Unit 7

Differential Equations

Slope fields, separation of variables, exponential and logistic models, and Euler’s method.

Topics in this unit

Modeling with Differential Equations · Verifying Solutions · Slope Fields · Reasoning with Slope Fields · Euler's Method · Separation of Variables · General & Particular Solutions · Domain of a Particular Solution · Exponential Growth & Decay · Newton's Law of Cooling · Logistic Growth · Second-Derivative Analysis

Slope Fields

Question 1

no calculatormedium

The slope field shown is for which differential equation?

Show the answer

The answer is B.

Slopes constant within each vertical column (fixed

x

) and zero along

x = 0

\Rightarrow

slope depends only on

x

\frac{d y}{d x} = x

Why not A:If the slope depended on $y$ , segments would match across each $row$ , not each column, and be horizontal along the $x$ -axis.
Why not C:Then slope would vary within a column (as $y$ changes); here columns are uniform.
Why not D:Then segments would be horizontal along both axes; here only the $y$ -axis is horizontal.

Question 2

no calculatormedium

Shown are four slope fields. Which one represents the differential equation

\frac{d y}{d x} = x

Show the answer

The answer is B.

For

\frac{d y}{d x} = x

the slope depends only on

x

: every segment in a vertical column is identical, and the segments are horizontal all along the

y

-axis (

x = 0

Why not A:This is $\frac{d y}{d x} = y$ : the slope depends only on $y$ , so rows are uniform and segments are horizontal along the $x$ -axis.
Why not C:This is $\frac{d y}{d x} = x + y$ : the slope changes within each column (it grows as $y$ rises), and segments are horizontal along the line $y = - x$ .
Why not D:This is $\frac{d y}{d x} = 1$ (a constant): every segment is parallel everywhere, with no zero-slope axis.

Reasoning with Slope Fields

Question 3

no calculatormedium

The slope field for a differential equation is shown. Let

y = f (x)

be the solution with

f (0) = 0

. Which of the following describes

f

for

x > 0

Show the answer

The answer is C.

Below the row

y = 3

the segments rise, and they flatten as

y

climbs toward

3

, where they are horizontal. So

f

increases with decreasing slope (concave down) and levels off toward

3

. (The field is for

\frac{d y}{d x} = \frac{3 - y}{2}

Why not A:The segments are not parallel: they are steep near $y = 0$ and nearly flat near $y = 3$ . The rate of increase shrinks.
Why not B:This reads the slopes as growing. Moving up toward $y = 3$ the segments get flatter, so the curve bends down and levels off.
Why not D:The row $y = 3$ is all horizontal, so it is an equilibrium solution. A solution starting below it cannot cross it.

Euler's Method

Question 4

BCno calculatorhard

Let

y = g (x)

satisfy

\frac{d y}{d x} = 6 - y

with

g (0) = 0

. Using Euler's method with

h = 0.5

g (1)

is approximated, and this approximation is

Show the answer

The answer is A.

Step 1: slope

6 - 0 = 6

y \approx 0 + 0.5 (6) = 3

. Step 2: slope

6 - 3 = 3

y \approx 3 + 0.5 (3) = 4.5

. Also

\frac{d ^{2} y}{d x ^{2}} = - \frac{d y}{d x} = - (6 - y) < 0

for

y < 6

, so the curve is concave down; tangent lines lie above it, making Euler an overestimate.

Why not B:The value $4.5$ is right, but $\frac{d ^{2} y}{d x ^{2}} = - (6 - y) < 0$ means the curve is concave down, so Euler overestimates (it lies above the curve), not under.
Why not C: $6$ is the limiting value as $x \to \infty$ , not $g (1)$ , and Euler is not exact here since $y^{''} \neq = 0$ .
Why not D: $3$ is the value after only the first step (at $x = 0.5$ ). A second step gives $4.5$ .

Separation of Variables

Question 5

no calculatormedium

The function

y = f (x)

satisfies

\frac{d y}{d x} = 2 x y

with

f (0) = 3

. What is the value of

f (1)

Show the answer

The answer is D.

Separate:

\frac{d y}{y} = 2 x d x

, so

ln ∣ y ∣ = x^{2} + C

and

y = A e^{x^{2}}

. The condition

f (0) = 3

gives

A = 3

, so

y = 3 e^{x^{2}}

. Then

f (1) = 3 e^{1} = 3 e

Why not A:That is $f (0)$ , the initial value, not $f (1)$ . Plug $x = 1$ into $3 e^{x^{2}}$ .
Why not B:You doubled the initial value instead of multiplying by $e^{x^{2}}$ . The growth is exponential, not linear.
Why not C:The $3$ is a coefficient, not an exponent. The solution is $3 e^{x^{2}}$ , so $f (1) = 3 e$ , not $e^{3}$ .

