Mathematics 51
April 3, 2017
Differential Equations
Department of Mathematics
12–1:20 pm
All sections
Exam II
Cabot Auditorium
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You must show work to receive full credit. You are required to sign your exam book. With your signature, you
pledge that you have neither given nor received assistance on this exam.
1. (15 points)
£ Compute
¤
(a) L (t + 2)2 e 7t ,
1
1
2
+4
+4
Solution: L [(t + 2)2 e 7t ](s) = L [t 2 + 4t + 4](s − 7) =
.
3
2
(s − 7)
(s − 7)
s −7
£
¤
3
(b) L −1 2
.
2
(s + 4)(s + 1)
2
¤
£ 2
¤
£ 1
£
1 ¤
1
3
−1 (s + 4) − (s + 1)
−1
=
L
=
L
−
= sin t − sin 2t .
Solution: L −1 2
(s + 4)(s 2 + 1) n
(s 2 + 4)(s 2 + 1)
s2 + 1 s2 + 4
2
0 t <2
2. (15 points) Solve
(D − 1)x = 3 t ≥ 2 ,
x(0) = 1.
3e −2s
1
3e −2s
, so L [x] =
+
.
s
s − 1 s(s − 1)
s − (s − 1)
1
1
1
=
=
− , so
Now do a partial fractions decomposition:
s(s − 1)
s(s − 1)
s −1 s
·
¸
·
¸
1
1
3e −2s
1
x(t ) = L −1
(t ) = e t + 3u 2 (t )L −1
(t − 2)
+
−
s − 1 s(s − 1)
s −1 s
½
¡ t −2
¢
¡ −2 t
¢
et
for t < 2
t
t
.
= e + 3u 2 (t ) e
− 1 = e + 3u 2 (t ) e e − 1) =
(1 + 3e −2 )e t − 3 for t ≥ 2
Solution: (s − 1)L [x] − 1 = L [3u 2 (t )] = 3e −2s L [1] =
[To check or as an alternate strategy, note that for t < 2 this is (D − 1)x = 0 with x(0) = 1, giving e t , and for
t ≥ 2 this is (D − 1)x = 3 with x(2) = e 2 , giving ce t − 3 with ce 2 − 3 = x(2) = e 2 , i.e., c = e −2 (e 2 + 3) = 1 + 3e −2 .]
3. (14 points)
¶
µ
³ ´
³ ´
1
1
e 2t + 1
(a) Determine which of the functions ~
h 1 (t ) = 1 , ~
h 2 (t ) = −1 and ~
h 3 (t ) = 2t
are solutions
e −1
³ ´
11
of the system D~
x= 1 1 ~
x of differential equations
(b) Determine whether
the general solution.
³ ´ those
³ ´ functions
³ ´ ³ that
´ ³ ´are solutions generate
³ ´
1
0
2
11 1
1
Solution: D~
h 1 (t ) = D 1 = 0 6= 2 = 1 1 1 , so ~
h 1 (t ) = 1 is not a solution.
³ ´ ³ ´ ³ ´³ ´
³ ´
1
0
11
1
1
D~
h 2 (t ) = D −1 = 0 = 1 1 −1 , so ~
h 2 (t ) = −1 is a solution.
¶
¶ µ
¶
µ
µ
¶ ³ ´µ
2e 2t
e 2t + 1
e 2t + 1
1 1 e 2t + 1
~
~
is a solution.
D h 3 (t ) = D 2t
=
=
, so h 3 (t ) = 2t
e −1
e − 1µ
2e 2t ¶ 1 1 e 2t − 1
1 e 2t + 1
W [~
h2 ,~
h 3 ](t ) = det
= e 2t − 1 + e 2t + 1 = 2e 2t 6= 0, so ~
h 2 and ~
h 3 generate the general solution.
−1 e 2t − 1
4. (6 points) [No credit unless every answer is correct]
µ1 0 0¶
0 5 0 , and if so, provide the
For each of the following vectors decide whether it is an eigenvector of
009
corresponding eigenvalue. For each part, your answer should be either “NO” or a number.
Please put all your answers on the inside front blue cover of your examination booklet.
µ0¶
µ2¶ µ0¶
µ−2¶
µ1¶
µ−1¶
µ1¶ µ2¶ µ1¶
µ0 ¶
µ−1¶
µ0 ¶
µ2 ¶
(a) 0 (b) 1 (c) 0 (d) 2 (e) 0 (f) 0 (g) 1 (h) 1 (i) 2 (j) 1 (k) 2 (l) 1 (m) 2
1
2
0
1
2
0
0
1
2
1
2
2
1
µ1¶ µ0¶
µ0 ¶
Solution: Clearly, the eigenvalues are 1, 5, and 9 with eigenvectors 0 , 1 and 0 , respectively. Since
0
0
1
we have 3 distinct eigenvalues and corresponding eigenvectors, we only need to check which vectors are
multiples of one of the 3 we already have; the answer is “No” in all cases (none of them contains 2 zeros).
5. (15 points) Check for linear independence:
1 2 −1
2  1   1 
(a) 3 , 4 , −1.
4
3
1
1
2
−1
2
1 2 −1 0
2 1  1 0
Solution: No: 3 − 4 − −1 = 0.
4
3
1
0
1
2
−1
0
µ0¶ µ4¶ µ2¶ µ2¶
(b) 2 , 1 , 4 , 2
1
2
1
1
Solution: No: 4 vectors of size 3.
µ1 −1 −1¶
6. (15 points) Find the eigenvalues of 0 0 −1 , and for each eigenvalue find as many (linearly indepen0 0 1
dent) corresponding eigenvectors as possible.
µ1¶
Solution: The matrix is triangular, so the eigenvalues are 0 and 1. By inspection, 1 is an eigenvector for
0
µ1 −1 −1¶
µ0 −1 −1¶
µ1¶
µ 0¶
0 0 −1 − I = 0 −1 −1 (“v 2 = −v 3 ”), so 0 and −1 are eigenvectors for eigenvalue 1.
eigenvalue 0.
0 0 1
0 0 0
0
1
µ1 0 0¶
7. (10 points) Find one nonzero solution of D~
x= 2 3 0 ~
x.
456
(Yes, just one solution is good enough, so long as it is not the zero function.)
µ0¶
µ0¶
6t
Solution: By inspection, 0 is an eigenvector for eigenvalue 6; the corresponding solution is ~
x (t ) = e 0 .
1
1
8. (10 points) Convert the differential equation x 000 + 17x 00 − t x 0 + πx = tan t to a system.
Solution:
x 10 =
x2
µ 0 1
µ
¶
0¶
0
0
1 ~
x2 =
x3
or
D~
x= 0 0
x+ 0
tan t
−π t −17
0
x 3 = −πx 1 + t x 2 − 17x 3 + tan t .