Direction Fields and Graphic Solutions of First-Order ODEs

A first-order equation y' = f(x,y) assigns a slope to every point of the plane. Drawing a short segment of that slope at each of many points produces a direction field, a picture of the flow that every solution must follow. This lets you see the shape of solutions before, or instead of, finding a formula for them.

Quick Reference

Idea	Statement
Direction field	A grid of short line segments whose slope at $(x, y)$ is $f (x, y)$
Integral curve	A solution curve; it is tangent to the field at every one of its points
Isocline	The curve $f (x, y) = m$ , along which every field segment has the same slope $m$
Nullcline	The isocline for $m = 0$ , where solution curves have horizontal tangents
Equilibrium solution	A constant solution $y = k$ ; it occurs when $f (x, k) = 0$ for all $x$
Semistable equilibrium	An equilibrium that attracts on one side and repels on the other, as $y = 0$ does for y' = y^2
Autonomous equation	y' = f(y), with no explicit $x$ ; its field is identical along every vertical line
Key property	Through each point of the region where $f$ is continuous, one integral curve passes

The Geometric Meaning of y' = f(x,y)

Recall from calculus that y' at a point is the slope of the tangent line to the graph of $y$ there. So the equation

\frac{d y}{d x} = f (x, y)

carries a purely geometric message: if a solution curve passes through the point $(x, y)$ , its tangent line there has slope $f (x, y)$ .

The equation therefore prescribes a slope at every point of the region $D$ where $f$ is defined, whether or not a solution has been found. This is a remarkable amount of information to have for free.

Definition 1.

The direction field, also called the slope field, of the equation y' = f(x,y) is obtained by drawing, at each point of a grid in $D$ , a short line segment whose slope equals $f (x, y)$ . Any solution curve, also called an integral curve, must be tangent to these segments at every point it passes through.

A direction field consisting of short slope segments on a grid across the xy-plane, with one solution curve drawn in red rising through the origin, reaching a maximum, and then falling. — A direction field with one integral curve traced through it in red. The curve is tangent to the little segments at every point, rising while the segments tilt upward, flattening where they are horizontal, and falling where they tilt downward.

How to Sketch a Direction Field

Procedure. Choose a grid of points covering the region of interest. At each point $(x_{0}, y_{0})$ , evaluate $m = f (x_{0}, y_{0})$ and draw a short segment through that point with slope $m$ . Keep every segment the same short length, so the picture shows direction rather than magnitude. Then trace curves that follow the segments.

Doing this point by point is tedious by hand. The shortcut is to draw whole families of segments at once using isoclines.

Definition 2.

An isocline of y' = f(x,y) is a curve on which the slope is constant, that is, the level curve $f (x, y) = m$ for a fixed value $m$ . Every field segment along an isocline has the same slope $m$ .

The word means "equal inclination". Choosing several convenient values of $m$ , sketching the corresponding isoclines, and then drawing segments of slope $m$ along each one fills the plane quickly.

Example 1.

Sketch the direction field of y' = x using isoclines, and identify the integral curves.

Solution

The isocline $f (x, y) = m$ is $x = m$ , a vertical line. So:

Along $x = 0$ , every segment is horizontal (slope $0$ ).
Along $x = 1$ , every segment has slope $1$ .
Along $x = - 2$ , every segment has slope $- 2$ .

The field does not depend on $y$ at all, so it looks identical along any horizontal shift: each vertical line carries segments of a single slope, tilting more steeply upward as you move right and more steeply downward as you move left.

Integrating directly, $y = \frac{x^{2}}{2} + c$ , a family of parabolas. Each one is a vertical translate of the others, which is exactly what the field's independence from $y$ predicted.

The direction field of y' = x. Every segment on a given vertical line has the same slope, and the integral curves are the vertical translates $y = \frac{x^{2}}{2} + c$ .

Example 2.

Sketch the direction field of y' = y and describe the behavior of solutions.

Solution

The isocline $y = m$ is a horizontal line, so:

Along $y = 0$ the segments are horizontal.
Along $y = 2$ every segment has slope $2$ , tilting up steeply.
Along $y = - 2$ every segment has slope $- 2$ , tilting down steeply.

