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Unformatted text preview: MAT 22B001: Differential Equations
Final Exam Solutions
Note: There is a table of the Laplace transform in the last page.
Name:
SSN:
Score of this page:
Total Score: throughout this exam). © §¥
§ ¤ ¨¦ ¤ $ " ©
%#! ¤ § , we have
$
&
§ © & ¥
¤ ¤ 0©
& ¤
)(©
& & Ans: Using the integration factor ¢
£¡
¡ Problem 1 (5 pts) Solve the following initial value problem ( represents ¤ '¥
§
& ¤
& So, we must have
$
& § © & That is,
$§ § 1 ©§ 2
& 2
& , so let’s take 8 IG¢ E! §
HF D © 1
& B
C¥ A9§
@8
© 6 § & 753
64 From this we have
D
8 H F ¢ D § © P¤ 8 H F ¢ ) Integrating this, we get
$ S¥ 8
D ©
Q¤ 8
D
R
H F¢
H F¢
"T
VU© . So, ﬁnally we have $8
YD T ©
¦'XW! § ¤
H F¢ R Using the initial condition, we get 1 . Now, (1) becomes: (1) Score of this page:
8§ § Problem 2 (15 pts) Consider two functions and .
¦" Vb
"T § .
p© §
¦" `9
"T
f
f
f
f
f
"f
T 8 § © f §
f©
ge 8 d§ c
§
§
8§ © 8§ except . $8§ 8
9§ h©
pi 9d§
8§ on a"
V`T c . So, these are linearly independent on .
¥ ¤ Thus "
a § (a) (8 pts) Show that they are linearly independent on
Ans: Consider the Wronskian of and over
¦" `9
"T (b) (7 pts) Can they be a fundamental set of solutions of some differential equations
on
? Answer yes or no and state your reasoning clearly.
[Hint: Check the Wronskian in this interval.]
Ans: They cannot be the solution of this differential equation. The reason is the following. Let , be a fundamental set of solutions of the differential equation of the
form
, where
. Then, from Abel’s theorem,
"
Wa § § 8 ¤ v¤
u
" Vb
"T
E© § 2 . Therefore, and 8§ W©xb
¤
¤
§ C¥ 9 § wq
s
8 ¤ v¤
u Rys
7£ q at ¦" Vb
"T ©¤
W9 § t¥ b § rq
¤
s a"
`VT ¥ ¤ 8
u¤
y § h©
WX ¤ v ©
0 8 d§
§
c c However, from (a),
tal set of solutions. are the fundamental set of solutions.
cannot be the fundamen $ $
A¦" $
¦" ¥
i§
" T
¢ eY
fQD 8 S¥ 8 9u B
D
B
¢ 3
©
! § g 88
u¤
¥ § ¤ S¥ § Af9u B
B ©
! § ¤
Finally the general solution is:
¥
j§
" T © "
U!¦vT
§ qu
" T
w¦" ¥
j§
T©
pe § 8 b § 8 h
¤ " u¤ u h ©
¥ § f9 § iEe § g Hence,
qu $ e bvT
D"
§
¢ T©
U! § 8 h " D
¢ 8 Y b" ¥
i§ t
" T©
# § ih
u Therefore, by integration by parts, we get:
e
o¢ D § T
Y
¢8 D§ n q
© " o§
n oS¢ 8 D
eD
fY rT
¢ ¢ D
q
¢ 8 sT
e
fY D T
n ¢D o§ T
p© o
¢ nuY efY D T
e
fY D
¢ D
¢ 8
¢8D n
© 8
o h
u h
n That is:
o§
$ © n 8
o h
u h
n efY D T
e
fY D o
¢ D
¢ 8 Q
¢8D n ¢ Representing these in a matrixvector notation, we have:
© 8 h f¦D
eY
¢
§ T u h 8
D
¢ 8
eY
E© h ¢ fQD u
¥ h¢8 D must satisfy the following 1st order differential equations:
$ 8
eY h ¢ f¦D T
88 ¤ h
88
¥ ¤ h
u ¤u
¥ 9ih 8h u
ih ©
§ u h¢8
u
¥ ¤
u
¥ ¤
u
¥ ¤ 8
u
u ¤ u
8 ¨T 8
¤
¤8
u ¤ ¤ ¥ h ¥ bf¨T ¤ ¥ ih
88
88
88
u ¤u
u ¤u h
¥ v9im ¥ ¤ h ¥ 9ih ¥ ¤ h ¥ ¤ h and 8¤8 h Therefore, the nonhomogeneous part © h
uh © uh © D
u g
© g
lT g ¥ k Ans: Let us ﬁrst solve the homogeneous equation:
. The characteristic
equation is
. Hence we have
, and
. Now,
consider the nonhomogeneous equation, and let the particular solution be of the following
form:
. Then,
. We choose ,
so that
. Thus,
, and
.
