integration solutions

Math 331.5: Homework 2
Solutions
1. Find the general solution of each di erential equation and draw several representative solutions.
(i) dy
dt = 􀀀y + 5
First divide by the right hand side.
1
􀀀y + 5
dy
dt
= 1
Next integrate with respect to t.
(1)
Z
1
􀀀y + 5
dy
dt

dt =
Z
1 dt
Now use u-substitution for the integral on the left hand side. Let u = 􀀀y + 5 so du =
􀀀dy
dt dt. The integral becomes
Z
1
􀀀y + 5
dy
dt

dt = 􀀀
Z
1
u
du
= 􀀀ln juj + C
= 􀀀ln j5 􀀀 yj + C
Then (1) becomes
􀀀ln j5 􀀀 yj = t + C
ln j5 􀀀 yj = 􀀀t + C
To solve for y, we exponentiate both sides.
eln j5􀀀yj = e􀀀t+C
j5 􀀀 yj = e􀀀teC
5 􀀀 y = eCe􀀀t
5 􀀀 y = Ce􀀀t
􀀀y = Ce􀀀t 􀀀 5
y = Ce􀀀t + 5
(ii) dy
dt = 2y + 5
Following the method outlined above, the general solution is y = Ce2t 􀀀 5
2 .
2
-20 -15 -10 -5 0 5 10 15 20
-12
-8
-4
4
8
12
Figure 1. y = Ce􀀀t + 5
-10 -7.5 -5 -2.5 0 2.5 5 7.5 10
-5
-2.5
2.5
5
Figure 2. y = Ce2t 􀀀 5
2
3
2. Solve the following initial value problems and draw the solutions.
(i) dy
dt = 3y 􀀀 7, y(0) = 1.
First nd the general solution as in problem 1. Divide by the right hand side.
1
3y 􀀀 7
dy
dt
= 1
Next integrate with respect to t.
(2)
Z
1
3y 􀀀 7
dy
dt

dt =
Z
1 dt
Now use u-substitution for the integral on the left hand side. Let u = 3y􀀀7 so du = 3dy
dt dt.
The integral becomes
Z
1
3y 􀀀 7
dy
dt

