K<4h
F-.rl.
24v1 )
t
E> o^,n"rr
M t {-[sr"rr
t6
L
Problem 1 Suppose that the following initial-value problem
form
has a powcr selies solution of the
g: L"."".
n:0
1a (10 points) Find the recurrence relation satisfied by
FCb
,,
<-
Y(X)=
rA
)-"
n-L
<
/.,
.'\-l)4'.X"--=
lnr
t\c
x
CYy\ -+
e)(rr\+\) O*
a._
ray\=9
\J 1rs; !:
=
n-L
'L
c\a ,\
qvn-\
6v
(-
X
)-^(n-r)4"'
vL= o
-_:-
a,,.
-5- -4 *^tt - I
+vy\=\
t\= o n+t
Y\^
{v.|
x
o
oa
=)
=)
)-
C nn
+r)Cv{ t z)
YYI=o
.
Q,-
-t
€
t-*
a.-ry'
w\rz
\tu\=
cmtrlen+z) 4"'*t I
$r\
x +f
m=z
oo qm-r
xs
Qo x + +
.rals t
=
x
|
oo
6 q3
x{'n
YV!
=
{
X
"
&
rlVrt t al
'nr\
:)
2az+(6qcl a- -\) "* #--tvn+r)(vn*")\.T
=)
@"
6a3+qr-l
=o
,
w\ C yr\ +
+
oa=t
Yto'y
Y(ol -{
-)
vq
=:=J' :o
e
oo
)r Z
1b (5 points) Determine the first five nonzero terms of the
Q.o
series
L
o.r,'
=6 t Q,--\ z A e =o I e3= N'r-o
b
t
&v r*=L
=-+=-h
a.,
eS = - Qa = O
G,)Lqt( s)
t:r =\ i o. t = - "a
- =
'
[4){sx6)
sn=S
e... =- art
=
/,. rl^-j:
/7
/z
(5)(c, (4 >
oo
)-q,"x'
I
x
4-+
6t
x+?r-'.'
n:.€
"+l
c
..,3 xt
vl
6!
a
+l
-
J"'
1c (5 points) Solve the recurrence relation you find in part a, i.e., express a", in terms of z.
cc
4 a=a
"r.r.^r.-
=)
a, =
|
2 c^'r-
/u.r.'
..+r-
:o
I-Esr^
*.
r----.*-_-
af = QB :- -- - ASr.i,
t^
=
a,o=
(
/
=-3l
I
Q1
= 91,
= -- 6r '
(
(t
t
-tr
t )ro
q3u+t) !
a3
o
--,---#
-1, ) q? = *, ,
-*""-*"., !(
4 3.. r, = (j-
ot
7
>
=o
1-
-
(-r )**'
(sL) !
lo
I
4r,
=
'4r
ra+t
Ilc.-r>
\
c=t')!
\ .-'r*
l-G";sn
t
Io
n30
n =3t<
.t
u
6=3h1
;
s
v ev-+
|
n=ttttZ
+
f,ce
o.
y'uloJ
f"\rt
Problem 2Let
f:
IR.
-+
lR.
2a (6 points) Show that
r,rr--*
be a differentiable function such that
/(f)
has a Laplace transform F'(s) for s
> 1.
-
_
d+ = S'- * Sc+rd t* !?-'t .t rt
f.-th*,
o
L9*tl----"r
+ .'s Jt'fe.-.\'..4 :) + is
C t'' -(s^
I.ts) = -Ll^_ ) e.
N-+ c,o z
1e<or,.
s> l-
FrS>
=l
l(t):"'fort>2.
2b (4 points)
Show
ca.
i: t:
that /'(f)
,6./ tl, z,
a,ts )
t) t
co.tirur-s.-
Jt -Al-
I.cs)
:f
L
frG) ca. ;s l> 2 \-€.,
- atJ{)
-'t
e-(s-r) N e
-(s-r7
n s-__==_
|
-Z(S-r)
Jl: o- !1 T-ocs; = g__-.'.-1rs-ro"-(s-r)
-F-r
U-r oo
- (S^r)N
.
has a Laplace transform for s
$,+, = "t
>
=)
1.
Sl*,
--
"t
t),
.z\
+' r-J 6a\ rvet5"" ,r.' fr 'Sc u"L cotn 4t-" -+A^
;
Pq{ 2a 4n
' =!ttn-{ "tt7'*'^ '*4 b\^L 'rxd
ar.-"r I
€,(irlo
fAor^r fuot
ft;'*fl*,tt
n/
r
,-\ At
,G- s>r
p/. e',.