Inserting these to the nonhomogeneous equation, we get:
88 ¤ h
u
ih 88
¥ ¤ h ¢ u ¤u
¥ 9ih
88 ¤ h eD
fY d© 8 ¤ g
u uh
¥ ¤ d© 8¤8 h
u ¤u ¥ Ijh ¥ D ©u
¢ 8 dev¤
SE¤
© ¤ T g
88
u ¤u h
W© 8 ¤ 8 h ¥ v¤ u h
u ¤ h ¥ 9i0© g
u uh
§ 8 9 § 8 h ¥ § f9 § i0e §
¤
u¤ u h ©
¥ v¤ E© ©
W! C¥ ¥ ¤ $§ © ¤
'T ¤
%¨T ©
(V¨T 8h g ¥
8 ¥
¤ Problem 3 (15 pts) Find the general solution to the following nonhomogeneous differential equation by the method of variation of parameters.
Score of this page: $u ¥x
q T 8x
" $ $ D
e
¢
" 4
D
D©
¥ ¢ 8 'T ¢ E § i¤ Taking the inverse Laplace transform simply yields:
T x
t T
Cx
¥ " " T "T
vrx
" © x
¦ g In other words, we have:
$ ©
" ©
R "T
VU© "
p©
Comparing the coefﬁcients of both sides, we have
R
¥
C¥ T
rx u
xR
q 0T 8 d S¥
¥ T x T x " T x
t%Cr¦t CTrx
¥ R
x R
¥ d %C¥
and rearranging terms, we have
$
¥ Multiplying both sides by "T
x
¥
Let’s represent the righthand side by the partial fractions:
$ T x " T x
%Ct¦vr T x " T x
%¨r¦vtP
T "
¥ x T x T x " T x
t%Cr¦t
" , we have
$"
%vT T
x
¥
x g © x
e¦t " T x " T x ©
%Ct¦vr(!%
% ¥ x T
8 g © x
!¦
¥ x T 8 t
x Since x
t Putting the initial conditions, we have:
$ T
rx g © x
e¦t " g ¤ T x
¥ d ilw¦ x
t T ¤ T ¤ x T x
w% '%'wQ . Taking the Laplace transform of both sides yields:
$ © ¤ ©
! 0! i¤ " D©¤
¢ e 0Q ¥ ¤ g x
8 g x } ¤z y © x
¦t~v{0!Q Ans: Let T ¤ Problem 4 (15 pts) Solve the following initial value problem using the Laplace transform.
Score of this page: . "
o VT
"
n ¢ Y
vQD
vY D 8 B T 9u B
¢D
$ o¢
vY D 8 C¥ bu
B
¢
¢D B n
o "VT
8
Do
uB
" n BS¥ ¢
n
uB
B
C¥ ¢ D u
¢ © 8 $ Y
vQD 8 ©
©
8 , we get o
© n
u o
"
U© 8 u
8
8
u
© n , we get .
o sT
T 8 ©
u u
8
u o
u o
$ . Taking © n
u
T by
¥ ©
p §
W© . Taking 8 8
¥ 8 u
XT
Tn
T
u U© , we have:
E©
u 8
T
u u
XT
n
u
"
U©
uT "
U©
f
f
2
XT f
f©
¦AD
e T , we need to compute . Now let’s compute the corresponding eigenvectors.
$ ©
PEe u 5
Therefore, we have the following general solution:
From this, we have
Similarly for From this, we have f "
¦vT
(© u
T
C¥ 8 © T
`'w%
¥
¦"
¥ f f ¥
T
© f " For Thus, ¥ Ans: The characteristic polynomial of this matrix is:
o
$ 8
o sT
u
n
"
VT 8
o
u © n
Pn Problem 5 (15 pts) Solve the following system of differential equations.