Reading the picture: above the $x$ -axis, slopes are positive and grow with $y$ , so curves rise faster and faster. Below the axis, slopes are negative and grow more negative, so curves fall faster and faster. On the axis itself the slope is zero everywhere, so the constant function $y = 0$ is a solution.

The general solution $y = c e^{x}$ confirms all of this: $c > 0$ gives runaway growth, $c < 0$ gives runaway decay, and $c = 0$ gives the equilibrium $y = 0$ .

The direction field of y' = y. Segments are horizontal on the line $y = 0$ and steepen as $| y |$ grows. The integral curves are $y = c e^{x}$ ; the equilibrium $y = 0$ separates the growing solutions from the decaying ones.

Example 3.

Use isoclines to sketch the direction field of y' = x^2 + y^2, and describe the solutions.

Solution

The isocline for slope $m$ is $x^{2} + y^{2} = m$ , a circle of radius $\sqrt{m}$ centered at the origin (and empty when $m < 0$ ). No isocline picture is easier to draw:

At the origin alone the slope is $0$ .
On the unit circle every segment has slope $1$ , a $45^{\circ}$ tilt.
On the circle of radius $2$ every segment has slope $4$ , about $76^{\circ}$ .

Three facts follow without any integration:

y' \geq 0 everywhere, with equality only at the single point $(0, 0)$ , so every solution is increasing.
The slopes grow without bound, and they do so fast enough that no solution survives for all $x$ . Once a solution passes $x = 1$ it satisfies y' \geq 1 + y^2, and comparison with $\tan$ forces it to reach $+ \infty$ at a finite value of $x$ .
The field is unchanged by the half-turn $(x, y) \mapsto (- x, - y)$ , since $f (- x, - y) = f (x, y)$ . So solutions come in pairs: if $y = ϕ (x)$ is a solution, so is $y = - ϕ (- x)$ . Running the same argument backward, each solution also comes up out of $- \infty$ at a finite $x$ . Every integral curve lives on a bounded interval and sweeps from $- \infty$ to $+ \infty$ across it.

It is interesting to know that this equation has no solution in elementary functions. It is a Riccati equation whose solutions are ratios of Bessel functions. The picture above cost nothing.

The direction field of y' = x^2 + y^2, with the isoclines $m = 1$ and $m = 4$ dashed. Every segment on a given circle carries the same slope. The three integral curves are nearly translates of one another, each rising from $- \infty$ to $+ \infty$ over a bounded interval of $x$ .

Equilibrium Solutions and Nullclines

Definition 3.

A constant function $y = k$ is an equilibrium solution (also called a stationary or constant solution) of y' = f(x,y) if $f (x, k) = 0$ for every $x$ in the interval considered.

Equilibrium solutions are the easiest solutions to find, since they require solving an algebraic equation rather than a differential one, and they organize the whole picture: no other integral curve can cross an equilibrium line where the uniqueness theorem applies, so the equilibria partition the plane into horizontal strips that trap all the other solutions.

Example 4.

Find the equilibrium solutions of the logistic equation $\frac{d P}{d t} = r P (1 - \frac{P}{K})$ , where $r$ and $K$ are positive constants, and describe the direction field.

Solution

Set the right-hand side to zero:

r P (1 - \frac{P}{K}) = 0 ⟹ P = 0 or P = K .

There are two equilibrium solutions, $P = 0$ and $P = K$ .

Now check the sign of P' in each strip:

For $0 < P < K$ , both $r P$ and $(1 - P / K)$ are positive, so P' > 0 and solutions rise toward $K$ .
For $P > K$ , the factor $(1 - P / K)$ is negative, so P' < 0 and solutions fall toward $K$ .
For $P < 0$ , which is not physically meaningful but is part of the field, P' < 0 and solutions fall away from $0$ .

So the field pushes every positive solution toward $P = K$ . We call $P = K$ a stable equilibrium and $P = 0$ an unstable one. All of this was read off the sign of $f$ , with no integration performed.

The direction field of the logistic equation (drawn with $r = 1$ , $K = 1$ ). Solutions starting between the equilibria climb toward $P = K$ , and solutions starting above it descend toward $P = K$ ; a solution starting below $P = 0$ runs away downward.