dt =
1
3
Z
1
u
du
=
1
3
ln juj + C
=
1
3
ln j3y 􀀀 7j + C
Then (2) becomes
1
3
ln j3y 􀀀 7j = t + C
ln j3y 􀀀 7j = 3t + C
To solve for y, we exponentiate both sides.
eln j3y􀀀7j = e3t+C
j3y 􀀀 7j = e3teC
3y 􀀀 7 = eCe3t
3y 􀀀 7 = Ce3t
3y = Ce3t + 7
y = Ce3t +
7
3
Now use the initial condition, y(0) = 1, to nd C. Setting t = 0 and y = 1,
1 = Ce3(0) +
7
3
C = 􀀀
4
3
So, the desired solution is
y = 􀀀
4
3e3t +
7
3
(ii) dy
dt = y 􀀀 1, y(0) = 1.
Observe that y = 1 () dy
dt = 0, so y = 1 is an equilibrium solution. Then the desired
solution to the ODE is y(t) = 1.
4
-10 -7.5 -5 -2.5 0 2.5 5 7.5 10
-5
-2.5
2.5
5
Figure 3. y = 􀀀4
3 e3t + 7
3
-5 -4 -3 -2 -1 0 1 2 3 4 5
-3
-2
-1
1
2
3
Figure 4. y = 1
5
3. Suppose an object with mass 10 kg and air drag coe cient 2 kg/s is dropped from above the
ground.
(i) Find the time that must elapse for the object to reach 95% of its terminal velocity.
The di erential equation for velocity is dv
dt = 9:8 􀀀 1
5v. Since the object is dropped, the
initial condition is v(0) = 0. In class, we saw that the solution for this initial value problem
is
v(t) = 􀀀49e􀀀0:2t + 49:
Letting t ! 1 in the above, the terminal velocity is 49 m/s. So, 95% of the terminal
velocity is 46.55 m/s. We wish to nd the time T when v(T) = 46:55. Substituting into
the equation,
46:55 = 􀀀49e􀀀0:2T + 49
􀀀2:45 = 􀀀49e􀀀0:2T
0:05 = e􀀀0:2T
ln(0:05) = ln(e􀀀0:2T )
ln(0:05) = 􀀀0:2T
T = 􀀀5 ln(0:05) 14:98
Therefore, the object reaches 95% of its terminal velocity after about 15 seconds.
(ii) What is the minimal height so that the object hits the ground at this speed?
We need to nd the equation for the displacement, x. Since dx
dt = v, the di erential
equation for displacement is dx
dt = 􀀀49e􀀀0:2t + 49. The rate function depends only on the
independent variable, t, so we can integrate to nd x.
Z
dx
dt
dt =
Z
(􀀀49e􀀀0:2t + 49)dt
x(t) = 245e􀀀0:2t + 49t + C
To nd C, we use the initial condition x(0) = 0. Substituting t = 0 and x = 0, we get
C = 􀀀245. So, our equation is
x(t) = 245e􀀀0:2t + 49t 􀀀 245
Setting t = 􀀀5 ln(0:05),
x = 245eln(0:05) + 49(􀀀5 ln(0:05)) 􀀀 245 501
Therefore, the minimal height is about 501 meters above ground.
4. Calculate the impact velocity if you jump o a 7 foot wall (ignore air drag).
Ignoring air drag (setting
 = 0), the di erential equation for velocity is dv
dt = 32:2, where g =
32:2ft/s2 is the acceleration due to gravity. Integrating,
Z
dv
dt
dt =
Z
32:2 dt
v(t) = 32:2t + C
6
Now use the initial condition v(0) = 0 to nd C. Substituting t = 0 and v = 0 we nd C = 0.
Then the equation for velocity is v(t) = 32:2t.
To compute the impact velocity, we need to know the time of impact. So, we nd the equation for
displacement, x. The di erential equation for displacement is dx
dt = v(t) = 32:2t. Integrating,
Z
dx
dt
dt =
Z
32:2t dt
x(t) = 16:1t2 + C
The initial displacement is x(0) = 0, so substituting t = 0 and x = 0 in the equation, we get C = 0.
Then the equation for displacement is x(t) = 16:1t2. Now we need to nd T when we hit the
ground, i.e. x = 7.
7 = 16:1T2
T =
p
7=16:1
We want T > 0, so we take the positive square root, T =
p
7=16:1. Putting this value back into
the equation for velocity, the impact velocity is v = 32:2
p
7=16:1 21:2ft/s.
5. The population of the United States was 8.6 million in 1820 and 40 million in 1897. Suppose
that the population increases at a rate proportional to the current population, with growth rate r.
(i) Write a di erential equation for the population of the U.S.
Let P = P(t) be the population of the U.S. in millions t years after 1820. Let r be the
growth rate (constant of proportionality). Then the di erential equation for P is dP
dt = rP.
(ii) Find the solution which satis es the conditions above.
First nd the general solution as in problem 1. Divide by the right hand side.
1
rP
dP
dt
= 1
Next integrate with respect to t.
(3)
Z
1
rP
dP
dt

dt =
Z
1 dt
Now use u-substitution for the integral on the left hand side. Let u = rP so du = r dP
dt dt.
The integral becomes
Z
1
rP
dP
dt

