Download Homework 2 solution.

MATH 307 I (Introduction to Differential Equations), Winter 2017
Solution 2
Problem 1. Section 2.3, problems 1, 6(c), 9, 16, 21(a)(b)
Sol.
1. Let Q(t) be the amount of dye in the tank at time t. Clearly Q(0) = 200g.
The differential equation governing the amount of dye is
Q0 (t) = −2Q(t)/200,
The solution of this separable equation is Q(t) = Q(0)e−t/100 = 200e−t/100 .
We need the time T such that Q(T ) = 2g. This means we have to solve
2 = 200e−T /100 and we obtain T = −100 ln(1/100) = 100 ln(100) ≈ 460.5
min.
6.(c) With A(h) = π, a = 0.01π, α = 0.6, the differential equation for the
water level h is
p
dh
= −0.006π 2gh,
π
dt
p
with solution h(t) = 0.000018gt2 − 0.006 2gh(0)t + h(0). Setting h(0) = 3
and g = 9.8, h(t) = 0.0001764t2 − 0/046t + 3, resulting in h(t) = 0 for
t ≈ 130.4s.
9. Let S(t) be the amount of the loan remaining at time t. Then dS/dt =
0.1S −k, S(0) = $8, 000. Solving for S(t) gives S(t) = 8000e0.1t −10k(e0.1t −
1). Setting S = 0 and substituting t = 3 gives k = $3, 086.64 per year. For
3 years this totals $9, 259.92, so $1, 259.92 has been paid in interest.
16. If T is the temperature of the coffee at any time t, then dT /dt =
−k(T − 70), T (0) = 200, TR(1) = 190. Then dT /dt + kT = 70k. Using
integrating factors, µ(t) = e kdt = ekt , and the integrating equations is
d
(T ekt ) = 70kekt .
dt
Integrating, we obtain T ekt = 70ekt + C, so T (t) = 70 + Ce−kt . Since
T (0) = 200, then C = 130 and T (t) = 70 + 130e−kt . To find k we use
the fact that T (1) = 190. Then 190 = 70 + 130e−k , so −k = ln(12/13)
and T (t) = 70 + 130eln(12/13)t . We want to find t such that 150 = T (t) =
70 + 130eln(12/13)t . Solving for t we find t = ln(8/13)/ ln(12/13).
21.(a) The differential equation for the motion is mdv/dt = −v/30 − mg.
Given the initial condition v(0) = 20 m/s, the solution is v(t) = −44.1 +
64.1e−t/4.5 . Letting v(t0 ) = 0, the ball reaches the maximum height at
1
2
t0 = 1.683 s. Integrating v(t), the position is given by x(t) = 318.45 −
44.1t − 288.45e−t/4.5 . Hence the max height is x(t0 ) = 45.78.
21.(b) Setting x(t1 ) = 0, the ball hits the ground at t1 = 5.128s.
Problem 2. A reservoir on a farm initially contains 10000 liters of water,
in which 200kg nitrate fertilizer is dissolved. The owner of the reservoir
decides that the amount of dissolved nitrate needs to be increased, and
therefore starts pumping in a 1 kg nitrate per liter water solution at a rate
of 100 liters per minute. However, the reservoir simultaneously develops a
leak, and starts draining at a rate of 200 liters per minute.
(a) Assuming the solution remains perfectly mixed at all times, set up a
differential equation for y(t), the mass of nitrate in the reservoir at time t.
Sol. Let V (t) be the volume of nitrate water in the reservoir at time t. Note
that V (t) = 10000 + (100 − 200)t = 10000 − 100t. The differential equation
for y is
dy
y
y
= 1(100) − 200 = 100 − 200
,
dt
V
10000 − 100t
or, in standard form,
2
dy
+
y = 100.
dt
100 − t
(b) Find the solution y(t).
Sol. This a linear first-order ODE, and the integrating factor is
R
1
e 2/(100−t) dt =
.
(100 − t)2
The integrating factor equation becomes
y
100
d
=
2
dt (100 − t)
(100 − t)2
and integrating with respect to t and simplifying gives
y = 100(100 − t) + C(100 − t)2 .
The initial condition y(0) = 200 becomes
200 = 100(100) + C(100)2
so that C = −0.98, and we get
y = 100(100 − t) − 0.98(100 − t)2 = −0.98t2 + 96t + 200
3
(c) What is the maximum amount of nitrate in the reservoir, and when does
it occur?
The maximal value of the function y is found when y 0 = 0, which gives
t ≈ 48.9796.
Problem 3. The corpse of King Joffrey is discovered in a room at midnight
and its temperature is 80◦ F, while the temperature of the room is kept
constant at 70◦ F. Two hours later, the temperature of the corpse dropped
to 75◦ F. Joffrey’s mother, Cersei, asks you to investigate the crime scene.
Even if you don’t like Joffrey and Cersei (who does?), you accept. Assuming
that the temperature of King Joffrey’s body just before dying was a healthy
98.6◦ F, can you find the time of death?
Sol. Let T (t) denote the temperature of Joffrey’s corpse t hours from midnight, so that T (0) = 80 and T (2) = 75. Newton’s law of cooling tells us
that
dT
= k(T − 70),
dt
a separable equation whose solution is
T = 70 + Cekt .
Using T (0) = 80 gives C = 10, so that
T = 70 + 10ekt .
Using T (2) = 75 then gives k = 0.5 ln 0.5, so that
T = 70 + 10e0.5t ln 0.5 .
We want t so that T = 98.6, i.e.
98.6 = 70 + 10e0.5t ln 0.5 .
Solving for t gives t ≈ −3.0320, so that Joffrey died approximately three
hours before midnight.