Sketching Slope Fields

A first-order differential equation

\frac{dy}{dx} = F(x, y)

gives the slope of a solution curve at every point the curve passes through. Drawing those slopes shows the shape of the solutions before any of them is found.

Slope fields

Each segment is a short piece of the tangent line to the solution curve through its point. A solution curve is tangent to the segment at each grid point it passes through, and between grid points it runs close to the nearby segments.

To sketch a slope field by hand, compute

\frac{dy}{dx}

at each point. Draw a horizontal segment where the slope is

0

, a rising one where it's positive, and a falling one where it's negative. Make a segment steeper when the slope is larger in absolute value. A free-response answer is scored on exactly these features at the points asked for.

Example 1

Consider the differential equation

\frac{dy}{dx} = y - e^x

(a) Sketch the slope field at the nine points with

x = -1

0

1

and

y = 0

1

2

(b) Find where the slope is

0

, where it's positive, and where it's negative.

(0, 2)

(a) Substitute each point into

y - e^x

, to two decimal places:

	$x = -1$	$x = 0$	$x = 1$
$y = 2$	$1.63$	$1$	$-0.72$
$y = 1$	$0.63$	$0$	$-1.72$
$y = 0$	$-0.37$	$-1$	$-2.72$

One segment is horizontal, the one at

(0, 1)

. Every segment in the column

x = 1

falls, and the steepest one is at

(1, 0)

, with slope about

-2.72

The slope field of dy/dx = y − eˣ at the nine points. Slopes of 1 and −1 are at 45°, and the segment at (1, 0) is the steepest.

(b) The slope is

0

at every point of the curve

y = e^x

, so any segment drawn on it is horizontal. Above that curve

y > e^x

and the slope is positive. Below it the slope is negative.

(c) Start at

(0, 2)

, where the slope is

1

, and follow the segments in both directions. Read from left to right, the curve rises, staying above the curve

y = e^x

. It meets that curve, levels off, and then falls below it, more and more steeply. So it has one relative maximum, on the curve

y = e^x

A denser slope field for dy/dx = y − eˣ, with the curve y = eˣ dashed. The slope is 0 on that curve, and the solution through (0, 2) has its horizontal tangent where it crosses it.

Interactive: Slope Fields

In part (c) the solution curve was followed by hand. In the field below, for

\frac{dy}{dx} = x - y

, the curve through the green point is drawn for you, so many starting points can be compared. To the right, every one of them closes in on the line

y = x - 1

, which is itself a solution.

Fields that depend on one variable

When

\frac{dy}{dx}

depends on

x

alone, every point on a vertical line has the same slope, so the columns of the field are identical. When it depends on

y

alone, the rows are identical. Checking for either pattern is a quick first step in reading a field.

Example 2

Consider the differential equation

\frac{dy}{dx} = \frac{x^2}{4} - 1

(a) Explain why the columns of its slope field are identical.

(b) Find the general solution, and describe how the solution curves are related to each other.

(a) The slope

\frac{x^2}{4} - 1

doesn't involve

y

. All points with the same

x

have the same slope, so each column repeats the same segment. The segments are horizontal on the lines

x = \pm 2

, where

\frac{x^2}{4} = 1

(b) The solutions are the antiderivatives of

\frac{x^2}{4} - 1

y = \frac{x^3}{12} - x + C.

Changing

C

shifts the graph vertically, so every solution curve is a vertical shift of

y = \frac{x^3}{12} - x

. A vertical shift doesn't change the slope at any

x

, so one column of segments fits every curve.

The slope field of dy/dx = x²/4 − 1 with the solution curves for C = −1.5, 0, and 1.5. Each curve is a vertical shift of the others, with horizontal tangents at x = −2 and x = 2.

Example 3

Consider the differential equation

\frac{dy}{dx} = \cos^2 y

(a) Describe the slope field.

(b) Show that

y = \arctan x

is a solution.

y = \arctan(x - a)

is also a solution, and find the one through

\left(3, \frac{\pi}{4}\right)

(a) The slope depends on

y

alone, so the rows are identical. Since

0 \le \cos^2 y \le 1

, no segment falls and none is steeper than

45^\circ

. The segments on the

x

-axis have slope

1

. They flatten as

y

approaches

\pm\frac{\pi}{2}

, where they're horizontal.

(b) For

y = \arctan x

\frac{dy}{dx} = \frac{1}{1 + x^2}

. For the right side,

\tan y = x

, so

\cos^2 y = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}.

The two agree for every

x

(c) For

y = \arctan(x - a)

\frac{dy}{dx} = \frac{1}{1 + (x - a)^2}

, and

\tan y = x - a

gives

\cos^2 y = \frac{1}{1 + (x - a)^2}

. So each horizontal shift is a solution for every

x

. Through

\left(3, \frac{\pi}{4}\right)

\arctan(3 - a) = \frac{\pi}{4}

gives

3 - a = 1

, so

a = 2

and the solution is

y = \arctan(x - 2)

The slope field of dy/dx = cos²y with the solutions y = arctan(x + 2), y = arctan x, and y = arctan(x − 2). Each is a horizontal shift of the others, by 2 units, and all level off toward y = π/2 and y = −π/2.