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MT305.01: Advanced Calculus for Science Majors
Examination 3
Answers
1.
(25 points) Solve the differential equation


0
d y
+ 4y = 1

dt2

0
2
0≤t<π
π ≤ t < 2π
2π ≤ t
with initial conditions y(0) = 1 and y 0 (0) = 1.
Answer: Rewrite the equation in the form y 00 + 4y = U (t − π) − U (t − 2π). Taking the Laplace transform
now yields:
s2 Y − s − 1 + 4Y =
e−πs − e−2πs
s
which can be rearranged to give
Y =
e−πs − e−2πs
s+1
+
s2 + 4
s(s2 + 4)
Partial fractions yields
1
1
=
2
s(s + 4)
4
and so
s+1
Y = 2
+
s +4
1
s
− 2
s s +4
e−πs − e−2πs
4
1
s
− 2
s s +4
Therefore,
−πs
s+1
s
e
− e−2πs
1
−1
y = L (Y ) = L
−
+L
s2 + 4
4
s s2 + 4
−πs
s
1 −1
2
e
− e−2πs
1
s
−1
−1
=L
+ L
+L
−
s2 + 4
2
s2 + 4
4
s s2 + 4
−πs −πs −2πs −2πs 1
1
e
e
s
e
e
s
−1
−1
−1
−1
= cos(2t) + sin(2t) +
L
−L
−L
+L
2
4
s
s2 + 4
s
s2 + 4
1
1
= cos(2t) + sin(2t) +
U (t − π)(1 − cos(2(t − π))) − U (t − 2π)(1 − cos(2(t − 2π)))
2
4
1
1
= cos(2t) + sin(2t) +
U (t − π)(1 − cos(2t)) − U (t − 2π)(1 − cos(2t))
2
4

1

0≤t<π
cos(2t) + 2 sin(2t)
= cos(2t) + 12 sin(2t) + 14 (1 − cos(2t)) π ≤ t < 2π


cos(2t) + 12 sin(2t)
2π ≤ t
−1
−1
2. (25 points) Suppose we write the solution to P
the differential equation y 00 + (x2 + 1)y = 0, with initial
0
conditions y(0) = 1 and y (0) = 2, in the form y =
cn xn . Write out the first 5 non-zero terms in the series
n≥0
expansion for y.
P
P
P
P
Answer: We have y 00 =
n(n − 1)cn xn−2 =
(n + 2)(n + 1)cn+2 xn and x2 y =
cn xn+2 = n≥2 cn−2 xn .
n≥0
n≥0
n≥0
So we have
X
n≥0
(n + 2)(n + 1)cn+2 xn +
X
n≥2
cn−2 xn +
X
cn xn = 0.
n≥0
We are given c0 = 1 and c1 = 2. The constant term in the series expansion now gives 2c2 + c0 = 0, and so
c2 = − 12 . The x-term gives 6c3 + c1 = 0, and so c3 = − 31 . Finally, the x2 -term gives 12c4 + c0 + c2 = 0, and
2
3
4
1
so c4 = − 24
. The first 5 non-zero terms are therefore 1 + 2x − x2 − x3 − x24 .
00
0
3. (25 points)
P The ntwo solutions of the differential equation 2xy + 5y + xy = 0 can both be written in the
r
form y = x
cn x , for two different values of r, with c0 = 1. Compute the two possible values of r.
n≥0
Answer: We have
y=
X
cn xn+r
xy =
X
cn xn+r+1
5y 0 =
X
5(n + r)cn xn+r−1
2xy 00 =
X
2(n + r)(n + r − 1)cn xn+r−1
2xy 00 + 5y 0 + xy =
X
2(n + r)(n + r − 1)cn xn+r−1 + 5(n + r)cn xn+r−1 + cn xn+r+1 = 0
The coefficient of xr−1 is c0 (2r(r − 1) + 5r), so we must have 2r2 + 3r = 0. This gives r = 0 and r = − 32 as
the two possible values for r.
4. (25 points) Write out the first 5 non-zero terms in the Fourier expansion of the function f (x) = | sin x|
for −π ≤ x ≤ π.
Answer: The function f (x) is even, and so the only terms which appear in the Fourier expansion are the
constant term and the ones involving cosines. We compute
Z
2 π
4
a0 =
sin x dx =
π 0
π
Z
π
2 π
1
a1 =
sin x cos x dx = sin2 x = 0
π 0
π
0
h
Z π
iπ
2
2
1
an =
sin x cos nx dx =
n
sin
x
sin
nx
+
cos
x
cos
nx
π
π
n2 − 1
0
0 2
1
=
((−1)n+1 − 1)
π
n2 − 1
−4
−4
−4
−4
and so a2 =
, a3 = 0, a4 =
, a5 = 0, a6 =
, a7 = 0, and a8 =
. We have
3π
15π
35π
63π
2
4 cos 2x cos 4x cos 6x cos 8x
| sin x| = −
+
+
+
+ ···
π π
3
15
35
63
Grade
85
78
77
75
68
65
64
60
48
43
Mean: 67.46
Standard deviation: 11.58
Number of people
1
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3
1
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