Not every equilibrium is cleanly stable or unstable. The next example shows the third possibility, and delivers a warning about how long solutions last.

Example 5.

Find the equilibrium solutions of y' = y^2, classify them, and describe the field.

Solution

Setting $y^{2} = 0$ gives the single equilibrium $y = 0$ . But the sign check behaves differently from the previous examples, because $y^{2}$ does not change sign:

For $y > 0$ we have y' > 0, so solutions rise away from $y = 0$ .
For $y < 0$ we have y' > 0 as well, so solutions rise toward $y = 0$ .

The equilibrium attracts from below and repels from above. Such an equilibrium is called semistable. In the picture the segments never tilt downward anywhere; they merely flatten out as they approach the $x$ -axis from either side.

Separating variables for $y \neq 0$ gives $- \frac{1}{y} = x + c$ , that is

y = \frac{1}{c - x},

together with the equilibrium $y \equiv 0$ . This confirms both halves of the reading. A solution starting at $y (0) = y_{0} < 0$ has $c = 1 / y_{0} < 0$ and increases to $0$ as $x \to \infty$ . A solution starting at $y (0) = y_{0} > 0$ has $c = 1 / y_{0} > 0$ and

y ⟶ + \infty as x \to {\frac{1}{y_{0}}}^{-} .

Finite escape time. Here 
  f
  (
  x
  ,
  y
  )
  =
  
    y
    2
  
 and 
  
    f
    y
  
  =
  2
  y
 are continuous on the entire plane, so a solution exists through every point. It does not follow that any of them exists for all 
  x
. The larger 
  
    y
    0
  
 is, the sooner the solution blows up: existence guaranteed by the theory is only local.

The direction field of y' = y^2. No segment ever tilts downward. Solutions below the axis climb toward $y = 0$ and solutions above it run off to $+ \infty$ at a finite $x$ , which is what makes $y = 0$ semistable.

Why Integral Curves Do Not Cross

The field already tells you what the theorem has to say. At every point of the region where $f$ is defined, the equation prescribes exactly one slope, so a solution arriving at that point has no choice about the direction in which it leaves: the direction field never branches. If that reading is right, two different solution curves can never pass through the same point. Making it right is the job of the following theorem, proved in the final chapter.

Theorem 1

Existence and uniqueness. If $f$ and $\partial f / \partial y$ are continuous on a rectangle $D$ containing $(x_{0}, y_{0})$ , then exactly one integral curve of y' = f(x,y) passes through $(x_{0}, y_{0})$ .

Its two halves are two statements about the picture:

Existence: the field is never blank. Some curve follows the segments through every point of $D$ , and continuity of $f$ alone is enough to guarantee it.
Uniqueness: the field never branches. Inside $D$ two distinct integral curves cannot cross, cannot touch, and cannot merge. This is the half that needs the condition on $\partial f / \partial y$ .

The picture gets you only halfway, and it is worth seeing exactly how far. Two curves crossing at an angle are ruled out by the field alone: at the crossing point the equation prescribes one slope, and they cannot both have it. But two curves that merely touch both carry the correct slope at the point of contact, so nothing in the picture forbids them. Ruling that out is what the theorem contributes.

Two solutions through one point $P$ . The heavy black segment is the slope the equation prescribes there. On the left the dashed curve is disqualified immediately, since it is not tangent to the segment. On the right both curves are tangent, and nothing in the picture forbids the split; it is the uniqueness theorem that does.

The consequence is one we have been using all along. An equilibrium solution is itself an integral curve, so no other curve may cross it. That is why a logistic population starting below $K$ stays below $K$ forever, and rises to a limit, without any need to solve the equation.

Where the hypotheses fail, curves really do cross. In y' = 3y^{2/3} the partial derivative $\partial f / \partial y = 2 y^{- 1 / 3}$ blows up on the line $y = 0$ , and through the origin pass both $y = 0$ and $y = x^{3}$ , plus infinitely many curves assembled from pieces of the two. The field there looks perfectly tame, which is a useful warning: a direction field shows what solutions do, not whether they are unique.

Exercises

Exercise 1.

Describe the direction field of y' = x + y by finding its isoclines. Where are the horizontal tangents?

