Worksheet (review for the Quiz (6.1 – 6.4 (differential equations and Euler’s method))
1.
What is a particular solution to the differential equation
!"
!#
= "# $ with the initial
condition #(2) = 1?
c
=
=
dyta
O
=
xdx
=y
ty = y
=
-
2.
The number of antibodies # in a patient’s bloodstream at time ( is increasing according to a
logistic differential equation. Which of the following could be the differential equation?
(A)
(B)
(C)
O
(D)
!"
!%
!"
!%
!"
!%
!"
!%
= 0.025(
= 0.025((5000 − ()
= 0.025(5000 − #)
= 0.025#(5000 − #)
3.
At time ( = 0 years, a forest preserve has a population of 1500 deer. If the rate of growth
of the population is modeled by -(() = 2000 / .$'% deer per year, what is the population at
time ( = 3?
223t
2000
-
0-23
(A) 3987
+C
=
1000
(B) 5487
(C) 8641
&
2000
O
(D) 10141
0
.
89
- e
23
0
.
-
7195
4.
A population of wolves is modeled by the function 4 and grows according to the logistic
!(
(
differential equation !% = 54 51 − )***6, where ( is the time in years and 4(0) = 1000.
Which of the following statements are true?
I.
lim 4(() = 5000
%→,
!(
~
is positive for ( > 0
~
III. !% ! is positive for ( > 0
X
II.
!%
!!(
d25(1-800 + 5005
=
&000-Food
(A) I only
(B) II only
O
(C) I and II only
(D) I and III only
=
5
-
70
5.
If
!"
!%
= −10/ -%⁄$ and #(0) = 20, what is the value of #(6)?
(A) 20/ -/
(B) 20/
O
-'
(C) 20/ -$
&
(D) 10/ -'
dy yoede
=
y
20 + y
=
=
-
g(b)
C 0
=
zoe
=
20
3
6.
!0
0
0
The function < satisfies the logistic differential equation !% = 1* 51 − 2)*6, where
<(0) = 105. Which of the following statements is false/
(A) lim <(() = 850 ~
%→,
!0
(B)
O !% has a maximum value when < = 105
(C)
!!0
!% !
= 0 when < = 425
!0
!!0
(D) When < > 425, !% > 0 and !% ! < 0
42
x
7.
?
x
−0.2
0
0.2
0.4
3 (")
0.8
1.2
1.7
2.3
The table above shows values of ? 3 , the derivative of a function ?, for selected values of ".
If ?(−0.2) = 1, what is the approximation for ?(0.4) obtained by using Euler’s method
with a step size of 0.2 starting at " = −0.2 ?
-
(A) 1.48
0
B
(B) 1.74
O
I
2
.
It 0 2 (0 8)
O
(C) 2.04
8
.
1
2
(D) 2.20
0 4
.
16 +
.
0 . 2x1-2
.
.
.
!"
Let # = ?(") be the solution to the differential equation !# = " − # with initial condition
?(2) = 8. What is the approximation for ?(3) obtained by using Euler’s method
with two equal length, starting at " = 2 ?
(A) 2
O
(C)
(D)
!
"
#!
$
2
61
.j
2
$
3
8
8+
57
0
.
5x( 6)
0
=
-
.
1
=
1 2)
.
.
1 4 + 0 2x1 7 %.
8.
(B)
16
-
.
.
5
5X-2 5
.
-
74
9.
The number of students in a cafeteria is modeled by the function 4 that satisfies the logistic
!(
1
differential equation !% = $*** 4 (200 − 4), where ( is the time in seconds and 4(0) = 25.
What is the greatest rate of change, in students per second, of the number of students in the
cafeteria?
d
O
(A) 5
(B) 25
⑳ (200-p) + 71 PC
= --
(C) 100
(D) 200
To For
p = 100
X100
(200-200
xioo Eg
10.
