Math 3400
Homework 1
1.2 # 1: Solve the initial value problem
dy
= −y + 5,
dt
(a)
Write as
dy/dt
= −1,
y−5
y(0) = y0 .
d
log(|y − 5|) = −1.
dt
[|y − 5| = eC1 e−t
log(|y − 5|) = −t + C1 ,
y − 5 = C2 e−t ,
y(t) = 5 + C2 e−t ,
dy
= −2y + 5,
dt
(b)
Write as
dy/dt
= −2,
y − 5/2
y(0) = y0 .
d
log(|y − 5/2|) = −2.
dt
[|y − 5/2| = eC1 e−2t
log(|y − 5/2|) = −2t + C1 ,
y − 5/2 = C2 e−2t ,
y(t) = 5/2 + C2 e−2t ,
dy
= −2y + 10,
dt
(c)
Write as
y(t) = 5 + [y0 − 5]e−t
dy/dt
= −2,
y−5
y(0) = y0 .
d
log(|y − 5|) = −2.
dt
[|y − 5| = eC1 e−2t
log(|y − 5|) = −2t + C1 ,
y − 5 = C2 e−2t ,
y(t) = 5/2 + [y0 − 5/2]e−2t
y(t) = 5 + C2 e−2t ,
y(t) = 5 + [y0 − 5]e−2t
The solutions are similar in that as t → +∞ the solution approaches a constant, which
is 5 in (a) and (c), but 5/2 in (b).
1.2 # 2: Solve the initial value problem
(a)
Write as
dy
= y − 5,
dt
dy/dt
= 1,
y−5
d
log(|y − 5|) = 1.
dt
log(|y − 5|) = t + C1 ,
y − 5 = C2 et ,
(b)
y(0) = y0 .
[|y − 5| = eC1 et
y(t) = 5 + C2 et ,
dy
= 2y − 5,
dt
1
y(t) = 5 + [y0 − 5]et
y(0) = y0 .
Write as
dy/dt
= 2,
y − 5/2
d
log(|y − 5/2|) = 2.
dt
[|y − 5/2| = eC1 e2t
log(|y − 5/2|) = 2t + C1 ,
y − 5/2 = C2 e2t ,
y(t) = 5/2 + C2 e2t ,
(c)
Write as
dy
= 2y − 10,
dt
dy/dt
= 2,
y−5
y(0) = y0 .
d
log(|y − 5|) = 2.
dt
log(|y − 5|) = 2t + C1 ,
y − 5 = C2 e2t ,
y(t) = 5/2 + [y0 − 5/2]e2t
[|y − 5| = eC1 e2t
y(t) = 5 + C2 e2t ,
y(t) = 5 + [y0 − 5]e2t
The solutions are similar in that as t → +∞ the solution grows without bound. As
t → −∞ the solution approaches a constant, which is 5 in (a) and (c), but 5/2 in (b).
1.2 # 7 The field mouse population satisfies
dp
p − 900
= 0.5p − 450 =
.
dt
2
(a) Find the time of extinction if p(0) = 850.
From the text or class notes
p(t) = 900 + Cet/2 ,
p(0) = 900 + C.
In this case with p(0) = 850,
p(t) = 900 + [850 − 900]et/2 = 900 − 50et/2 .
The solution hits zero if
0 = 900 − 50et/2 ,
t = 2 ∗ log(18) = 5.78 months.
(b) Find the time of extinction if p(0) = p0 . Then
p(t) = 900 + [p0 − 900]et/2 .
The solution hits zero if
0 = 900 + [p0 − 900]et/2 ,
t = 2 ∗ log(
900
) months.
900 − p0
(c) Find p0 if extinction is in one year. By part (b), using units of months
12 = 2 ∗ log(
2
900
).
900 − p0
Then
900[e6 − 1]
= 900[1 − e−6 ] ' 898.
e6
1.2 # 8 Suppose (time units are now days)
p0 =
dp
= rp.
dt
(a) Find r if the population doubles in 30 days. The general solution is p(t) = Cert . Then
p(0) = C,
p(30) = 2C = Ce30r .
We want
e30r = 2,
r=
log(2)
.
30
(a) Find r if the population doubles in N days. The general solution is p(t) = Cert . Then
p(0) = C,
p(N ) = 2C = CeN r .
We want
eN r = 2,
r=
log(2)
.
N
1.1 # 22 A spherical raindrop evaporates at a rate proportional to surface area. Find a
differential equation for the volume V (t).
Since volume is V = 43 πr3 and SA = 4πr2 , surface area is proportional to V 2/3 . Thus
dV
= kV 2/3 .
dt
Note that since V is decreasing due to evaporation, k < 0.
1.1 # 23 Write an equation for Newton’s law of cooling. Let T (t) be the object temperature,
and Tb the background temperature. (If the rate constant is positive we should use Tb − T so
the temperature is decreasing if T exceeds Tb . The text is not clear on this point.)
dT
= k[Tb − T ].
dt
If Tb = 70 and k = 0.05 then
dT
= 0.05[70 − T ].
dt
1.1 # 24 (a) Let A(t) be the amount of drug in the body. Then
dA
= rate in − rate out.
dt
Given the problem statement
dA
= 100 ∗ 5 − 0.4 ∗ A.
dt
3
(b) How much drug is present in the long run? The quick solution is to assume in the
long run we see dA/dt ' 0. Then
A = 500/0.4 = 1250 mg.
More systematically, solve the equation
dA/dt
= −0.4,
A − 1250
log(|A − 1250|) = −0.4t + C1 ,
A(t) = 1250 − Ce−0.4t .
As t → ∞ we have A(t) → 1250.
1.3 # 1) linear,
2) nonlinear,
3) linear,
4) nonlinear,
5) nonlinear,
1.3 # 15
y 0 + 2y = 0.
Try y = ert to get
rert + 2ert = 0,
r + 2 = 0,
r = −2.
1.3 # 16
y 00 − y = 0.
Try y = ert to get
r2 ert − ert = 0,
r2 − 1 = 0,
r = ±1.
1.3 # 17
y 00 + y 0 − 6y = 0.
Try y = ert to get
r2 ert + rert − 6ert = 0,
r2 + r − 6 = 0,
r = 2, −3.
1.3 # 18
y (3) − 3y (2) + 2y (1) = 0.
Try y = ert to get
r3 ert − 3r2 ert + 2rert = 0,
r3 − 3r2 + 2r = 0,
4
r = 0, 2, 1.
6) linear