Solution

The isocline for slope $m$ is $x + y = m$ , that is, the line $y = - x + m$ . So all the isoclines are parallel lines of slope $- 1$ , and along the line $y = - x + m$ every segment has slope $m$ .

Horizontal tangents occur where $m = 0$ , on the line $y = - x$ . Above that line y' > 0 and solutions rise; below it y' < 0 and solutions fall.

One curiosity: the isocline for $m = 1$ is $y = - x + 1$ , whose own slope is $- 1$ , not $1$ , so it is not a solution. But the isocline for $m = - 1$ is $y = - x - 1$ , whose slope $- 1$ matches, so $y = - x - 1$ is an integral curve. An isocline is a solution exactly when its own slope equals the slope it carries.

The direction field of y' = x + y. The isoclines are the parallel lines $y = - x + m$ ; the dashed one, $y = - x$ , carries the horizontal segments. The heavy line $y = - x - 1$ is itself an integral curve, and the curves $y = c e^{x} - x - 1$ peel away from it.

Exercise 2.

Find all equilibrium solutions of y' = \left(y - 1\right)\left(y + 2\right) and classify each as stable or unstable.

Solution

Setting the right-hand side to zero gives $y = 1$ and $y = - 2$ .

Check the sign of y' in each strip:

$y > 1$ : both factors positive, so y' > 0 and solutions move up, away from $y = 1$ .
$- 2 < y < 1$ : the first factor is negative and the second positive, so y' < 0 and solutions move down, away from $y = 1$ and toward $y = - 2$ .
$y < - 2$ : both factors negative, so y' > 0 and solutions move up, toward $y = - 2$ .

Therefore $y = - 2$ is stable (arrows point toward it from both sides) and $y = 1$ is unstable (arrows point away from both sides).

The direction field of y' = (y-1)(y+2). Because the equation is autonomous the picture repeats along every vertical line. Solutions leave $y = 1$ in both directions and approach $y = - 2$ from both sides.

Exercise 3.

Without solving, decide whether the solution of y' = y^2 - x passing through $(0, 0)$ is increasing or decreasing at that point, and find where its tangent is horizontal.

Solution

At $(0, 0)$ we have y' = 0^2 - 0 = 0, so the tangent is horizontal there.

Horizontal tangents occur on the nullcline $y^{2} = x$ , a parabola opening to the right. To the right of it, where $x > y^{2}$ , we get y' < 0 and solutions decrease; to the left, where $x < y^{2}$ , we get y' > 0 and solutions increase.

Since the point $(0, 0)$ sits on the nullcline itself, look just past it: moving right from the origin along a solution puts us in the region $x > y^{2}$ , so the curve turns downward. The origin is a local maximum of that solution.

The direction field of y' = y^2 - x. The dashed parabola $x = y^{2}$ is the nullcline; segments tilt upward to its left and downward to its right. The heavy curve through the origin reaches its maximum exactly on the nullcline and falls afterward.

Exercise 4.

An object falling through air experiences a drag force proportional to its speed, so its velocity satisfies

\frac{d v}{d t} = 9.8 - \frac{v}{5},

with $v$ in metres per second and $t$ in seconds. Read the terminal velocity off the direction field and describe how a body released from rest approaches it.

Solution

The equation is autonomous, so the field repeats along every vertical line and the whole story is in the sign of $9.8 - v / 5$ .

Setting it to zero gives the one equilibrium

9.8 - \frac{v}{5} = 0 ⟹ v = 49 m/s,

the terminal velocity. For $v < 49$ the right side is positive and the object accelerates; for $v > 49$ it is negative and the object slows down. Arrows point toward $v = 49$ from both sides, so the equilibrium is stable and every solution tends to $49$ m/s, whether the object is dropped, thrown down hard, or thrown upward.

Two more readings come free. At $v = 0$ the slope is exactly $9.8$ : at the instant of release the drag contributes nothing and the acceleration is $g$ . And since the field is steepest at $v = 0$ and flattens monotonically as $v$ climbs toward $49$ , the graph from rest is concave down throughout, with no inflection.

