BC 3
Differential Equations Practice
(1)
Name: Key
A population of bacteria grows at a rate proportional to the number present. After
2 hours, there are 300 bacteria, and after 8 hours, there are 1800. Write the
differential equation that models this situation and find how many bacteria were
present at time t = 0?
IVP: P  kP, P(2)  300, P(8)  800 . The solution to this differential equation is given
by: P(t )  Aekt . Now, P(2)  300, P(8)  800  300  Ae2k and 800  Ae8k . Dividing the
8
1 8
second equation by the first gives:  e6 k  ln    k .Then
3
6 3
1 8
ln  
3  3
300 300 3 3

 150 3 3 .
2
8
3
3
3
Thus, P(0)  150 3  216 .
300  Ae
(2)
 A
AP BC 1991: 6 (non-calculator) A certain rumor spreads through a community
dy
 2y(1 y) , where y is the proportion of the population that has
at the rate of
dt
heard the rumor at time t.
(a)
What proportion of the population has heard the rumor when it is
spreading the fastest?
When dealing with the logistic equation, the rate is always maximum at ½ the
carrying capacity; in this case that is ½.
(b)
If at time t = 0, ten percent of the people have heard the rumor, find y as a
function of t.
We need to solve:
dy
 2 y (1  y ), y (0)  0.1 . Separating variables, we get:
dt
dy
 2dt. Using partial fraction decomposition,
y (1  y )
1
A
B
 
 A  1 and B  1 . So
y (1  y ) y (1  y )
dy
 y(1  y)   2dt  
dy
dy
y

 2t  C  ln y  ln 1  y  2t  C  ln
 2t  C
y (1  y )
1 y
.
IMSA BC 3
Asst. DE p.1
Fall ‘10
BC 3
Differential Equations Practice
Name: Key
Exponentiation leads to:
y
Ae2t
 e2t C  Ae2t  y  Ae2t  y  Ae2t  y 
. Since y (0)  0.1 ,
1 y
1  Ae2t
1 2t
e
Ae0
A
1
e 2t
9
.
So,
.
.1 


A

y


1 2t 9  e 2t
1  Ae0 1  A
9
1 e
9
(c)
At what time t is the rumor spreading the fastest?
We know y = ½ when the rumor is spreading fastest, so
1
e 2t
1

 9  e2t  2e2t  e2t  9  t  ln  9   1.099
2t
2 9e
2
IMSA BC 3
Asst. DE p.2
Fall ‘10
BC 3
Differential Equations Practice
Name: Key
3. When the valve at the bottom of a cylindrical tank is opened, the depth of the liquid in the
tank drops at a rate proportional to the square root of the depth of the liquid. If y(t) is the
depth of the liquid t minutes after the valve is opened, then this can be modeled by the
differential equation
y(t )  k y(t )
a. Describe in words what the differential equation says about the change in the
depth of the liquid.
As the depth decreases, the rate of flow also decreases.
b. Does the level of the liquid drop faster when the tank is full or when the tank is
half full? Explain.
c. Suppose that for a certain tank y(0) = 9 feet, and y(20) = 4 feet. Solve for y(t),
then sketch the graph of y vs t.
y(t )  k y (t ) 
dy
dy
 kdt  
 kdt
y
y 
1
2
 kt  C  . Since y (0)  9,
4
1
1
2
9  C 2  C  6. Since y (20)  4, 4   20k  6  
4
4
1
k  .
10
 2 y  kt  C  y 
1 1

