Download Exam 1 Sol

APPM 2360 — Section Exam 1
Wednesday September 24, 7:00pm – 8:30pm, 2014
ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your student ID number, (3)
recitation section (4) your instructor’s name, and (5) a grading table. Text books, class notes, and
calculators are NOT permitted. A one-page one-sided crib sheet is allowed.
Problems 1d and 4 are given a small number of points in relation to the amount of work involved.
You may want to save these for last.
Statistics: Mean = 64.4, Median = 65, Max = 99.
Problem 1: (28p) These questions are worth 7 points each. No justification is necessary. If you do
submit your work, then please box your final answer and know that the work will not be graded.
(a) Which of the following equations are linear differential equations in y = y(t):
(i) d2 y/dt2 + 2 sin(t) dy/dt + y = 0.
(ii) d2 y/dt2 + 2 dy/dt + y − sin(t) = 0.
(iii) d2 y/dt2 + 2 dy/dt + sin(y) = 0.
(iv) d2 y/dt2 + 2 sin(y) dy/dt + y = 0.
(b) Consider the differential equation
(
dy/dt = 1/(t + y 2 ),
y(0) = 2.
Specify the approximation to y(1/2) that you obtain from a single step of the Euler method.
(c) Specify the general solution to
y 00 (t) + 4 y(t) = 1.
Hint: The general solution to the equation z 00 (t) + 4 z(t) = 0 is z(t) = A cos(2t) + B sin(2t),
where A and B are arbitrary constants.
(d) Find one solution to
y 0 (t) cos(t) − y(t) sin(t) = cos (2t).
Solution:
(a) (i) and (ii) are linear.
(b) We have h = 1/2 and f (t0 , y0 ) = 1/(0 + 22 ) = 1/4, so y(1/2) ≈ y0 + h f (t0 , y0 ) = 2 + 1/8 = 17/8.
(c) For instance, y(t) = A cos(2t) + B sin(2t) + 1/4.
(d) For instance, y(t) = sin(t).
Note 1: Problem (c) was intentionally hard.
Note 2: Problem (d) can be done systematically by first dividing through by cos(t) to get it
to standard form, y 0 − tan(t) y = 2 sin(t), and then using, e.g., the method of an integrating
factor. However, there is a quick solution if you simply observe that y 0 (t) cos(t) − y(t) sin(t) =
d
1
dt (cos(t) y(t)). Then upon integration, you get cos(t) y(t) = 2 sin(2t). Now use that sin(2t) =
2 sin(t) cos(t), and divide through by cos(t).
Problem 2: (20p) Consider the differential equation
2
dy/dx = x y 2 e−x .
(1)
(a) (5p) For what initial conditions y(x0 ) = y0 does a (possibly local) solution exist? Why?
(b) (5p) If the solution exists, is it unique? Why or why not?
(c) (10p) Solve the initial value problem for eq. (1) with y(0) = 4.
Solution:
(a) The solution exists for any initial condition because the right hand side of the DE f (x, y) =
2
xy 2 e−x is continuous for all x and y so we can apply the first part of Picard’s Theorem.
(b) The solution is unique for any initial condition because the partial derivative of the DE
2
right hand side ∂f /∂y = 2xye−x is continuous for all x and y so we can apply the second
part of Picard’s Theorem.
(c) The equation is separable so
Z
Z
dy
dy
2
2 −x2
= xy e
⇒
= xe−x dx
⇒
dx
y2
Applying the initial condition y(0) = 4
1
1
1
− =− +C ⇒ C = .
4
2
4
Then the solution is
4
y(x) = −x2
.
2e
−1
−
1
1
2
= − e−x + C
y
2
Problem 3: (20p) Consider a country whose population at time t is A(t) for t ≥ 0. At time t = 0,
you know that the population is precisely one million. Each year, the mortality rate is twenty out
of every thousand, and the birth rate is ten for every thousand. The country also experiences a net
migration of s people every year (s can be either positive or negative).
(a) (8p) Write down a differential equation that determines A(t).
(b) (6p) Solve the differential equation.
(c) (6p) Specify the long term behavior of the population of the country for different values of s.
Solution:
(a) Use “persons” as the unit for A, and “years” as the unit for t. The change dA in a time period
dt is then
20
10
A(t) dt −
A(t) dt + s dt.
dA =
1000
1000
Setting r = 0.01 and A0 = 1 000 000 we find that the equation is

 dA = − r A + s,
dt
 A(0) = A .
0
(b) The solution is
s −rt s
A(t) = A0 −
e
+ .
r
r
(c)
Case 1 — s ≥ 0: In this regime, we find that A(t) → s/r as t → ∞.
Case 2 — s < 0: In this regime, we formally find that A(t) → s/r as t → ∞. In practice, A(t)
cannot go negative, so what happens is that the population dies out in finite time.
Note: For the case s < 0, the “extinction time” is text =
specify this to get full credit.
1
r
log 1 −
rA0
s
. It was not necessary to
Problem 4: (20p) Consider the direction field
1.5
1
y
0.5
0
−0.5
−1
−1.5
−4
−3
−2
−1
0
1
2
3
4
t
for a differential equation dy/dt = f (t, y). For all questions pertaining to this problem, utilize only
the information pictured in this portion of the direction field.
(a) (3p) Based on the domain and range pictured, is the DE autonomous? separable? linear?
(b) (7p) Identify any equilibria of the DE and classify their stability.
(c) (7p) Given an initial condition y(0) = y0 ∈ [−1.5, 1], determine the long term behavior of
the solution by assuming that the direction field pictured extends unchanged for all t. Your
answer will depend on y0 .
(d) (3p) Write down a DE that exhibits the properties identified in (a)-(c) and could correspond
to the direction field given. There are many possible choices.
Solution:
(a) The DE is autonomous because the direction field is independent of t. This also implies
that the DE is separable. The DE is nonlinear because the variation in the slopes with y
is nonlinear.
(b) Equilibria are {−1, 0, 1}. -1 is stable. 0 is unstable. 1 is semistable.
(c) See table
y0
limt→∞ y(t)
(−1.5, 0)
-1
0
0
(0, 1]
1
(d) For example y 0 = y(y + 1)(y − 1)2 .
Problem 5: (12p) Find the general solution to the equation
y 0 (t) +
y(t)
1
=p
t
t y(t)
in the regime t > 0 and y > 0. Hint: Try to change the dependent variable to v(t) = y(t)
3/2
.
Solution:
Observing that y = v 2/3 , we find
2 −1/3 0
v
v.
3
The differential equation, expressed in v, is then
1
2 −1/3 0 1 2/3
v
v + v
= 1/2 v −1/3 ,
3
t
t
which simplifies to
31
3 1
v0 +
v = 1/2 .
2t
2t
The integrating factor is
R 31
3
3/2
u(t) = e 2 t = e 2 log t = elog t = t3/2 ,
so the equation can be written
3
d 3/2
t v(t) = t.
dt
2
Integrating, we get
3
t3/2 v(t) = t2 + C,
4
which we rewrite as
3
C
v(t) = t1/2 + 3/2 .
4
t
It only remains to convert back to y:
3 1/2
C 2/3
y(t) =
t + 3/2
.
4
t
y0 =
Note 1: Before you use the technique of an integrating factor, it is important to bring the equation
√
back to its standard form y 0 + p y = q. In other words, for the equation (2/3)y 0 + (1/t) y = 1/ t,
we have p = (3/2t) and the integrating factor is t3/2 . It is not the case that p(t) = 1/t!
Note 2: If you believe that is is generally the case that (a + b)2/3 = a2/3 + b2/3 , then please please
review the rules for exponents!