Reasoning with Slope Fields

A slope field and its differential equation describe the same solutions. Each solution is a function, and together they form a family. Many features of those solutions can be read from the field or the equation, without a formula for any of them.

Matching a field to its equation

To decide which equation a field belongs to, compare features that are quick to check.

In the slope field	In the equation
horizontal segments along a line or curve	$\frac{dy}{dx} = 0$ there
segments rising in a region	$\frac{dy}{dx} > 0$ there
segments identical all the way up each column	$\frac{dy}{dx}$ depends on $x$ alone
segments identical along each row	$\frac{dy}{dx}$ depends on $y$ alone
no segment at a point	$\frac{dy}{dx}$ undefined there

Example 1

The slope field below belongs to one of these three equations. Identify it, and give a reason that rules out each of the other two.

\begin{aligned} &\text{(A)}\quad \frac{dy}{dx} = y\left(1 - x^2\right) \\[6pt] &\text{(B)}\quad \frac{dy}{dx} = x\left(1 - y^2\right) \\[6pt] &\text{(C)}\quad \frac{dy}{dx} = x\left(y^2 - 1\right) \end{aligned}

A slope field on −2 ≤ x ≤ 2, −2 ≤ y ≤ 2, with segments every half unit.

Along the

y

-axis every segment is horizontal (they cross the axis like tick marks), so

\frac{dy}{dx} = 0

whenever

x = 0

. In (A),

\frac{dy}{dx} = y(1 - 0) = y

when

x = 0

, which is

1

(0, 1)

. So the field isn't the field of (A).

The rows

y = 1

and

y = -1

are horizontal too, and (C) is

0

on both of them as well. But at

(1, 0)

, (C) gives

\frac{dy}{dx} = 1(0 - 1) = -1

, while the field's segment there rises. So the field isn't the field of (C) either.

Equation (B) gives

\frac{dy}{dx} = 0

when

x = 0

and when

y = \pm 1

, which matches all three lines of horizontal segments. At

(1, 0)

it gives

1(1 - 0) = 1

, the rising segment. The field belongs to

\frac{dy}{dx} = x\left(1 - y^2\right)

Equilibrium solutions

When

\frac{dy}{dx} = g(y)

with

g

continuous, the equilibrium solutions come from the zeros of

g

. Between two neighboring equilibria the sign of

g

doesn't change, so the solutions there all increase or all decrease.

Example 2

Consider the differential equation

\frac{dy}{dx} = \frac{y^3 - 4y}{4}

(a) Find the equilibrium solutions.

(b) Let

y = f(x)

be the solution with

f(0) = 1.8

. Describe

f

for

x \ge 0

, and find

\lim_{x \to \infty} f(x)

\frac{d^2y}{dx^2}

in terms of

y

. Find the value of

y

at which the graph of

f

has a point of inflection.

(a) Factor the right side:

\frac{y^3 - 4y}{4} = \frac{y(y - 2)(y + 2)}{4}.

It's

0

when

y = -2

0

, or

2

, so the equilibrium solutions are

y = -2

y = 0

, and

y = 2

(b) For

0 < y < 2

the factors

y

and

y + 2

are positive and

y - 2

is negative, so

\frac{dy}{dx} < 0

. At

y = 1.8

the slope is

\frac{5.832 - 7.2}{4} = -0.342

, so

f

starts by decreasing.

