Take-Home Quiz # 1 Solutions
Math 81, Spring 2017
Instructor: Dr. Doreen De Leon
1. A new species of animal has just been discovered, and the rate of change in the size
3
of its population P is proportional to P 2 . If the initial population is 100 animals, and
after 3 months, the population has grown to 400 animals, find the size of the population
at any time t.
Solution: We have that
3
P 0 = kP 2 ,
where k is the constant of proportionality. In addition, we are given that P (0) = 100
and P (3) = 400. So, we have the following IVP
3
P 0 = kP 2 , P (0) = 100.
First, find the general solution using the method of separation of variables.
3
dP
= kP 2
dt
3
Separate variables by multiplying both sides by P − 2 and dt, obtaining
3
P − 2 dP = k dt.
Integrate both sides.
Z
P
− 23
Z
dP =
k dt
1
−2P − 2 = kt + c1 (c1 ∈ R)
1
1
1
1
P − 2 = − kt − c1 = − kt + c (c ∈ R)
2
2
2
√
1
P = 1
(c ∈ R)
− 2 kt + c
1
P (t) =
2 (c ∈ R).
− 21 kt + c
Next, use the initial condition to find c.
100 = P (0) =
1
c2
1
c= ,
10
100 =
1
1
− 21 k(0) + c
2
where we take the positive square root because above we had the positive square root
of P before squaring both sides. So,
P (t) =
1
− 12 kt +
=
1 2
10
100
.
(−5kt + 1)2
Next, use P (3) = 400 to solve for k.
400 = P (3) =
100
(−5k(3) + 1)2
10
−15k + 1
1
2=
−15k + 1
1
−15k + 1 = .
2
1
−15k = −
2
1
k= .
30
20 =
Therefore, the population size at any time t is given by
P (t) =
100
− 61 t + 1
2 animals,
or more nicely formatted
P (t) =
3600
animals.
(6 − t)2
2. A tank initially contains 50 gal of brine in which 5 pounds of salt is dissolved. Brine
containing 0.5 pounds of salt per gallon enters the tank at the rate of 4 gal/s, and the
well-mixed brine in the tank flows out at 6 gal/s. How much salt does the tank contain
at any time t before the tank is empty?
Solution: Let y(t) = amount of salt in the tank at time t (in pounds). The model for
a mixing problem is
y 0 = rate in − rate out.
So, we need to determine the “rate in” and the “rate out.”
rate in = 4 gal/s × 0.5 lb/gal = 2 lb/s
rate out = 6 gal/s×
y(t) lb
6y(t)
3
=
lb/s =
y(t) lb/s
50 gal + 4 gal/st s − 6 gal/st s
50 − 2t
25 − t
2
We obtain the initial value problem
3y
, y(0) = 5.
25 − t
y0 = 2 −
General solution: This is a linear differential equation.
First, write the equation in standard form:
y0 +
3
y = 2.
25 − t
Then, find the integrating factor. Since
p(t) =
3
,
25 − t
we have
R
µ(t) = e
3
25−t
dt
= e−3 ln |25−t| = (25 − t)−3 .
Multiply both sides of the equation by µ(x) = (25 − t)−3 and solve.
(25 − t)3 y 0 + 3(25 − t)−4 y = 2(25 − t)−3
d (25 − t)−3 y = 2(25 − t)−3
dt
Z
Z
d −3
(25 − t) y dt = 2(25 − t)−3 dt
dt
(25 − t)−3 y = (25 − t)−2 + c (c ∈ R)
y(t) = 25 − t + c(25 − t)3 .
Find c: Use y(0) = 5.
5 = y(0) = 25 − 0 + c(25 − 0)3
253 c = −20
20
4
c=− 3 =−
.
25
3125
Therefore, the tank contains
y(t) = 25 − t −
4
(25 − t)3 pounds of salt.
3125
3