Download Math 217 Final Exam Dec 11, 2015 Instructions: 1. There are three

Math 217 Final Exam
Dec 11, 2015
Instructions:
1. There are three parts in this exam. Part I is multiple choice, Part II is True/False, and Part
III consists of hand-graded problems.
2. The total number of points is 100.
3. You may use a calculator.
4. The scantron and Part III will be collected at the end of the exam. You may take Part I and
Part II with you at the end of the exam.
Taylor series:
x
e =
∞
X
xn
n!
n=0
cos x =
∞
X
n=0
sin x =
=1+x+
x2 x3
+
+ ...
2
3!
x2 x4
(−1)n x2n
=1−
+
− ...
(2n)!
2
4!
∞
X
(−1)n x2n+1
n=0
∞
X
cosh x =
n=0
(2n + 1)!
=x−
x3 x5
+
− ...
3!
5!
x2n
x2 x4
=1+
+
+ ...
(2n)!
2
4!
∞
X
x3 x5
x2n+1
sinh x =
=x+
+
+ ...
(2n + 1)!
3!
5!
n=0
ln(1 + x) =
∞
X
(−1)n+1 xn
n=1
n
=x−
x2 x3
+
− ...
2
3
∞
X
1
=
xn = 1 + x + x2 + x3 + . . .
1−x
n=0
Trigonometry:
cos(x + y) = cos x cos y − sin x sin y
sin(x + y) = sin x cos y + sin y cos x
2 cos x cos y = cos(x − y) + cos(x + y)
2 sin x sin y = cos(x − y) − cos(x + y)
2 sin x cos y = sin(x + y) − sin(x − y)
cos(2x) = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x
et + e−t
2
et − e−t
sinh t =
2
eit = cos t + i sin t
cosh t =
1
Laplace transforms:
f (t) = L−1 {F (s)}
1
eat
tn ,
n = positive integer
tp ,
p > −1
sin at
cos at
sinh at
cosh at
eat sin bt
eat cos bt
tn eat ,
n = positive integer
uc (t)
uc (t)f (t − c)
ect f (t)
f (ct)
Z t
f (t − τ )g(τ )dτ
F (s) = L{f (t)}
1
, s>0
s
1
, s>a
s−a
n!
, s>0
n+1
s
Γ(p + 1)
, s>0
sp+1
a
, s>0
2
s + a2
s
, s>0
2
s + a2
a
, s > |a|
2
s − a2
s
, s > |a|
2
s − a2
b
, s>a
(s − a)2 + b2
s−a
, s>a
(s − a)2 + b2
n!
, s>a
(s − a)n+1
e−cs
, s>0
s
e−cs F (s)
F (s − c)
1 s
,
F
c
c
c>0
F (s)G(s)
0
δ(t − c)
e−cs
f (n) (t)
sn F (s) − sn−1 f (0) − sn−2 f 0 (0) − · · · − f (n−1) (0)
(−t)n f (t)
F (n) (s)
2
Part I. Multiple Choices
5 × 10 = 50 points
1. Find the solution of the initial value problem:
(1 + x)y 0 + y = cos x,
A. y = 1 + x
B. y =
C. y =
D. y =
cos x
1+x
sin x
1+x
1+sin x
1+x
ex
1+x
E. y =
F. none of the above
3
y(0) = 1.
(1)
2. Find the (implicit) solution of the exact differential equation
(x + y)y 0 = −(x + arctan y)(1 + y 2 ),
A.
x2
2
x2
2
+ x arctan y = 0
+ x arctan y + 12 ln(1 + y 2 ) = 0
C. y = tan − x2
B.
2
D. y 2 = e−x − 1
2
E. y = e−x − 1
F. none of the above
4
y(0) = 0.
(2)
3. Find the solution to the initial value problem:
2y 000 − 3y 00 − 2y 0 = 0,
y(0) = 1,
1
A. − 27 + 4e− 2 t + 12 e2t
1
B. 1 − 2e− 2 t + 2e2t
C. cosh t + 12 sinh(2t)
D. cos t + 12 sin(2t)
E. eit + 3e−it
F. none of the above
5
y 0 (0) = −1,
y 00 (0) = 3.
(3)
4. Which of the following is NOT a solution of the following differential equation?
y 00 + 4y = cos 3t
A. − 51 cos 3t
B. cos 2t − 51 cos 3t
C. sin 2t − 15 cos 3t
D. e2ti + e−2ti − 15 cos 3t
E. cos 2t + sin 2t
F. all of the above are solutions of (4)
6
(4)
5. For the differential equation (5), what is the lower bound of the radius of convergence for
the power series solution around 0?
(1 + x2 )y 00 − cos x · y 0 + ex y = 0.
A. 0
B. 1
√
C. 2
D. π
E. ∞
F. none of the above
7
(5)
6. If
y(x) =
∞
X
an xn
n=0
is the Taylor series around 0 for the function
cos(1 + x) + sin(1 + x),
then what is the value of a2015 ?
A.
a2015 =
sin 1 − cos 1
2015
a2015 =
cos 1 − sin 1
2015!
a2015 =
sin 1 − cos 1
2015!
B.
C.
D.
a2015 =
π
2015!
a2015 =
π 2015
2015!
E.
F. none of the above
8
7. Find the inverse Laplace transform of
F (s) =
e2−2s (2s2 + π 2 − 1)
.
(s − 1)(s2 + π 2 )
A.
f (t) = et + e2 cos(πt) +
B.
C.
D.
e2
sin(πt)
π
e2
t
2
f (t) = u2 (t) e + e cos(πt) +
sin(πt)
π
e2
t−2
2
sin(πt)
f (t) = u2 (t) e
+ e cos(πt) +
π
e2
t
2
sin(t − 2)
f (t) = u2 (t) e + e cos(t − 2) +
π
E.
f (t) = u2 (t) et−2 + e2 cos(t − 2) + e2 sin(t − 2)
F. none of the above
9
8. Find the Laplace transform of the piecewise