Therefore, y (t )    t  6 
4  10

2
d. Is this a realistic model? Explain.
It seems reasonable – as the tank empties, there is less pressure pushing the
water out of the tank. Hence, the rate at which the water flows out is slowing.
IMSA BC 3
Asst. DE p.3
Fall ‘10
BC 3
Differential Equations Practice
. 4.
Name: Key
Solve each IVP.
z   0.15 z , z (0)  5000
a.
dz
dz
dz
z   0.15 z 
 .15 z 
 .15dt     .15dt  ln z  .15t  C . Since
dt
z
z
z (0)  5000
ln 5000  C . Solving ln z  .15t  C for z gives z  e.15t C  eC e.15t  5000e.15t .
y  0.2( y  68), y (0)  202
dy
dt
dt
y  0.2( y  68) 
 0.2( y  68) 
 0.2dt  
 0.2dt  ln y  68  0.2  C
dt
( y  68)
( y  68) 
Solving ln y  68  0.2  C for y gives y  68  e.2t C  eC e.2t  y  Ae.2t  68 .
b.
Since y(0)  202, 202  Ae0  68  A  134 . therefore y  134e.2t  68
c.
5.
P  0.3P(200  P), P(0)  25 P = 0.3 P (200 – P), P(0) = 25
Show that the function y  Ae3t  Bet , where A and B are constants, is a solution to
the differential equation y  2 y  3 y  0 .
If y  Ae3t  Bet , then
d2
d
Ae3t  Be  t   2  Ae3t  Be  t   3  Ae3t  Be  t 
2 
dt
dt
3t
t
  9 Ae  Be   2  3 Ae3t  Be  t   3  Ae3t  Be  t 
y  2 y  3 y 
 9 Ae3t  6 Ae3t  3 Ae3t  Be  t  2 Be  t  3Be  t
0
6.
Find real numbers m and n so that the functions y  e2 x and y  e5x are both
solutions to the differential equation my  ny  y  0
If y  e2 x , my  ny  y  4me2 x  2ne2 x  e2 x  e2 x  4m  2n  1  0  4m  2n  1  0 ,
if y  e5x ,
my  ny  y  25me5 x  5ne5 x  e5 x  e5 x  25m  5n  1  0  25m  5n  1  0 .
 4m  2 n  1  0
7
1
 (m, n)   ,  
Solving for (m,n): 
 10 10 
25m  5n  1  0
IMSA BC 3
Asst. DE p.4
Fall ‘10
BC 3
Differential Equations Practice
7.
Name: Key
Coffee is cooling in a ceramic cup. The initial temperature of this coffee is 190° F,
and the room temperature is 32°F (because your room happens to be a tent you are
using on a camping trip). After 5 minutes, the temperature of the coffee is 130° F.
a.
Write the IVP that represents this situation. (No solving of equations to
determine any constants that might be present in your IVP!)
y  k ( y  32); y (0)  190 and y(5)  130 , where y is the temperature( F ) of the
coffee after t minutes.
Determine how long (to the nearest tenth of a minute) the coffee’s temperature
will stay above 98.6° F?
Solving this DE or remembering the solution to a Newton’s law of cooling gives:
y  Aekt  32. Since y (0)  190 , A  158 . Then
1  49 
y (5)  130  130  158e k 5  32  ln    k
5  79 
b.
So, on calculator we solve 98.6  158e
8.
1  49 
ln  t
5  79 
 32 . So t  9.0 min
x
and y (2)  3 .
2y
Use Euler’s method with a step size of 0.5 to estimate the value of y(4).
Consider the IVP y  
a.
From Calculator: y (4)  1.94946
b.
Solve this IVP and use your solution to determine the exact value of y(4).
Your solution should be of the form where y is a function of x. No TI-89 or
integral tables, please.


y  
x
dy
x


 2 ydy   xdx  2 ydy   xdx . Therefore,
2y
dx
2y
y2  
x2
 C. Since y(2)  3,9  2  C  C  11 . Then,
2
y 
x2
 11 (do you know why we choose the negative square root?)
2
So, y (4)   3  1.732
IMSA BC 3
Asst. DE p.5
Fall ‘10
BC 3
Differential Equations Practice
9.
Name: Key
Find a general solution to each of the following differential equations. Each
solution should be of the form where y is a function of x. No TI-89 or integral
tables, please.
y  3 x 2 e  y
a.
b.
y  3x 2e y  e y dy  3x 2 dx 



IMSA BC 3



dy
dy
 dx 
 dx
1 y
1 y
 x  C 
  Ln 1  y   x  C  1  y  e
y  1  y 
e y dy  3x 2 dx  e y  x3  C 
y  Ln x3  C
y  1  y
 y  1 e
Asst. DE p.6
 x  C 
Fall ‘10
BC 3
Differential Equations Practice
10.
Name: Key
In a state park in Kenya, the population of topi in the year 2000 was 150. One year
later, the topi population was 180. This state park can hold no more than 900 topi
at any one time. It is assumed that the topi population in the park is growing
logistically.
a.
Write the IVP that represents this situation. (No solving of equations to
determine any constants that might be present in your IVP!)
P  kP(900  P); P(0)  150 and P(1)  180 , where P(t) is the number of topi
t years after the year 2000.
b.
Determine a logistic function that models the topi population.
We need to solve:
we get:
dP
dP
 kP(900  P ); P (0)  150, P (1)  180. . Separating variables,
dt
 P(900  P)   kdt
Using partial fraction decomposition,
1
A
B
1
1
. So
 
 A
and B 
P(900  P) P (900  P)
900
900
dP
1 dP
dP
 P(900  P)   kdt  900  P  (900  P)  kt  C 
.
P
ln P  ln 900  P  900kt  C  ln
 900kt  C.
900  P
P
150
1
 Ae900 kt . Since P(0)  150 ,
 A A .
900  P
900  150
5
900 kt
P
1
900e
 e900 kt  P 
So,
. Since P (1)  180 ,
900  P 5
5  e900 kt
900e900 k
900
1
5
180 
 900  180e900 k  900e900 k 
 e900k  k 
Ln   .
900 k
5e
720
900  4 
Therefore,
Therefore, P 
900e
5e
c
5
ln  t
4
5
ln  t
4
What is the topi population when it is growing most rapidly?
When P = 450.
IMSA BC 3
Asst. DE p.7
Fall ‘10
BC 3
Differential Equations Practice
d.
In what year is the topi population growing most rapidly?
450 
900e
5e
year 2007.
IMSA BC 3
Name: Key
5
ln  t
 4
5
ln  t
 4
 5e
5
ln  t
 4
 2e
5
ln  t
 4
5e
Asst. DE p.8
5
ln  t
 4
t
5
    5  t  7.21 0r the
4
Fall ‘10