=
The temperature of a solid at time ( ≥ 0 is modeled by the nonconstant function A and
!4
increases according to the differential equation !% = 2A + 1, where A(() is measured in in
degrees Fahrenheit and ( is measured in hours. Which of the following must be true?
(A) A = A$ + ( + C
%
(B) ln|2A + 1| = $ + C
dit
(C) ln|2A + 1| = ( + C
-
(D) ln|2A + 1| = 2( + C
O
2H+
=
[In (2H 1)
+
(n(214 + 11
dt
=
=
et C
2ttC
11.
Let g be a function such that g(#) > 0 for all #. Which of the following could be a slope
!"
Field for the differential equation !# = (" $ − 1)g(#) ?
+
X
+0
=
①
+
X71
X
1
=
-
1 to
X
X
12.
!"
Let # = ?(") be the solution to the differential equation !# = " + 2# with initial
condition ?(0) = 2 . What is the approximation for ?(−0.4) obtained by using Euler’s
method with two steps of equal length starting at " = 0 ?
(A) 0.76
8
O
2
(B) 1.20
=
(C) 1.29
0
2) (4)
2
2 + z0
4
1 2 + 70 2) (22)
.
.
=
12
(D) 3.96
-
0
.
.
.
=
0
.
76
13.
!"
If !# = 2 − # , and if # = 1 when " = 1 , then # =
(A) 2 − / #-1
(B) 2 − /
o
1-#
Eydy
=
dx
(C) 2 − / -#
(D) 2 + / -#
-
(n(2-y)
2
=
In12yl
2
=
y
2 -
E
=
1
C
=
2
x +2
=x
-
-
-
c
g
Ce
Cex
=
y
T
=
2
=
2
-
ex
14.
The population 4 of rabbits on a small island grows at a rate that is jointly proportional to
the size of the rabbit population and the difference between the rabbit population and the
carrying capacity of the population. If the carrying capacity of the population is 2400
rabbits, which of the following differential equations best models the growth rate of the
rabbit population with respect to time ( , where G is a constant?
(A)
(B)
(C)
!(
!%
!(
!%
!(
!%
= 2400 − G4
= G(2400 − 4)
1
= G ( (2400 − 4)
(D) !% = G4(2400 − 4)
①
!(
15.
In a national park, the population of mountain lions grows over time. At time ( = 0, where ( is
measured in years, the population is found to be 20 mountain lions.
(a) One zoologist suggests a population model 4 that satisfies the differential equation
!(
!%
1
= 5 (220 − 4). Use separation of variables to solve this differential equation for 4
With the initial condition 4(0) = 20.
(b) A second zoologist suggests a population model H that satisfies the differential equation
!6
!%
1
!6
= )** H(220 − H). Find the value of !% at the time when H grows most rapidly.
(C) For the population model H introduced in part (b), use Euler’s method, starting at ( = 0
with two steps of equal size, to approximate H(10). Show the computations that led to your
answer.
D )
a)
.
=
-
In 1220-p1
it
-
220
-p
=
e
2--50
c
-
= Q
220-Cert
220
C
-
p
=
=
200
-
220
-
2009
4
110
d = Go
20
=
-
=
=
=
C
p
+
500
+C
=
do t = (220Q) +1
t
24 2
.
15.
In a national park, the population of mountain lions grows over time. At time ( = 0, where ( is
measured in years, the population is found to be 20 mountain lions.
(a) One zoologist suggests a population model 4 that satisfies the differential equation
!(
!%
1
= 5 (220 − 4). Use separation of variables to solve this differential equation for 4
With the initial condition 4(0) = 20.
(b) A second zoologist suggests a population model H that satisfies the differential equation
!6
!%
1
!6
= )** H(220 − H). Find the value of !% at the time when H grows most rapidly.
(C) For the population model H introduced in part (b), use Euler’s method, starting at ( = 0
with two steps of equal size, to approximate H(10). Show the computations that led to your
answer.
.)
c
O
5
10
5
20
20 +
Q(220
5x5
.
200
60 + 5x160
=
156
-
a)
=
60