t
f (t) = 2 − t


0
A.
F (s) =
continuous function
if 0 ≤ t < 1
if 1 ≤ t < 2
if t ≥ 2
2
1
1
− 2+ 2
2
s
s
s
B.
F (s) = 1 − e−s + e−2s
C.
F (s) = 1 − 2e−s + e−2s
D.
F (s) =
1
2e−s e−2s
+
+ 2
s2
s2
s
F (s) =
1
2e−s e−2s
−
+ 2
s2
s2
s
E.
F. none of the above
10
(6)
9. If y(t) is the solution of the initial value problem:
y 00 + 4y = δ(t − 2014π),
y(0) = 0,
then what is the value of y(2015π)?
A. 0
B. −1
C. 1
D. π
E.
π
2
F. none of the above
11
y 0 (0) = 0,
(7)
10. Find the inverse Laplace transform of
F (s) = ln
A.
s2 + 1
s2 + 4
2 +1
f (t) = et
.
2 +4
+ et
B.
f (t) = e2t + e−2t
C.
f (t) = u1 (t) cos t + u2 (t) sin(2t)
D.
E.
2
f (t) = (2 cos t + 1)(cos t − 1)
t
2
f (t) = (cos t − cos(2t))
t
F. none of the above
12
(8)
Part II. True/False
5 × 2 = 10 points
Choose A if the statement is true; choose B if the statement is false.
11. There is a unique solution to the initial value problem:
1
y 0 − y 2 = 0,
y(0) = 0.
12. If p and q are continuous functions, and y1 and y2 are solutions of the differential equation
y 00 + p(t)y 0 + q(t)y = 0,
then we have
−
W (y1 , y2 )(t) = e
Rt
t0
p(τ )dτ
.
for some constant t0 .
13. If p and q are continuous real-valued functions, and u(t)+iv(t) are complex-valued solutions
of the differential equation
y 00 + p(t)y 0 + q(t)y = 0,
(9)
then both u and v are also solutions of (9).
14. For the power series
∞
X
an xn ,
n=0
if the radius of convergence is ρ and |x1 | ≥ ρ, then the series
∞
X
an (x1 )n
n=0
does not converge.
15. If f is continuous function on [0, ∞), then the Laplace transform
L{f (t)} = F (s)
exists for s > 0.
13
Math 217 Final Exam
Dec 11, 2015
Part III will be collected separately. Please write your NAME and your STUDENT NUMBER.
Student Number:
Name:
For graders:
16a + 16b. + 16c. + 16d.
16e.
17.
Total:
14
Part III. Hand-graded problems
10 + 10 + 20 = 40 points
16a. (4 points)
Find the general solution of the homogenous differential equation:
y 00 − 4y = 0.
15
(10)
16b. (6 points) Use the method of undetermined coefficients to find the general solution of the
non-homogeneous differential equation:
y 00 − 4y = sinh 2t.
16
(11)
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17
16c. (6 points)
Use the method of variation of parameters to find the general solution of the non-homogeneous
differential equation:
y 00 − 4y = sinh 2t.
(12)
18
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19
16d. (4 points)
Use the results in 16b. or 16c. to find the unique solution of the initial value problem:
y 00 − 4y = sinh 2t,
y(0) = 0,
20
y 0 (0) = 0.
(13)
16e. (10 points)
Use the method of Laplace transform to find the solution of the initial value problem:
y 00 − 4y = sinh 2t,
y(0) = 0,
21
y 0 (0) = 0.
(14)
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22
17. Find the solution of the initial value problem:
y 00 + y = f (t),
where
(
1
f (t) =
0
y(0) = 0,
y 0 (0) = 1
if 0 ≤ t < π
if t ≥ π
23
(15)
(16)
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24
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25