Score of this page: Score of this page:
Problem 6 (30 pts) Consider a system of differential equations: " "
© ©
Wh © is a real number and . where
(a) (10 pts) Use an inductive argument to show that
8
$
u u
Y u
Y
8
Y
Y
© s
[Hint: Induction means that you need to do the following: 1) Prove the statement is
true for
and
; 2) Assume the statement is true for and prove the statement
.]
holds for
, we have
Ans: Let’s proceed the induction. For
"
U© $
© " " © " " ¥
W©
W© So, this is ﬁne.
For
, we have
"
U© $
" "
©
u
© u
Y
8
u Y
Y
¡
u £
u u
8
Y ¥ u
Y
u ¢
u¡
"
Y
u u
¢
u¡
¦" ¥
Y 8
Y ¡
u ¢
¦" ¥
¥
8
¡
¢
¡
u ¢
¦" ¥
£
ty
© " " © ¡
u ¢
© ¡
u ¢
© 6 . ¥ Therefore, this formula is correct for all " So, this is ﬁne too.
Now, suppose this relationship is correct for . Then, consider the case . ¯¬
°
®¢
¯ u Y
¬
²±
b® Y ¢
¬
¯ 8 °
³ F bG ¢
±® «
¨ ¢
« ¨
u £
§¦
¨
8 8 ¢
¦
F¢
« ¦ 7
¬
¢D
¬
¬
¢D
¢¬ D
¬ 8§
¬
D
¢ D§ ¢ D
F¢
¢
«
¯¬
¢
¨
«
«
ª
®¢
¯¬
A £
¯ u Y ° u ¢
¨
¨
¬
¦
²b5 ¢ §
±®
®
G ¢
¦
¦
Y
u
Y
8
u
8
Y u
Y © © ©
© ¨ ¢
§ § ¦ Using (a) and the hints, we have
$ ¨ £ ©
§ § © ¤
¥¢ D ¦ Ans: Recall the deﬁnition of a matrix exponential:
$
© %CT
8 Y§ ¨
8 £
§
© 8Y 8§ "
¦vT
©
§ ¦ ¨ ¢
§
¦ ¨
u ¢
"
© ¦T
Y§
u
© §§ Y
u and ¨ ¢ ©
§ § ¦ ¦
¤
¥¢ D (b) (8 pts) Compute
[Hint: . © "
7b
´ § ´ ©
e ¤ u
Y ¢
D©
0 § ´ ´ $ Y ¢
¬
Y
D
¶u ¬ ¢
D"
9¦vT §
u
Y ¢
¶¬
D
© ¤ u
Y 8
u .
¶¬
Y
D
u ¬ ¢
Y
D"
b¦T §
u ¢
¶¬
8
Y
D
Y
u ¢
F u ¢
© "
!¦b
´ ¢
D©
dµ §
´ It is easy to verify that ´ Using (b), we clearly have: © "
x¦9 . Note that (c) (6 pts) Find the fundamental matrix
satisfying
This is different from
.
Ans: For this initial condition, it is clear that ; ´ ©
e § "
"
" 9
¶¬
Y
D
u ¢ ¬
Y
D
$
§
u
¶¬¢
Y
D ¡8
F
u ¢
u G¢
¶¬
Y
D
¬ u ¢
Y
·
{
D "
Pd¦T § ¥ b
"
u ¢
ª
¶¬
8
Y
D¸
"
Y
¥ ¦vT § ¥ "
F u ¢
³
u ¢
¶¬
Y
D
u ¬ ¢
¬
Y
Y
D"
b¦T
D
§
u ¢
¶¬
¶u ¬ ¢
¶¬
8
Y
Y
Y
D
D"
bT §
D
Y
u ¢
F u ¢
u ¢
u ¢
, we have the following solution: "
b § § ´ © © ©
p § Ans: Using the fundamental matrix
Therefore, . is given. Then, solve this IVP.
"
" © "
!¦9 " (d) (6 pts) Suppose that the initial condition: u ou T
Tn
© 8
or "
o VT
"
VT . Here, .