Solving confirms it: $v = 49 + C e^{- t / 5}$ , and from rest

v = 49 (1 - e^{- t / 5}),

which reaches $95 %$ of terminal velocity at $t = 5 \ln 20 \approx 15$ seconds. Note that $v = 49$ is approached but never attained.

The direction field of v' = 9.8 - v/5, with $t$ in seconds and $v$ in metres per second. The lowest curve is the release from rest; the others start at $21$ and $70$ m/s. All three flatten onto the terminal velocity $v = 49$ .

Exercise 5.

Sketch the direction field of y' = -\dfrac{x}{y} and identify the integral curves. What goes wrong on the $x$ -axis?

Solution

The isocline for slope $m \neq 0$ is $- x / y = m$ , that is the line $y = - x / m$ through the origin. Its own slope is $- 1 / m$ , and

m \cdot (- \frac{1}{m}) = - 1,

so every field segment is perpendicular to the ray it sits on. A family of curves crossing all rays from the origin at right angles can only be the family of circles centered there.

Separating variables confirms it: $y d y = - x d x$ gives $x^{2} + y^{2} = r^{2}$ .

Reading the special lines: on the $y$ -axis, $x = 0$ makes the slope $0$ , so the circles have horizontal tangents there. On the $x$ -axis, $f$ is not defined at all, and the segments turn vertical as $y \to 0$ .

That last point is the interesting one. A circle is not the graph of a function, so a single solution is only a semicircle,

y = \sqrt{r^{2} - x^{2}} or y = - \sqrt{r^{2} - x^{2}},

each defined on the open interval $(- r, r)$ and each running out of road exactly where it meets the $x$ -axis. The uniqueness theorem never promised otherwise: it requires $f$ to be continuous, and $f (x, y) = - x / y$ fails that on the whole line $y = 0$ .

The direction field of y' = -x/y. Each segment is perpendicular to the ray joining it to the origin, so the integral curves are circles. The field is undefined on the $x$ -axis, where the segments stand vertical and each solution semicircle ends.

Exercise 6.

Find all equilibrium solutions of y' = \sin y, classify them, and describe the behavior of the solution with $y (0) = 1$ as $x \to \pm \infty$ .

Solution

The equation is autonomous, and $\sin y = 0$ exactly when

y = n π, n = 0, \pm 1, \pm 2, \dots

so there are infinitely many equilibria, evenly spaced.

Check the sign of $\sin y$ on one strip and the pattern is fixed. On $(0, π)$ , $\sin y > 0$ , so solutions rise; on $(π, 2 π)$ , $\sin y < 0$ , so solutions fall. Both strips push toward $y = π$ and away from $y = 0$ and $y = 2 π$ . Since $\sin$ alternates sign from strip to strip:

$y = (2 k + 1) π$ is stable, for every integer $k$ .
$y = 2 k π$ is unstable.

Because integral curves cannot cross the equilibrium lines, a solution is trapped in the strip it starts in, and inside that strip $\sin y$ keeps one sign, so the solution is monotone. It therefore runs from the unstable edge of the strip to the stable one.

For $y (0) = 1$ , which lies in $(0, π)$ : the solution increases, and

y \to π as x \to + \infty, y \to 0 as x \to - \infty .

Every solution of this equation is bounded, in sharp contrast to y' = y or y' = y^2. Separating gives $y = 2 \arctan (A e^{x})$ on the strip $(0, π)$ ; here $A = \tan \frac{1}{2} \approx 0.546$ .

The direction field of y' = \sin y. The equilibria $y = n π$ alternate between unstable and stable, and every other solution is trapped in one strip, climbing or falling monotonically from the unstable edge to the stable one. No solution is unbounded, and no segment is steeper than $45^{\circ}$ .

Exercise 7.

The section observed that y' = 3y^{2/3} admits more than one solution passing through the origin. Sketch its direction field and exhibit infinitely many such solutions.

Solution

No $x$ appears on the right-hand side, so the slope the equation prescribes at a point depends only on the height $y$ , and the field repeats identically along every vertical line. It depends, in fact, only on $| y |$ : writing $y^{2 / 3} = {(\sqrt[3]{y})}^{2}$ , the squaring erases the sign of the cube root. The field is therefore symmetric about the $x$ -axis, and no segment ever tilts downward, since $3 y^{2 / 3} \geq 0$ everywhere, with equality precisely on $y = 0$ .

