MATH 3400
Name:.....................................................................
10/28/15
Sample solutions for Test 2
No books, notes, or calculators allowed. This test is worth 60 points,
which is 20% of your final grade. READ EACH QUESTION CAREFULLY.
SHOW ALL WORK. BE SURE TO DO THE QUESTIONS ON THE
BACK OF EACH PAGE AS WELL.
(1) (12 points) Find the general solution of the following DE using the
method of undetermined coefficients.
y 00 − 3y 0 + 2y = 2x + e2x .
The characteristic polynomial m2 − 3m + 2 = (m − 1)(m − 2)
has two real roots m1 = 1 and m2 = 2. It follows that the general
solution of the complementary homogeneous DE
y 00 − 3y 0 + 2y = 0 is given by yc (x) = c1 ex + c2 e2x .
The tentative form yptentative = Ax + B + Ce2x of the particular
solution contains a term Ce2x that is part of yc (x) and associated
with the root m2 = 2 of multiplicity k2 = 1. Thus we get the
following form of the particular solution:
yp = Ax + B + Cxe2x .
Differentiate it twice and plug the results back into the DE:
yp0 = A + Ce2x + 2Cxe2x .
yp00 = 4Ce2x + 4Cxe2x .
4Ce2x + 4Cxe2x − 3A + −3Ce2x −6Cxe2x + 2Ax + 2B + 2Cxe2x =
2x + e2x .
The terms containing xe2x cancel out, and by setting corresponding coefficients equal to each other we obtain the following system
of linear equations:
2A = 2
−3A + 2B = 0
C=1
The solution is A = 1,
B = 1.5,
C = 1.
It follows that the general solution of the nonhomogeneous DE is
y(x) = yc (x) + yp (x) = c1 ex + c2 e2x + x + 1.5 + xe2x .
1
2
(2) (12 points) Express the general solution of the following DE in terms
of integrals using the method of variation of parameters. Find
formulas for the integrands, but do not try to evaluate the integrals.
y 00 + 6y 0 + 25y = ln x
Since the characteristic polynomial m2 + 6m + 25 = (m + 3)2 + 16
has roots m1 = −3i + 4 and m2 = −3i − 4, we find that y1 (x) =
e−3x sin 4x and y2 (x) = e−3x cos 4x are two linearly independent solutions of the complementary homogeneous DE y 00 + 6y 0 + 25y = 0.
The Wronskian for these solutions is
e−3x sin 4x
e−3x cos 4x
4e−3x cos 4x − 3e−3x sin 4x −4e−3x sin 4x − 3e−3x cos 4x
W = det
= −4e−6x (sin2 4x + cos2 4x) = −4e−6x .
Substituting y1 (x) = e−3x sin 4x, y2 (x) = e−3x cos 4x, W = −4e−6x ,
and g(x) = ln x in the formula
R
R
dx + y2 (x) y1 (x)g(x)
dx we get:
y(x) = y1 (x) −y2 (x)g(x)
W
W
R e−3x sin 4x ln x
R
−e−3x cos 4x ln x
y(x) = e−3x sin 4x
dx+ e−3x cos 4x
dx
−4e−6x
−4e−6x
R
R
cos 4x ln x
sin 4x ln x
y(x) = e−3x sin 4x
dx + e−3x cos 4x
dx
4e−3x
−4e−3x
(3) (12 points) Find the following inverse Laplace transform
s−4
−1
L
.
s3 + 4s2 + 4s
A partial fraction decomposition gives:
A
B
C
A(s2 + 4s + 4) + B(s2 + 2s) + Cs
s−4
=
+
+
=
s3 + 4s2 + 4s
s
s + 2 (s + 2)2
s(s2 + 4s + 4)
(A + B)s2 + (4A + 2B + C)s + 4A
=
s3 + 4s2 + 4s)
A+B =0
4A + 2B + C = 1
4A = −4
A = −1, B = 1, C = 3.
n
n
o
n
o
o
s−4
1
1
−1
−1 1 +L−1
By linearity, L−1 s3 +4s
=
−L
+3L
.
2 +4s
s
s+2
(s+2)2
For the first two terms we use basic formulas, for the third term
the First Translation Theorem for F (s) = s12 and a = −2.
n
o
s−4
L−1 s3 +4s
= −1 + e−2t + 3te−2t .
2 +4s
3
(4) (12 points) Find the Laplace transform of the solution y(t) of the
following IVP. Do not attempt to simplify your result as much as
possible or to find the solution y(t) itself.
y 00 + 7y 0 + 10y = t4 + 5 cos 3t,
y(0) = −1,
y 0 (0) = 2.
Taking the Laplace transform of both sides of the DE, using linearity, and our formulas for L{t}, L{y 0 (t)}, L{y 00 (t)} we get:
L{y 00 (t) + 7y 0 (t) + 10y(t)} = L{t4 + 5 cos 3t}
L{y 00 (t)} + 7L{y 0 (t)} + 10L{y(t)} = L{t4 } + 5L{cos 3t}
s2 L{y(t)}−sy(0)−y 0 (0)+7sL{y(t)}−7y(0)+10L{y(t)} =
4!
s
+5 s2 +3
2
s5
Substituting the initial condition and solving for L{y(t)} we get
the formula for the Laplace transform of the solution of the IVP:
24
+ s25s+9
s5
5s
−s−5
s2 +9
(s2 + 7s + 10)L{y(t)} + s + 5 =
(s2 + 7s + 10)L{y(t)} = 24
+
s5
1
24
5s
L{y(t)} = s2 +7s+10
+
−
s
−
5
5
2
s
s +9
This is not required here, but one can simplify this result to:
L{y(t)} =
−s8 −5s7 +5s6 −45s5 +24s2 +216
s5 (s2 +9)(s2 +7s+10)
4
(5) (12 points) A cup of tea is brought into a room where it cools.
Let T (t) denote the temperature of the tea at time t, where time is
measured in minutes and temperature is measured in degrees Celsius.
Assume that the ambient temperature stays constant at 20 degrees
Celsius.
(a) Express the rate of change of the temperature of the tea as a DE
that embodies Newton’s Law of Cooling.
dT
dt
= k(T − 20).
(b) Express the general solution of the DE that you found in point (a)
by an explicit formula. You can use a formula that you have memorized, or you can derive the formula using the methods that you
have learned in this course.
T (t) = 20 + cekt .
(c) The DE that you found in part (a) contains a parameter that
embodies physical properties of the tea, the surrounding air, and the
tea cup. Assuming that it took the tea 15 minutes to cool from 90
degrees Celsius to 60 degrees Celsius, express the numerical value of
this parameter in terms of the measured values that are given here.
We can translate the given information, together with the result
of point (b), into T (0) = 90 and T (15) = 60.
T (0) = 90 = 20 + cek0 = 20 + c and T (15) = 60 = 20 + ce15k .
Thus c = 90 − 20 = 70, and by substituting this value in the
second equation we find that
60 = 20 + 70e15k , which in turn gives e15k = 40
70 .
4
After taking logarithms: 15k = ln 7 = ln 4 − ln 7.
We find that k =
ln 4−ln 7
.
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