8 n "
¾ VT
Y D o
¢8
"
VT n 8 n
© 8 $ 8 must satisfy
u
8
8
B
$ 8 Y¦D o § S¥ B
¢
8
8
§ B ¥ B
8
8
B
o B T § %
8 B T § 8 B
Y
¦D
¢8
¥
j§ "T
`© o
$ ©
n ©
n
u . Thus, 8 u . $
W©
%
8
¥ © ¥ u %
B
¥ uBn © TB
u %
T
u B
n ©
»© 8
8
8 o
u
u
© o
n YD
¢8 ©
© o
$ 8 ©
¼ § i¤
u YD
¢8
u
u
u
©
u u ¥8 u
T n
T 8
u 8 T
u
u u
º
n
u
¥
E©
8
sT
u o
u
u
© n
(© u ¥ 8 º
T
T 8
Q
©
8
u
¥ ¢ 8 Y D u
§ © 8
Y
¢8 D8 . The vector
o
o ½`8
o
u
BS¥ ¢ 8 Y D
n
n B
8 S¥ §
uB
B
u
8 o@"
© , we get , we get . Taking . Taking . Let’s compute the eigenvector
u 10
Thus, the general solution is:
can take other values. For example, the other choices lead to
From this, we have , i.e., To compute
, let us assume that
the following equation:
From this, we have sTp©
f
f
f
©fu So, we have repeated eigenvalues
f
T f
f º uT ¹ f
¥ Ans: The characteristic polynomial of this matrix is:
o
$ 8
u
n º
T ou ¹
T
u n 8
o
u n
© Problem 7 (20 pts) Solve the following system of differential equations.
Score of this page: Score of this page:
©
#Á¤ ¥ ¤ S¥ ¤
B
À are positive. There are three
and
Ç
ÈÀ À¥
8 ©
¿
Å ¿
p© ¥
B . Let B
¥ BÀ
¢¢ B
l¥ À¥
8
À¥
lr8 ¿ ¿ ©
0! Ç
VB ¿ Ans: The characteristic equation is clearly
consider the roots of
. BÀ
¢¢ ¿ (a parabola) when Â
Ã§ [Hint: Think the graph of
cases you need to check.] approach zero as Ä Bonus Problem (20 pts) Show that all solutions of
if
are all positive real constants. Å Ç
`B T À
8 u © , i.e., two distinct real roots. In this case, because
,
,
, both roots are negative. (Consider the graph of
which is a convex parabola.)
Let these two roots be
and
with
and
. Then, the general
solution is:
a
Í#Ë ÍpÉ
a
Å
¢
$ as AÀ Ñ©
@T tend to Â
7§ ¢ ÏD Ï
D 8 S¥ 9u B
ÎD
B
¢
¢ and Ä Ë
Ì© ÎD É
Ê© Æ Ç ¿ ©
! § ¤
take, both
as
. Ä B
8 Â
!§ and uB
Â § ¤ No matter what values
and
. Thus, ¿ Case I: because a
iË B
Ð© T 8À a
'É u © Case II:
. In this case, we have repeated roots
, which is
negative since
and
. The graph of
in this case is a parabola whose
apex is touching on axis at
. The general solution in this case is:
Å AÀ p©
@T ¿ $ I8 fQD § 8 S¥ I8 f¦bu B
B
Ó G¢ Ò Y
Ó G¢ Ò Y D
H
H
AÀ T
@ Ä Â
`§
Ä Â
e§ . ¿ since
Ía and grows TB
¿
Â § i¤
ÓHÒ
I8 G¢ fY D § ©
¿ ¿ ©
! § ¤
ÖÕ Ô
SeÀ T
u Æ . In this case, we have two complex roots,
8À ¿ Ç tend to as
as Ò
ÓH
I8 G¢ D Ç
À ÓH
I8 G¢
va B § Ò
fY D ¿ u T À
8
© Æ Case III: ¿ But both
and
much faster than . Thus, So, we have the general solution:
o TB
¿ ¿ u Ön 4
8
Õ jI8 G¢ fQD S¥
B
ÙÓ HÒY
Ä Â
§ . 8À Ä Â
e§ o 8À
§ TB
¿
Â § ¤
¿ u Ön ÓHÒ
I8 G¢ fY D
Â § i¤ 11 as as § ÙØ× Ó H ÒYD
%jI8 G¢ fQ9u B ©
! § ¤
Therefore, in all cases, we established , $ Again, because of the factor
dies out eventually.) . (The solution oscillates but Çx Çx ¿ FÓ ¡FÛ
Ó
§ FÓ ¡FÛ
Û
Ó
u Çx Å ¿
§ YÛ àß
² £Û
¯Ü Çx
d § 12
Û
Çx
u
y © x
{0!QÁÚ ¿
4 ÕÙ ÙØ
%×
¢ÓD £ y Ý Ü
tÞ£I§ 1
x Ú
d¦Á Y E! §
y©
u Å Table of Elementary Laplace Transforms ...
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 Fall '10
 Hunter
 Math, Differential Equations, Equations

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