This makes $y \equiv 0$ an equilibrium solution. Away from the $x$ -axis, separation of variables gives

\int y^{- 2 / 3} d y = \int 3 d x ⟹ 3 y^{1 / 3} = 3 x + C ⟹ y = (x - a)^{3},

a one-parameter family of cubics, each a horizontal translate of $y = x^{3}$ . That the translates must all be solutions is worth checking directly: if $y = ϕ (x)$ solves the equation and $ψ (x) = ϕ (x - a)$ , then

\psi'(x) = \phi'(x-a) = 3\bigl[\phi(x-a)\bigr]^{2/3} = 3\bigl[\psi(x)\bigr]^{2/3} ,

so $ψ$ solves it too. Sliding a solution sideways changes nothing, precisely because the right-hand side never consults $x$ .

We now assemble solutions through the origin. Both $y \equiv 0$ and $y = x^{3}$ qualify, and so does any hybrid: choose $b \leq 0 \leq a$ and set

y = \begin{cases} (x-b)^3, & x < b, \\ 0, & b \leq x \leq a, \\ (x-a)^3, & x > a . \end{cases}

Each piece satisfies the equation, and at each join both the function and its derivative take the value $0$ , so the assembled function is $C^{1}$ and hence a genuine solution. Distinct pairs $(a, b)$ give distinct solutions, and there are infinitely many of them through the origin.

The culprit is 
  
    
      
        ∂
        f
      
      
        ∂
        y
      
    
  
  =
  2
  
    y
    
      −
      1
      /
      3
    
  
, which blows up as 
  y
  →
  0
. The uniqueness hypothesis fails along the entire line 
  y
  =
  0
, which is exactly where a solution may linger for an arbitrary time before peeling off. Note how tame the field itself looks: the branching is invisible in the picture, a good reminder that a direction field shows what solutions do, not whether they are unique.

The direction field for y' = 3y^{2/3}. The equilibrium $y = 0$ and the cubics $y = (x - a)^{3}$ are all solutions, and their pieces glue together smoothly wherever they meet the $x$ -axis; the marked origin lies on infinitely many solution curves.

Frequently Asked Questions

What is a direction field?

It is a picture of the equation y' = f(x,y) made by drawing, at each point of a grid, a short segment whose slope is the value $f (x, y)$ prescribes there. Because any solution must be tangent to the field everywhere, tracing a path that always follows the segments sketches a solution curve.

What is the difference between a direction field and a vector field?

A direction field records only slope, so its segments are drawn with a uniform short length and carry no arrowhead. A vector field also records magnitude and orientation. For a first-order equation the slope is all the information the equation supplies, so the direction field is the natural picture.

What is an isocline, and how does it help?

An isocline is the set of points where the slope takes a given value $m$ , that is, the curve $f (x, y) = m$ . It helps because you can draw an entire family of identical segments along one isocline at once, instead of computing $f$ at every grid point separately. Sketching four or five isoclines is usually enough to see the shape of the whole field.

Can two solution curves ever cross?

Not in a region where $f$ and $\partial f / \partial y$ are continuous, since exactly one integral curve passes through each point there. Crossings signal that the uniqueness hypotheses have failed. The standard example is y' = 3y^{2/3}, where infinitely many solutions pass through the origin.

Why bother with a direction field if I can solve the equation?

Because most equations cannot be solved in elementary functions. The field is available directly from $f$ with no integration at all, and it answers the questions people usually care about: does the solution grow without bound, settle to a limit, or oscillate. Even when a formula exists, the field is often the faster route to the qualitative answer.

How does a direction field relate to numerical methods?

Euler's method is nothing but following the direction field in discrete steps: from the current point, move a short distance along the segment, then recompute the slope at the new point and repeat. Sketching a field by hand and running Euler's method are the same idea, one done by eye and one by arithmetic.

Geometric Interpretation of First-Order Differential Equations. Direction Fields

Quick Reference

The Geometric Meaning of y' = f(x,y)

How to Sketch a Direction Field

Equilibrium Solutions and Nullclines

Why Integral Curves Do Not Cross

Exercises

Frequently Asked Questions

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