C Roettger, Fall 13
Math 267 - Exam 2A - solutions
Problems 1, 3 are worth 15 points. Problems 2, 4, 5 are worth 10 points for
60 points possible. As threshold for 100% I have decided to use 55 points
again.
Problem 1 a) Find the general solution of
x00 − 6x0 + 10x = 0
b) Find a particular solution of
x00 − 6x0 + 10x = 5 + et .
(1)
c) What is the general solution of Equation (1)?
Solution. a) We look at the auxiliary equation
(5+5+5=15 points)
r2 − 6r + 10 = 0
which has solutions r = 3 ± i, corresponding to the general solution
xc (t) = e3t (c1 cos t + c2 sin t).
b) There is no overlap of terms in xc with terms 5 and et on the right-hand
side, so we try
xp = A + Bet .
We substitute x0p and x00p into the original equation, getting
Bet − 6Bet + 10(A + Bet ) = 5 + et .
This means 10A = 5 and 5B = 1, so we put A = 1/2 and B = 1/5, and get
the particular solution
1 et
xp (t) = + .
2
5
c) We just add up,
x(t) = xc (t) + xp (t) = e3t (c1 cos t + c2 sin t) +
1 et
+ .
2
5
Problem 2 Set up a trial solution for (do NOT solve)
(10 points)
y 00 + 10y 0 + 16y = 3te−2t + cos(8t).
Solution.
We do need to check the auxiliary equation
r2 + 10r + 16 = (r + 2)(r + 4) = 0
so r = −2 and r = −4 are roots, and the complementary function is
yc = c1 e−2t + c2 e−4t .
We notice the overlap with e−2t on the right, so we need to have an extra
factor of t for that, putting
yp (t) = t(A + Bt)e−2t + C cos 8t + D sin 8t.
Problem 3 A shell of mass 0.5 kg is shot upward with an initial velocity
of 120 m/s, then allowed to fall under the influence of gravity. Assume that
the force of air resistance is −6v, where v is the velocity of the shell. If the
shell is 100 m above the ground at time t = 0, determine when it will hit
the ground, assuming that the exponential term has died off. Use the value
g = 10 N/kg for the magnitude of the acceleration due to gravity. (15 points)
Solution. You can think of Newton’s Second Law, ma = F with the total
force being gravity + friction. But in standard form, you will want to rewrite
this as
0.5v 0 + 6v = −0.5g,
or
v 0 + 12v = −10.
So the integrating factor is e12t , and
Z
5
12t
e v = −10 e12t dt = − e12t + C,
6
which we solve for v,
5
v = − + Ce−12t .
6
From v(0) = 120, we get
C = 120 +
5
725
=
.
6
6
Then the position is (note that we have to use a new variable instead of t,
because t is already in use as the upper limit of the integral)
Z t
5 725e−12s
5
725(e−12t − 1)
− +
ds = − t −
+ D.
x(t) =
6
6
6
72
0
From x(0) = 100, we determine D = 100. The exponential becomes negligible very soon, and if we neglect it, we can solve
5
725
0 = x(t) ≈ − t +
+ 100
6
72
which gives the time when the shell hits the ground as
t=
1585
≈ 132.1
12
(in seconds).
Problem 4 A cart of mass 4 kg is connected to the wall by a spring with
stiffness k = 36 N/m . It moves horizontally without friction.
a) Let the displacement of the cart from rest position be x(t). Determine the
equation of motion for x(t).
b) Given that x(0) = 0 and x0 (0) = 5, find x(t) for arbitrary time t.
(5+5=10 points)
Solution.
a) The equation is simply
4x00 + 36x = 0.
b) The auxiliary equation
4r2 + 36 = 0
has roots r = ±3i, so a general solution is
x(t) = c1 cos 3t + c2 sin 3t.
The initial conditions tell us c1 = 0 and c2 = 5/3.
Problem 5 Find the general solution of
(10 points)
t2 x00 − 5tx0 + 5x = 0.
Solution.
equation
This is a Cauchy-Euler equation. We form the characteristic
r(r − 1) − 5r + 5 = 0
which has roots r = 1 and r = 5, leading to solutions t1 and t5 . The general
solution is then
x(t) = c1 t + c2 t5 .