dt =
1
r
Z
1
u
du
=
1
r
ln juj + C
=
1
r
ln jrPj + C
Then (3) becomes
1
r
ln jrPj = t + C
ln jrPj = rt + C
7
To solve for P, we exponentiate both sides.
eln jrPj = ert+C
jrPj = erteC
rP = eCert
rP = Cert
P = Cert
Now use the initial condition, P(0) = 8:6, to nd C. Setting t = 0 and P = 8:6,
8:6 = Cer(0)
C = 8:6
Then P(t) = 8:6ert. To nd r we use the information P(77) = 40.
40 = 8:6e77r
40=8:6 = e77r
ln(40=8:6) = ln(e77r)
ln(40=8:6) = 77r
r =
ln(40=8:6)
77 0:02
So, P(t) = 8:6e0:02t.
(iii) Use this model to calculate the population of the U.S. in 2003. If the actual population of
the U.S. in 2003 was 291 million, how good is this estimate?
P(183) = 8:6e0:02(183) 332
(iv) According to this model, when would the population of the U.S. double?
We wish to nd the time T when the population is twice its initial value, i.e. P(T) =
2P(0) = 17:2.
17:2 = 8:6e0:02T
2 = e0:02T
ln 2 = ln(e0:02T )
0:02T = ln 2
T = (ln 2)=0:02 35
Therefore, according to this model, the population of the US doubled by 1855.
8
6. Suppose that a building loses heat in accordance with Newton's law of cooling (see homework
1, problem 8) and that the rate constant is 0:15=hr. Assume that the interior temperature is 70 F
when the heating system fails. If the external temperature is 10 F, how long will it take for the
interior temperature to fall to 32 F?
Let T be the temperature of the building in F and t be time in hours. The surrounding temperature
is 10 F, so by Newton's law, the di erential equation for T is
dT
dt
= 0:15(10 􀀀 T):
The building starts o at 70 F before the heating system fails, so rst we need to solve the initial
value problem
(4)
(
dT
dt
= 0:15(10 􀀀 T)
T(0) = 70
Solving this IVP by the method outlined in problem 2, we nd T = 10 + 60e􀀀0:15t.
To nd how long it will take for the temperature to fall to 32 F, we set T = 32 and solve for
t.
32 = 10 + 60e􀀀0:15t
22 = 60e􀀀0:15t
22=60 = e􀀀0:15t
ln(22=60) = ln
􀀀
e􀀀0:15t
ln(22=60) = 􀀀0:15t
t =
ln(22=60)
􀀀0:15 6:69
So, the building reaches freezing in about 6.7 hours.
7. Consider an electric circuit containing a capacitor, resistor and battery. The charge Q(t) on the
capacitor satis es the equation
(5) R
dQ
dt
+ Q
C
= V
where R is the resistance, C is the capacitance, and V is the constant voltage supplied by the
battery.
(i) If Q(0) = 0, nd an expression for Q in terms of the constants R;C and V .
We need to solve the initial value problem
(6)
(
R
dQ
dt
+ Q
C
= V
Q(0) = 0
First nd the general solution. We will rewrite the equation as
dQ
dt
= V
R
􀀀
Q
CR
9
Next we divide by the right hand side.
1
V
R 􀀀 Q
CR
dQ
dt
= 1
Next integrate with respect to t.
(7)
Z 
1
V
R 􀀀 Q
CR
dQ
dt
!
dt =
Z
1 dt
Now use u-substitution for the integral on the left hand side. Let u = V
R 􀀀 Q
CR so du =
􀀀 1
CR
dQ
dt dt. The integral becomes
Z 
1
V
R 􀀀 Q
CR
dQ
dt
!
dt = 􀀀CR
Z
1
u
du
= 􀀀CRln juj + K
= 􀀀CRln

V
R
􀀀
Q
CR

+ K
Then (7) becomes
􀀀CRln

V
R
􀀀
Q
CR

= t + K
ln

V
R
􀀀
Q
CR

= 􀀀
1
CR
t + K
To solve for Q, we exponentiate both sides.
elnj V
R􀀀 Q
CRj = e􀀀 1
CRt+K

V
R
􀀀
Q
CR

= e􀀀 1
CRteK
V
R
􀀀
Q
CR
= eKe􀀀 1
CRt
V
R
􀀀
Q
CR
= Ke􀀀 1
CRt
􀀀
Q
CR
= Ke􀀀 1
CRt 􀀀
V
R
Q = Ke􀀀 1
CRt + CV
Now use the initial condition, Q(0) = 0, to nd K. Setting t = 0 and Q = 0,
0 = Ke0 + CV
K = 􀀀CV
Then Q(t) = 􀀀CV e􀀀 1
CRt + CV .
(ii) Find the limiting value QL that Q(t) approaches as t ! 1.
As t ! 1, e􀀀t=CR ! 0. So,
QL = lim
t!1
Q(t) = CV:
10
(iii) Suppose that at some time T the battery is removed and the circuit is closed again. If
Q(T) = QL, nd Q(t) and sketch its graph.
V is the voltage supplied by the battery, so if the battery is removed, the voltage is zero.
Then for t > T, the di erential equation for Q becomes
(8) R
dQ
dt
+ Q
C
= 0
At time T, Q(T) = QL = CV . So, we want to solve the IVP
(9)
(
R
dQ
dt
+ Q
C
= 0
Q(T) = CV
From part i, the general solution of (8) is
Q = Ke􀀀 1
CRt:
Using the initial condition Q(T) = CV , we nd
CV = Ke􀀀 1
CRT
or
K = CV e
1
CRT
Then for t > T,
Q(t) = CV e
1
CRT e􀀀 1
CRt = CV e(T􀀀t)=CR:
Now note that Q(t) = CV is an equilibrium solution of the original di erential equation
(5). So, if Q(T) = CV at time T, we must have had Q(t) = CV for all times t up to T.
Hence, the solution Q with this behavior is de ned by
Q(t) =

CV if t T
CV e(T􀀀t)=CR if t T
0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 4.4 4.8
0.25