AM33 Homework assignment # 2
1. (2.2.11)
xdx + ye−x dy = 0
y(0) = 1
Solution:
xex dx = −ydy
therefore
x
xe dx = −
−y 2
+c
2
ydy
xdex
= xex − ex dx
=
= xex − ex
so
y2
= c + ex − xex
2
since y(0) = 1, so c = − 21 , i.e,
y 2 = −1 + 2ex − 2xex
y=
√
2
2ex − 2xex − 1
the interval of the solution is defined on
2ex − 1 − 2xex ≥ 0
2. (2.2.12)
dr/dθ = r2 /θ,
Solution:
dr
dθ
=
r2
θ
1
r(1) = 2
therefore,
1
− + c = ln|θ|
r
because r(1) = 2, − 12 + c = 0,
so c =
1
2
1 1
− + = ln|θ|
r 2
so,
r=
the solution is defined on:
1
2
1
2
1
− ln|θ|
− ln|θ| ≥ 0
so, |θ| ≤ e1/2
3. (2.2.25)
y(0) = −1
y = 2 cos 2x/(3 + 2y),
Solution:
2 cos 2x
dy
=
dx
3 + 2y
(3 + 2y)dy = 2 cos 2xdx
(3 + 2y)dy = 2 cos 2xdx
3y + y 2 + c = sin 2x
because y(0) = 1
3y + y 2 + 2 = sin 2x
which yields,
y=
−3 +
√
4 sin 2x + 1
,
2
, for
sin 2x ≥ −
the only place where y may achieve its maximum is at
dy
dx
1
4
= 0, i.e.
2 cos x
=0
3 + 2y
, that is: cos 2x = 0. since our solution only exists when sin 2x ≥ −1/4,
so the only “possible” place is at sin 2x = 1, i.e. x = π4 + kπ, and by verify
the second derivative, we know that y does achieve its maximum there.
2
4. (2.2.28.a)
y =
ty(4 − y)
,
1+t
y(0) = y0 > 0
Solution:
ty(4 − y)
dy
=
dt
1+t
tdt
dy
=
y(4 − y)
1+t
dy
1 dy
(
+
) =
4 y
4−y
=
1
)dt
1+t
dt
dt −
1+t
(1 −
so,
lg |y| − lg |y − 4| − lg |y0 | + lg |y0 − 4| = 4(t − lg(1 + t))
so,
|
e−c+4t
y
|=
,
y−4
(1 + t)4
where c = lg |y0 − 4| − lg |y0 |
when y > 4; we can derive
4e4t+c
,
− (1 + t)4
y=
e4t+c
as t → ∞, y → 4.
when 0 < y < 4, we can derive
y=
4e4t+c
(1 + t)4 + e4t+c
, and when t → ∞, y → 4. ✷
5. (2.2.35)
x + 3y
dy
=
dx
x−y
1 + 3( xy )
dy
=
dx
1 − xy
let v = f racyx,in other word, y = vx
)
dy = vdx + xdv
3
1 + 3v
vdx + xdv
=
dx
1−v
)
)
1 + 3v
vdx + xdv
=
dx
1−v
)
xdv
dx
=
=
)
)
1 + 3v
−v
1−v
(v + 1)2
1−v
1−v
dx
dv =
(v + 1)2
x
dx
2dv
dv
=
−
2
(1 + v)
(1 + v)
x
−2(1 + v)−1 − lg |1 + v| = lg |x| + c
plug v = xy back,
−2(1 +
y −1
y
) − lg |1 + | = lg |x| + c
x
x
✷
6. (2.3.6)
Solution:
The Model:
dS
= rS
t
Solution with the initial value S0 :
S(t) = S0 ert
here r is the rate of return compound.
so set ert = 2, get
lg 2
r
lg 2
r=
t
T =
lg 2
0.07
lg 2
8
(b): T =
(c) r =
4
(1)
(2)
7. (2.3.12)
Solution:
Model:
dS(t)
t
= rS − 800(1 +
)
dt
120
Solve the equation:
ds − rsdt = −800dt −
20
tdt
3
Multiply both side with e−rt , get:
d(e−rt S) = −800e−rtdt −
20 −rt
te dt
3
Integration both side,
e
−rt
800
20
(1 − e−rt ) −
S − S(0) = −
r
3
t
ue−ru du
0
It turns out to be:
S = ert S(0) −
20 1
800 rt 800 20 t
20 1 rt
e +
+
+
−
e
2
r
r
3 r
3 r
3 r2
(3)
here, S(0) = 100, 000, r = 0.09.
one can use (3) numerically solve for (a) and (b). ✷.
