Download Math 246 - Exam 2 Solutions

Math 246 - Exam 2 Solutions
Monday, June 23rd, 2014
(1) In the following problem D =
d
.
dt
(a) Give a general solution to the following differential equation
D3 (D2 + 9)2 (D2 − 2D − 7)2 (D2 + 2D + 10)y = 0.
(b) What is the order of the differential equation in (a)?
Solution:
(a) The characteristic polynomial is
p(z) = z 3 (z 2 + 9)2 (z 2 − 2z − 7)2 (z 2 + 2z + 10)
√
√
√
√
which has roots 0, 0, 0, 3i, 3i, −3i, −3i, 1 + 2 2, 1 + 2 2, 1 − 2 2, 1 − 2 2, 1 +
3i, 1 − 3i. Therefore a general solution is given by
Y (t) = c1 + c2 t + c3 t2 + c4 cos 3t + c5 sin 3t + c6 t cos 3t + c7 t sin 3t
√
√
√
+ c8 e(1+2 2)t + c9 e(1−2 2)t + c10 te(1+2
+ c12 et cos 3t + c13 et sin 3t.
2)t
+ c11 te(1−2
√
2)t
(b) The order is 13.
(2) Find the natural fundamental set of solutions (associated with t = 0) for each of the
following equations:
(a) y 00 − 4y 0 + 13y = 0
(b) y 00 + 8y 0 + 16y = 0
Solution:
(a) The characteristic polynomial is p(z) = z 2 − 4z + 13, which has roots 2 ± 3i.
Therefore a general solution is
Y (t) = c1 e2t cos 3t + c2 e2t sin 3t.
We apply the general initial conditions Y (0) = y0 , Y 0 (0) = y1 . Which gives
(
y 0 = c1
y1 = 2c1 + 3c2 .
1
This has a solution c1 = y0 and c2 = (y1 − 2y0 )/3. Substituting this into the
general solution and grouping by y0 and y1 gives
1
2 2t
2t
Y (t) = y0 e cos 3t − e sin 3t + y1 e2t sin 3t.
3
3
Therefore the natural fundamental solutions are
1
N0 (t) = e2t cos 3t − e2t sin 3t,
3
1
N1 (t) = e2t sin 3t.
3
(b) The characteristic polynomial is p(z) = z 2 +8z +16 = (z +4)2 , which has a double
root at −4. The general solution is given by
Y (t) = c1 e−4t + c2 te−4t .
We apply the general initial conditions Y (0) = y0 , Y 0 (0) = y1 . Which gives
(
y0 = c1
y1 = −4c1 + c2 .
This has solutions c1 = y0 and c2 = y1 + 4y0 . Substituting this into the general
solutions and grouping by y0 and y1 gives
Y (t) = y0 (e−4t + 4te−4t ) + y1 te−4t .
The natural fundamental solutions are therefore
N0 (t) = e−4t + 4te−4t ,
N1 (t) = te−4t .
(3) Find a particular solution to the following differential equation
d2 y
+ 4y = sec (2t).
dt2
Solution 1: Since the forcing is not in characteristic form we can solve this by Green
function method. The characteristic polynomial is p(z) = z 2 + 4, with roots ±2i.
Therefore the Green function can be written as a general homogeneous solution,
g(t) = c1 cos 2t + c2 sin 2t.
Apply the conditions g(0) = 0, g 0 (0) = 1, we find
(
0 = c1
1 = 2c2
2
Therefore c1 = 0 and c2 =
1
2
and the Green function is
g(t) =
1
sin 2t.
2
A particular solution is then given by
Z
1 t
Yp (t) =
sin 2(t − s) sec 2s ds
2 0
Z t
Z t
cos 2s
sin 2s
1
1
= sin 2t
ds − cos 2t
ds
2
2
0 cos 2s
0 cos 2s
1
1
= t sin 2t + cos 2t (log (| cos 2t|)) .
2
4
(4) Suppose that a mass of weight 2 kg in a viscous fluid hangs at rest from a spring and
streches the spring by 0.2 m due to the force of gravity (you may assume the spring
has zero length when unstretched). The viscous fluid also pushes on the mass with a
force of 48 N (N = kg·m/s2 ) when the mass is moving 2 m/s. Assume that the spring
force is proportional to the displacement and the damping force is proportional to the
velocity. There is no external forcing. At t = 0 the mass is then displaced downward
from its rest position 1 m and kicked downwards with a velocity of 1.5 m/s.
(a) What is the spring constant k and damping coefficient γ associated with this
system? Recall that the acceleration due to gravity is g = 9.8 m/s2 .
(b) Write down (but do not solve) the initial value problem (IVP) which describes
the displacement of the mass from its rest position.
(c) Is the motion under-damped, over-damped, or critically damped? State why.
Solution:
(a) We know that at rest the spring force and the force of gravity are equal and
opposite mg = −kyr , where yr = −0.2 is the rest displacement. This gives
k=−
2(9.8)
mg
=
= 98 N/m.
yr
0.2
The drag force is given by −48 = Fdrag = −γ dy
= −2γ. Therefore
dt
γ=
48
= 24 kg/s.
2
(b) The initial value problem is
2
d2 h
dh
+ 24
+ 98h = 0,
2
dt
dt
with initial conditions h(0) = −1, h0 (0) = −1.5.
