow exact decay. Refine the steps to recover the shape.

Decay rate k1Initial value y(0)1Steps to t = 24

y′ = −1y, y(0) = 1. At t = 2: exact 0.135, Euler 0.063, signed error -0.073. Step h = 0.5; update factor = 0.5. Euler is in its decay stability range. Green: exact; orange: Euler; gray: slope field. Vertical scale adjusts to include the approximation.

This explorer compares Euler steps with the exact solution of $y'=-ky$ on $0\le t\le2$ . Change the decay rate, starting value or step count and compare error with the stability factor.

Practice

Where are the slopes zero for $y'=1-y^2$ ?
What is the slope at $(2,3)$ for $y'=t-y$ ?
Can a positive solution of $y'=-y$ cross zero?

Show worked solutions

At the equilibrium levels $y=1$ and $y=-1$ .
$2-3=-1$ .
No. Its exact form is positive, and uniqueness prevents crossing the zero solution.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.