8. Problem 15, page 59 The population of mosquitoes in a certain area increases at a rate proportional to the current population and, in the absence
of other factors, the population doubles each week. There are 200,000
mosquitoes in the area initially, and predators eat 20,000 mosquitoes/day.
Determine the population of mosquitoes in the area at any time
Solution
First we need to find the rate of growth of the population. We know that
, in the absence of other factors the population is: 20000ert where r is the
rate of growth. The population doubles each week. Then we have:
400000 = 200000e7r → r = ln(2)/7
Now we found the rate lets right the full ODE that takes into account the
predators:
dP
= P ln 2/7 − 20000
dt
P (0) = 200000
This is a linear DE and can be solved by using an integrating factor. The
solution is:
5
P (t) =
140000 log(2) t
140000
+ (200000 −
)e 7
log(2)
log(2)
9. Problem 20, page 60
Consider a lake of constant volume V containing at time t an amount Q(t)
of polutant, evenly distributed throughout the lake with concentration
c(t) = Q(t)/V . Assume that water containing concentration k of polutant
enters the lake at a rate r, and that water leaves the lake at the same rate.
Suppose that polutants are also added directly to the lake at a constant
rate P .
(a) If at time t = 0 the concentration of pollutant is c0 , find an expression for the concentration c(t) at any time. What is the limiting
concentration as t → ∞?
Solution The ODE that governs Q(t) is:
Q = rk + P −
Q
r,
V
Q(0) = c0 V
The solution of this initial value problem is:
Q(t) = V k +
P −rt
V
P + V (c0 − k − )e V
r
r
Thus c(t) is :
c(t) = k +
P
P −rt
+ (c0 − k − )e V
r
r
The limiting concentration is
lim t → ∞k +
P −rt
P
P
+ (c0 − k − )e V = k +
r
r
r
(b) If the addition of the pollutants to the lake is terminated ( k = 0
and P = 0 for t > 0) determine the time interval T that must elapse
before the concentration of pollutants is reduced to 50% of its original
value; 10% of its original value.
Solution In this case the ODE that governs Q(t) is :
Q = −r
Q
V
which has the solution:
−r
Q(t) = Q0 e V
t
Then we can find T corresponding to 50% reduction by solving:
−r
Q0 e V
T
=
Q0
2
→T =
V ln 2
r
Similarly for 10% reduction T can be found to be
6
V ln 10
r
(c) Table 2.3.2 contains data for several of the great lakes. Using these
data determine the time T necessary to reduce the contamination of
each of these lakes to 10% of their current value by stopping pollutants to enter the lake.
Solution For lake Superior V = 12200km3,r = 65.2km3 /year. Thus
ln 10
= 430.851812 years!
T = 12200
65.2
10. Problem 27, page 62: A body falling in a relatively dense fluid, is acted
on three forces: gravity, buoyant force, and a resistive force. The resistive
force is R = 6πµa|v| where v is the velocity of the object. Bouyant force
is the weight of the fluid displaced by the object.
Find the limiting velocity of a solid sphere of radius a and density ρ falling
freely in a medium of density ρ2 and coefficient of viscosity µ
Solution:
The mass of the solid sphere is volume x density :
Msphere =
4 3
πa ρ
3
Buoyant force on the sphere is: volume of the sphere x density of the
fluid x gravity. However since the buoyant force is always in the opposite
direction of gravity we add a minus sign in front:
4
B = − πa3 ρ2 g
3
Resistive force on the sphere is given above, however since resistance
should always be in the opposite direction of the motion we modify that
equation to:
R = −6πµav
Finally the weight of the sphere is :
Msphere g =
4 3
πa ρg
3
Hence total force acting on the sphere is:
F =
4 3
4
πa ρg − πa3 ρ2 g − 6πµav
3
3
By the Newton’s second law:
dv
4
4
Msphere = F = πa3 ρg − πa3 ρ2 g − 6πµav
dt
3
3
We can rewrite this equation as:
9µ
ρ − ρ2
dv
+
v=
dt
2ρa2
ρ
7
This is a linear first order equation and we can solve this by multiplying
2
with an integrating factor e9µt/2ρa
And the solution is:
v(t) = g(ρ − ρ2 )
−9µ
2a2
t
+ Ce 2ρa2
9µ
In this last equation if we let t → ∞ we see that the limiting velocity is:
g(ρ − ρ2 )
8
2a2
9µ