3
(c) We see that
r
γ
k
µ=
= 6, ω0 =
=7
2m
m
Since ω0 > µ, the oscillations dominate the motion and the system is underdamped.
(5) Suppose that x and x log x solve the following homogeneous differential equation (you
don’t need to check this)
x2
d2 y
dy
+ y = 0,
−x
2
dx
dx
for x > 0.
(a) Show that these solutions are linearly independent for x > 0.
(b) Solve the initial value problem
x2
d2 y
dy
−x
+ y = x,
2
dx
dx
y(1) = 1, y 0 (1) = 2.
Solution:
(a) To see that the solutions are linearly independent, we check the Wronskian,
x x log x
W [x, x log x](x) = det
= x 6= 0, when x > 0.
1 log x + 1
Therefore they are linearly independent for x > 0.
(b) Write the equation in normal form
d2 y
1 dy
1
1
−
+ 2y =
2
dx
x dx x
x
To construct a general solution we use variation of parameters and assume that
the general solution is given by
Y (x) = u1 (x)x + u2 (x)x log x.
We impose the conditions
(
u01 x + u02 x log x = 0
u01 1 + u02 (log x + 1) =
This has solutions
u01 (x) = −
Therefore
Z
u1 (x) = −
log x
,
x
1
x
u02 (x) =
1
.
x
log x
1
dx + c1 = − (log x)2 + c1
x
2
4
and
Z
u2 (x) =
1
dx + c2 = log x + c2 .
x
Therefore the general solution is
1
Y (t) = c1 x + c2 x log x + x(log x)2 ,
2
x > 0.
Imposing the initial conditions
(
1 = c1
2 = c1 + c2 ,
gives c1 = 1, c2 = 1. Therefore the solution to the IVP is
1
Y (t) = x + x log x + x(log x)2 .
2
(6) Consider the following MATLAB code
ode = ‘D2y + 3*Dy + 2*y = 2*t + 5’;
dsolve(ode, ‘y(0) = 1’, ‘Dy(0) = 2’,‘t’)
(a) What is the initial value problem that is being solved?
(b) Give the output of the dsolve command. You do not need to write it in MATLAB
output format.
Solution:
(a) The initial value problem being solved is
d2 y
dy
+ 3 + 2y = 2t + 5,
2
dt
dt
y(0) = 1, y 0 (0) = 2.
(b) To solve this problem we note that the characteristic polynomial is p(z) = z 2 +3z+
2, which has roots −1, −2. The forcing is in characteristic form with µ + iν = 0,
d = 1, m = 0. We will use key identity evaluations
L (1) = p(0) = 2
L (t) = p(0)t + p0 (0) = 2t + 3.
We can write the second equations as
L (t) + 2 = 2t + 5,
5
However since the first equation tells us that
L (1) = 2,
we find
L (t + 1) = 2t + 5.
Therefore a general solutions is
Y (t) = c1 e−t + c2 e−2t + t + 1.
Applying the initial conditions
(
1 = c1 + c2 + 1
2 = −c1 − 2c2 + 1,
which has solution c1 = −1, c2 = 1. Therefore
Y (t) = −e−t + e−2t + t + 1
(7) Give a general (real valued) solution to the following differential equation
y 00 + 4y 0 + 5y = 2e−2t cos t.
Solution: The characteristic polynomial is p(z) = z 2 + 4z + 5, which has roots −2 ± i.
The forcing is characteristic with µ + iν = −2 + i, d = 0, m = 1. We use key identity
evaluations
L(te(−2+i)t ) = p(−2 + i)te(−2+i)t + p0 (−2 + i)e(−2+i)t = 2ie(−2+i)t
Muliplying the equation by −i (or equivalently dividing by i) gives
L(−ite(−2+i)t ) = 2e(−2+i)t .
Taking the Real part of both sides gives
L Re −ite(−2+i)t
= 2e−2t cos t.
Therefore a particular solution is
Yp (t) = Re −ite(−2+i)t = te−2t sin t.
The general solution is
Y (t) = c1 e−2t cos t + c2 e−2t sin t + te−2t sin t.
6
(8) Extra credit Suppose that one hundred ants break away from, their current colony to
start a new one. Assume that their population P grows according to a logistic model
dP
= r(1 − P/K)P.
dt
The new colony has a carrying capacity of one million ants. Suppose that once the
colony is started t = 0 it is measured that the one hundred ants can produce about 10
ants per day.
(a) What are the parameters r and K? You can round to the nearest decimal place.
(b) Solve the logistic equation with P (0) = 100 for any r and K.
(c) What is the population size after 10 weeks for the r and K computed in (a)?
Keep only one significant figure.
Solution:
(a) r = 10/(100 ∗ (1 − 10−4 ) ≈ .1 (days)−1 . K = 1 × 106 ants.
(b) This equation is autonomous so we write
Z
P
1
dz
100 r(1 − z/K)z
Z P
1
1
+
dz
=
rz
100 r(K − z)
1
P (K − 100)
= log
r
100(K − P )
t=
We can take the exponential of both sides,
P (K − 100)
= ert ,
100(K − P )
We can then solve for P to obtain
100Kert
K + 100(ert − 1)
100K
.
=
−rt
Ke + 100(1 − e−rt )
P =
(c) 10 weeks = 70 days. So
P (70) =
e−7
100
≈ 9.8104 .
+ 10−4 (1 − e−7 )
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