The right side and its derivative with respect to

y

are continuous, and for such an equation two solution curves never meet. This course uses that fact without proof. So

f

can't cross the equilibrium

y = 0

. It keeps decreasing, and its slope approaches

0

only as

y

approaches

0

. So

f

decreases toward

0

, and

\lim_{x \to \infty} f(x) = 0

(c) Differentiate the equation with respect to

x

, using the chain rule on the right side, and substitute for

\frac{dy}{dx}

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{3y^2 - 4}{4} \cdot \frac{dy}{dx} \\[4pt] &= \frac{(3y^2 - 4)(y^3 - 4y)}{16}. \end{aligned}

For

0 < y < 2

the factor

y^3 - 4y

is negative. The factor

3y^2 - 4

is positive when

y > \frac{2}{\sqrt{3}}

and negative when

y < \frac{2}{\sqrt{3}}

. So the graph of

f

is concave down while

y > \frac{2}{\sqrt{3}}

and concave up after

f

falls below that value. The inflection point is where

y = \frac{2}{\sqrt{3}} \approx 1.155

The slope field of dy/dx = (y³ − 4y)/4. Its rows of horizontal segments at y = −2, 0, and 2 are the equilibrium solutions. The solution through (0, 1.8) decreases toward 0, with its inflection point (amber) at y = 2/√3. The solution through (0, −1) increases toward 0, and the one through (0, 2.2) increases without bound.

Concavity and tangent lines from the equation

The equation gives

\frac{dy}{dx}

at any point, so it gives the tangent line to the solution curve through that point. Differentiating the equation implicitly gives

\frac{d^2y}{dx^2}

. Substituting the equation for

\frac{dy}{dx}

puts the result in terms of

x

and

y

, and its sign gives the concavity of the solution curves.

Example 3

Let

y = f(x)

be the solution of

\frac{dy}{dx} = x\left(1 - y^2\right)

, the equation of Example 1, with

f(1) = 0.8

(a) Write an equation for the line tangent to the graph of

f

x = 1

. Use it to approximate

f(1.25)

(b) Find

\frac{d^2y}{dx^2}

in terms of

x

and

y

. Determine whether the approximation in (a) is an overestimate or an underestimate.

(0, -0.5)

and

(0, 1.5)

. For each one, determine whether it has a relative minimum, a relative maximum, or neither at

x = 0

(a) At

(1, 0.8)

the equation gives

\frac{dy}{dx} = 1\left(1 - 0.64\right) = 0.36

. The tangent line is

y = 0.8 + 0.36(x - 1)

, so

f(1.25) \approx 0.8 + 0.36(0.25) = 0.89.

(b) Differentiate

\frac{dy}{dx} = x\left(1 - y^2\right)

with respect to

x

, using the product rule and the chain rule, then substitute for

\frac{dy}{dx}

\begin{aligned} \frac{d^2y}{dx^2} &= \left(1 - y^2\right) - 2xy\,\frac{dy}{dx} \\[4pt] &= \left(1 - y^2\right) - 2x^2y\left(1 - y^2\right) \\[4pt] &= \left(1 - y^2\right)\left(1 - 2x^2y\right). \end{aligned}

(1, 0.8)

this is

(0.36)(1 - 1.6) = -0.216 < 0

. For

x \ge 1

, the solution rises from

0.8

toward

1

, since

\frac{dy}{dx} > 0

while

0 < y < 1

. There

1 - y^2 > 0

and

1 - 2x^2y \le 1 - 2(1)(0.8) < 0

, so

\frac{d^2y}{dx^2} < 0

. The graph of

f

is concave down on

[1, 1.25]

, so it lies below its tangent line, and

0.89

is an overestimate.

(c) When

x = 0

the equation gives

\frac{dy}{dx} = 0\left(1 - y^2\right) = 0

, so both solutions have a horizontal tangent at

x = 0

. Use the Second Derivative Test with the formula from (b).

(0, -0.5)

\frac{d^2y}{dx^2} = (1 - 0.25)(1 - 0) = 0.75 > 0

, so that solution has a relative minimum at

x = 0

(0, 1.5)

\frac{d^2y}{dx^2} = (1 - 2.25)(1 - 0) = -1.25 < 0

, so that solution has a relative maximum at

x = 0

The solution through (1, 0.8) and its tangent line (amber), which lies above the curve where the curve is concave down. The solutions through (0, −0.5) and (0, 1.5), in blue, have a minimum and a maximum at x = 0.