General & Particular Solutions

Question 6

no calculatormedium

The general solution of

\frac{d y}{d x} = \frac{y}{x}

x > 0

is which family of curves?

Show the answer

The answer is B.

Separate:

\frac{d y}{y} = \frac{d x}{x}

, so

ln ∣ y ∣ = ln ∣ x ∣ + C_{1}

and

y = C x

. Check:

\frac{d y}{d x} = C

and

\frac{y}{x} = \frac{C x}{x} = C

. They match.

Why not A:For $y = x + C$ , $\frac{d y}{d x} = 1$ but $\frac{y}{x} = 1 + \frac{C}{x} \neq = 1$ . Not a solution.
Why not C:That solves $\frac{d y}{d x} = y$ , not $\frac{d y}{d x} = \frac{y}{x}$ . Integrating $\frac{d x}{x}$ gives $ln ∣ x ∣$ , leading to $y = C x$ .
Why not D:For $y = C / x$ , $\frac{d y}{d x} = - C / x^{2}$ but $\frac{y}{x} = C / x^{2}$ ; they differ by a sign, so it solves $\frac{d y}{d x} = - \frac{y}{x}$ .

Exponential Growth & Decay

Question 7

calculatorhard

Coffee cooling in a

7 0^{\circ}

F room follows

T (t) = 70 + 130 e^{k t}

, and

T (5) = 18 0^{\circ}

F. To the nearest tenth, what is

T (15)^{\circ}

Show the answer

The answer is A.

T (5) = 180 \Rightarrow e^{5 k} = \frac{110}{130} = \frac{11}{13}

. Then

T (15) = 70 + 130 (e^{5 k})^{3} = 70 + 130 (\frac{11}{13})^{3} \approx 148. 8^{\circ}

Why not B:Linear extrapolation ( $- 4^{\circ}$ /min) overshoots; cooling slows, giving about $148. 8^{\circ}$ .
Why not C:That's $70 + 130 \cdot \frac{11}{13} = 180$ , which is $T (5)$ again (exponent $\times 1$ ). Over $15$ min the exponent triples to $(11/13)^{3}$ , giving $\approx 148. 8^{\circ}$ F.
Why not D:Recompute $70 + 130 (\frac{11}{13})^{3} \approx 148. 8^{\circ}$ .

Logistic Growth

Question 8

BCno calculatormedium

For the logistic model

\frac{d P}{d t} = 0.03 P (1 - \frac{P}{1200})

with

P (0) = 100

t \to \infty lim P (t) =

Show the answer

The answer is B.

In a logistic model the population approaches the carrying capacity

M

, the value where

\frac{d P}{d t} = 0

with

P > 0

. Here

1 - \frac{P}{1200} = 0

gives

P = 1200

, so

lim_{t \to \infty} P (t) = 1200

Why not A: $600 = M /2$ is where growth is fastest, not the long-run limit. The population keeps rising past $600$ up to the capacity.
Why not C: $100$ is the initial population; since $P (0) < M$ the population grows toward $1200$ .
Why not D:Logistic growth is bounded by the carrying capacity, so $P$ levels off at $1200$ rather than growing without bound.

Second-Derivative Analysis

Question 9

no calculatorhard

The point

(1, - 1)

lies on the graph of a solution

y = f (x)

of the differential equation

\frac{d y}{d x} = x + y

. Which of the following is true at

(1, - 1)

Show the answer

The answer is D.

(1, - 1)

\frac{d y}{d x} = 1 + (- 1) = 0

, a critical point. Differentiate the DE:

\frac{d ^{2} y}{d x ^{2}} = 1 + \frac{d y}{d x} = 1 + (x + y)

, which at

(1, - 1)

1 + 0 = 1 > 0

. Concave up at a horizontal tangent

\Rightarrow

local minimum.

Why not A: $\frac{d ^{2} y}{d x ^{2}} = 1 + \frac{d y}{d x} = 1 + (x + y)$ , which at $(1, - 1)$ equals $1 > 0$ (concave up), so the critical point is a minimum, not a maximum.
Why not B:At $(1, - 1)$ , $\frac{d ^{2} y}{d x ^{2}} = 1 \neq = 0$ , so the graph is not changing concavity there; with $\frac{d y}{d x} = 0$ and concave up it is a local minimum.
Why not C: $\frac{d y}{d x} = 1 + (- 1) = 0$ there, so it IS a critical point; since $\frac{d ^{2} y}{d x ^{2}} = 1 > 0$ it is a local minimum.

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