Slope fields

Drawing and reading slope fields (direction fields) for first-order differential equations, sketching solution curves through a point, matching DEs to slope fields, and spotting constant solutions.

The idea

A DE $\dfrac{dy}{dx} = f(x, y)$ tells you the gradient of the solution curve at every point. A slope field shows this with a short line segment at each point of a grid, drawn with that gradient. A solution curve follows the segments, like a leaf floating on a stream.

Slope field for dy/dx = x + y

Each segment has gradient x + y. The solution curve y = Ce^x − x − 1 follows the field. Move C to see the family.

Try this

Which value of $C$ gives a straight-line solution?

Reading a slope field

If $\dfrac{dy}{dx} = f(x)$ only, the segments in each vertical column are parallel.
If $\dfrac{dy}{dx} = g(y)$ only, the segments in each horizontal row are parallel.
Where $\dfrac{dy}{dx} = 0$, the segments are horizontal. If $g(c) = 0$, then $y = c$ is a constant solution.

Slope field for dy/dx = y(2 − y)

The rows y = 0 and y = 2 have horizontal segments, so they are constant solutions. Curves between them rise towards y = 2.

Try this

What happens to a solution that starts above $y = 2$?

Example 1●●

For $\dfrac{dy}{dx} = x - y$, find the gradients at $(0, 0)$, $(2, 1)$ and $(1, 3)$, and the curve where the segments are horizontal.

At $(0, 0)$: $0$. At $(2, 1)$: $1$. At $(1, 3)$: $-2$.

The segments are horizontal where $x - y = 0$, that is, along the line $y = x$.

Flashcards

Slope field

Short segments showing the gradient $\dfrac{dy}{dx}$ at grid points.

$dfrac{dy}{dx} = f(x)$ only

Segments in each vertical column are parallel.

$dfrac{dy}{dx} = g(y)$ only

Segments in each horizontal row are parallel.

Constant solution of $dfrac{dy}{dx} = g(y)$

$y = c$ wherever $g(c) = 0$.

Practice

Find the gradient of the slope field of $\dfrac{dy}{dx} = x + y$ at $(1, 2)$.
Find the gradient of the slope field of $\dfrac{dy}{dx} = \dfrac{x}{y}$ at $(2, -1)$.
Find the gradient of the slope field of $\dfrac{dy}{dx} = 2x - y$ at $(3, 2)$.
Find the gradient of the slope field of $\dfrac{dy}{dx} = 1 + x^2$ at the origin.
Find both constant solutions of $\dfrac{dy}{dx} = y(2 - y)$. Enter the two $y$-values separated by &.
For $\dfrac{dy}{dx} = x - y$, the segments are horizontal along:
In a slope field, every row has parallel segments. The DE has the form:
For $\dfrac{dy}{dx} = y(2 - y)$, a solution starting at $y = 3$: