Review Problems for Exam 1, Math 308-503, Fall 2014
Notes: The exam will take place in class on Friday, October 10. No calculators, notes, textbooks
or other aids will be allowed. You will have the entire class period to work on the exam, but
exams will be collected promptly at the end of the period, so please be sure to arrive on time
to ensure that you have sufficient time to complete the exam. There will be no makeup exam
unless it is for a pre-excused absence or a legitimate emergency. The exam will cover material
included on the first five homework assignments. No explicit questions will be asked concerning
material from Chapter 1, so the covered sections are 2.1-2.7 and 3.1-3.8. The exam will consist
of approximately five questions, each having length similar to that of an average homework
question. Homework questions will also serve as an excellent study guide.
You will be expected to know integrals, derivatives, and basic properties of standard functions:
polynomials and powers, exponential (ex ), logarithmic (ln x), and basic trig (sin x, cos x, tan x).
In addition, you should be able to use basic differentiation rules (product, quotient, chain rule)
and “undo” these rules in order to calculate integrals.
In addition to solving the various types of ODEs covered in the last several weeks, you may also
be asked questions that will require you to recall, understand, and apply the hypotheses and
conclusions of relevant theorems. These are:
1. Theorem 2.4.1 (existence and uniqueness of solutions to linear first-order IVP’s)
2. Theorem 2.4.2 (existence and uniqueness of solutions to general first-order IVP’s)
You don’t necessarily need to memorize these theorems word for word, but you do need to be
able to cite them to answer questions (examples are below). I will give you the above list on the
exam.
Expectations for Sections 3.7 and 3.8 are as follows. You should know the ODE that models a
spring-mass system, be able to obtain the needed parameters (m, γ, k), and solve the homogeneous problem. You should also be able to distinguish damped versus undamped and forced
versus unforced vibrations, and know what the transient and steady-state solutions to a forced
vibration are along with their behavior as t → ∞.
Euler’s method. You will not be allowed to use calculators on the exam, so I won’t expect
you to take large number of timesteps using Euler’s method. You should be able to:
1. Take one or two timesteps by hand for simple ODE’s (I’ll make sure the arithmetic isn’t too
complicated).
2. Draw a sketch illustrating how Euler’s method works.
3. Explain why Euler’s can give spurious results in some situations, as for example in homework
problem 2.7.15.
Study hint: Calculus students sometimes feel that they study hard and know the material
well, but can’t solve problems under time pressure when taking an exam. My advice is to first
do some practice problems at a more leisurely pace to help you understand the material, then do
some problems with a timer set for 10 minutes (that’s about how long you’ll have to complete
each problem on the exam).
1. For each of the following problems:
i) Identify the equation as linear or separable.
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ii) Find the general solution (if no initial condition is given) or the solution to the initial value
problem (if an initial condition is given). If the equation is separable and no initial condition is
given, the general solution may be represented implicitly.
a) y 0 + 2ty = t.
2
Answer: Linear; y(t) = 1/2 + Ce−t .
sin x
b) y 0 = y1+y
2 , y(0) = 1.
2
Answer: Separable; y2 + ln|y| = − cos x + 32 . Note that no explicit solution formula is possible
here; the instructions may have been a little confusing on this point.
−x
dy
c) dx
= x−e
y+ey .
2
2
Answer: Separable; y2 + ey = x2 + e−x + C.
d) y 0 = cos t − (cos t)y.
Answer: Linear; y(t) = 1 + Ce− sin t .
e) 2x + y cos xy + (2y + x cos xy)y 0 = 0.
Answer: Exact; x2 + y 2 + sin(xy) = C.
2. Solve problems 5a and 11a in Section 2.3 of the text.
Answers: See back of book.
3. Consider the problem y 0 + p(t)y = g(t), y(1) = 3. State a condition on p and g which will
ensure that the above problem has a unique solution on the interval −1 < t < 4. Please cite an
appropriate theorem as justification for your answer.
Answer: Note that the given ODE is linear and the initial time 1 lies in the interval −1 < t < 4.
According to Theorem 2.4.1 (existence and uniqueness of solutions to linear equations), the
above linear equation is guaranteed to possess a unique solution for −1 < t < 4 if p(t) and g(t)
are both continuous for −1 < t < 4.
4. Write down all initial values (t0 , y0 ) for which you can guarantee that the initial value problem
3t2 +1
y 0 = y−4t
2 , y(t0 ) = y0 has a unique solution. Please cite an appropriate theorem as justification
for your answer.
3t2 +1
Solution: This is a nonlinear equation with f (t, y) = y−4t
2 , so we must apply Theorem 2.4.2.
2
3t +1
Note that f (t, y) is NOT continuous when (t, y) lies on the parabola y = 4t2 . Also, fy = − (y−4t
2 )2
is discontinuous at the same points as f (t, y). Thus according to Theorem 2.4.2, the IVP will
have a unique solution whenever y0 6= 4t20 .
5. (a) Use Euler’s method with h = .1 to approximate y(0.1), where y solves the initial value
1
, y(0) = 0.
problem y 0 = sin(y)+0.5
1
Answer: y0 = t0 = 0 and f (t, y) = sin(y)+0.5
. Thus the Euler approximation to y(0.1) with
1
h = 0.1 is y1 = y0 + hf (t0 , y0 ) = 0 + 0.1 sin(0)+0.5 = 0.1 × 2 = 0.2.
(b) Now suppose you use Euler’s method with step size h (taken to be as small as you like!) to
approximate y for 0 ≤ t ≤ T . Do you expect the approximate solution to be valid no matter
what you choose T to be, or will there be a problem at some point?
Answer: We expect y to have a vertical asymptote when sin y = −0.5, which occurs at an
infinite number of points. If y approaches such a value, the numerical method will not detect
the impending vertical asymptote and will return an incorrect approximate solution.
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6. Find the Wronskian of two solutions of the equation
t2 y 00 − ty 0 + (cos2 t + esin t )y = 0.
assuming that W (2) = 1. Note: You should also review the homework problems from Section
3.2, as there are other skills in that section that you may be test on. These for example include
computing the Wronskian of two given functions and using that information to check whether
they form a fundamental solution set for a given ODE, etc.
Solution: Rewriting in the form y 00 + p(t)y 0 + q(t)y = 0 yields y 00 − 1t y 0 +
R
R
1
dt
t
esin t
y
t2
= 0. Thus
= Ct. This formula is valid for
p(t) =
and by Theorem 3.3.2, W = Ce− p(t)dt = Ce
any two solutions. The condition W (2) = 1 implies that C = 1/2, so W (t) = t/2.
− 1t ,
7. Find the general solution to the following two equations, and describe the behavior of the
solutions as t → ∞:
a) y 00 + y 0 − 6y = 0
Solution: This equation has characteristic equation r2 + r − 6 = 0, which in turn has roots −3
and 2. Thus the general solution is y(t) = c1 e−3t + c2 e2t . The solution behavior depends on c1
and c2 . If c2 6= 0, then grows without bound as t → ∞. If c2 = 0, then y(t) → 0 as t → ∞.
b) y 00 − 2y 0 + 2y = 0.
√
Solution: The roots of the characteristic equation r2 − 2r + 2 = 0 are 2± 24−8 = 1 ± i. Thus the
general solution is y(t) = et (c1 sin t + c2 cos t). The solution oscillates with amplitude increasing
to ∞ as t → ∞.
8. Find the solution to the given initial value problem:
y 00 + 4y 0 + 4y = 0, y(0) = 0, y 0 (0) = 2.
Solution: The characteristic equation r2 + 4r + 4 = 0 has the double root −2, so the general
solution is y(t) = c1 e−2t + c2 te−2t . y(0) = c1 = 0, so y(t) = c2 te−2t and y 0 = c2 (e−2t − 2te−2t ).
Then 2 = y 0 (0) = c2 , so y(t) = 2te−2t .
9. Find the general solution to the following equations:
a) y 00 + y = e2t + t2 .
Solution: We use the method of undetermined coefficients. We first solve the homogeneous
equation y 00 + y = 0, which has general solution c1 cos t + c2 sin t. We then find specific solutions
to the nonhomogeneous subproblems Y100 + Y1 = e2t and Y200 + Y2 = t2 . To solve the first
subproblem, we guess Y1 = Ce2t and calculate that Y100 + Y1 = 4Ce2t + Ce2t = e2t , so C = 1/5
and Y1 = 51 e2t . For the second subproblem, we guess that Y2 (t) = A0 t2 + A1 t + A2 and compute
that Y200 + Y2 = 2A0 + (A0 t2 + A1 t + A2 ) = t2 . Thus A0 t2 + A1 t + (2A0 + A2 ) = t2 , so A0 = 1,
A1 = 0, and 2A0 + A2 = 0 implies that A2 = −2. Thus Y2 (t) = t2 − 2. Collecting all terms, the
general solution to the original nonhomogeneous equation is y(t) = 51 e2t +t2 −2+c1 cos t+c2 sin t.
t
e
b) y 00 − 2y 0 + y = 1+t
2.
Solution: We use the method of variation of parameters. The homogeneous equation y 00 − 2y 0 +
y = 0 has fundamental solution set y1 = et , y2 = tet , and W (y1 , y2 )(t) = e2t . Then a specific
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solution to the nonhomogeneous equation is
Z
Z
tet et
et et
t
Y (t) = −et
dt
+
te
e2t (1 + t2 )
e2t (1 + t2 )
Z
Z
1
t
dt + tet
dt
= −et
2
1+t
1 + t2
1
= −et ln(1 + t2 ) + tet tan−1 t
2
t
e
= − ln(1 + t2 ) + tet tan−1 t.
2
Thus the general solution is y(t) = c1 et + c2 tet −
et
2
ln(1 + t2 ) + tet tan−1 t.
10. A spring-mass system has a spring constant of 3 N/m. A mass of 2 kg is attached to the
spring, and the motion takes place in a viscous fluid that offers a resistance of 1 N sec/m. The
spring is set in motion by displacing it 25 cm in the positive direction and releasing it with no
initial velocity.
a) Assuming that no forcing takes place, write down an initial value problem describing the
displacement u(t) of the mass.
Solution: All units are compatible except for the initial displacement, which we rewrite as
25cm = 41 m. Our initial value problem is 2u00 + u0 + 3u = 0, u(0) = 1/4, u0 (0) = 0.
b) Without first solving the equation, describe what will happen to the displacement u(t) as
t → ∞. Explain your answer by referring to the characteristics of the physical situation being
modeled.
Solution: The above system is a damped oscillator with no forcing term, so u(t) → 0 as t → ∞.
In slightly different terms, the energy put into the system by the initial displacement of the mass
will be dissipated through damping, and since no further energy is infused into the system, the
motion will die out over time.
c) Find u(t).
√
Solution: The characteristic equation is 2r2 + r + 3 = 0, which has roots − 14 ± i 423 , i.e., λ = − 14
√
√
√
23
−t/4 (c cos 23 t + c sin 23 t). u(0) = 1/4 implies that c = 1 .
1
2
1√
4 . Thus u(t) =√e
4
4
√
√
√ 4
c1 23
c2 23
23
23
23
1
1
−t/4
0
0
u (t) = − 4 u(t) + e
(− 4 sin 4 t + 4 cos 4 t), so 0 = u (0) = − 4 u(0) + c2 4 =
√
√
1
23
− 16 + c2 4 . Thus c2 = 4√123 , and u(t) = e−t/4 ( 14 cos 423 t + 4√123 sin 423 t).
and µ =
d) Determine the steady-state response of the system when an external force f (t) = 3 cos(t)N
is also applied to the mass.
Solution: We use the method of undetermined coefficients to find a particular solution to 2u00 +
u0 + 3u = 3 cos t. Setting u = A cos t + B sin t, we have u0 = −A sin t + B cos t and u00 =
−A cos t−B sin t. Thus 2u00 +u0 +3u = −2(A cos t+B sin t)−A sin t+B cos t+3(A cos t+B sin t),
so −2A + B + 3A = 3 and −2B − A + 3B = 0, that is, A + B = 3 and −A + B = 0. Thus
A = B, and A = 32 = B, so the steady-state response is U (t) = 32 (cos t + sin t).
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