C'--t+
rr = -_:tt-t'
J e
ic*r
s_ r
L
f1
v\l
s)r
, t-Lt'.*rI ea"rb
2c (5 points) Use the definition of the Laplace transform to show that the Laplace transform
of /'(l) is given by sF(s) - /(0) for s > 1, i.e., L{f' (t)} : s f{/(t)} - /(0).
f- [ ti,,]
a"t ji+,tt
=
(r@r
=
n,) - lJr't 1,+,J.t1
)tic"f
o
-Sl
.I
t&
a
1, *)t . Ac+t dt
{tu}.u, - $..) + sti++'1
= J;
N-t a
=
l
"
"!F.+,
;t*{ * sFrl - fr., :
- -&:
N-loo L- o becatt- S>l
s Frsl--S"'1
a
Problem 3 Let
/
and g be differentiable functions satisfyirrg
f'(t)-2s(t):ur(t),
/r'\
f(t):
()\
s'(t) +
/(o)
r,
: g(o) :0,
/a\
\ U,/
where z2(t) denotes the step function satisfying u2(t):0lor t <2 and u2(t) :1for.J > 2.
Suppose that /(l) and 9(r) have respectively Laplace transforrns, F(s) and G(s), for s > 0.
3a (10 points) Take the Laplace transform of both sides of Equations (1) and (2) and
result to determine F(s) and G(s).
L\ti+,\= sfis>-f'o1= sQs)
tGtsr - $co1 -- r 6ts/
=
t l9'r*,1
t -st
,\rrn*l
elt*urc+)J+
=)e Jt
Lraz,+5=J
t-
use
thc
^1/.
;^€
-Ls
t
{".
=
s)o
rlt
=)sF(s)-26(s) = {: s
e) =7S6ts;+Fts) = *
(r)
vil
[:
.
=l
_
t* [ ":
."\i- -:\
I
.-es
e- -t
E
4@
L
3r
s?+ L
-2_S
-=#
- z.s
J-{ [s 'rr,
L\
s\a
t
t
s
t-1
-"=J
_-:-j"*-**---
L
-as
jl[:::, l=Li.
-zs
fts) -t s Gtr) =
t- ,:3"Sttu
15
L
*frGl
_zs
\_
st4 z
:-----'
s(sht)
3b (15 points) Use your response to Problem 3a to obtain
$.+
r
-\
Ll (
p
= J:'1n,,]
=
f'{ #
-*
:- \
r-'l+
k
#.\=
:LS
s'*
st+ a
a
(-tc^ (-l-ect-<.)) trz(a) Jt
=\
t,r,..,
;,
=
p-l f
Lt
J
In r
+
s-r"!E:tJt
=l
-o
{-:-
'ra+
c o,t
=J
1
{
c^[.F.0 * 1t+l
t<z
-
C
r*
( 1=I-,{+.+=\=
.
!
-i- 5aaz
s(slrr J
?\
=-)
: Lt.<{; x c^(fzt)'
\==t
l.=.
.[..-
and g(t).
.dtJ*l
{lS1.*
*o,,\=
-' L
1c-
t1)"*q
/(/)
i:::'-t;;,,--l;'
'
'fa
t.I(EJ''l
'f-z
tl:$r!i1
' {-r-
=- ":-tla:a)\:= 5!j9!
*-+, = f, ,...rr s'''"L{1c{-?)l + t- e'4('i1t)
}d*t = L-l l6c.r] =
-_ t*l
r'f #. - :.3".,\
I.l*3' !'= t)
: :'\g*)
{-i
.-'l
1
+
;E;
1-,[ -rI-,\
" t \u..+,\
s.,...,({?
rts'\
-+.
=)
t
s.\','"ISZ
(!*.g11
9lt
!?-
l-(.ct
t
=.1
{z
)
C-r C{:ic.c-€r)
rl
J
^t- =
r::.
(,- *L{ic{-zrf
)
s;cs:"tl -
>+
u.ct)
'4
t <2,
{i. :,," (?l_t:--l]
r t- ,{
.f-rz
c-^
({-zct-ar:}
r)r-
Problem 4 (15 points) Find the general solution
x(t)=
o{ the following system of equations.
l1 01 -4.l
0
l0
l1o r lx(t).
l
j
, rt-r o -q I
(t-r1-1{tr-r)
&
o )=O =1
\L -t r-r
J
t-f
o
=") (r-r) fcv-rr'+ cl3(c =, [ : =;.,
f=
1i
o
L"t
o
l'= f,: = t+ri
-zi
u rt.9'l
o
d
.T
,L t-(rt2i)
a
lo
-.1{
-ztf --
y =f, 3l-ai
r^(tt zi)
?.'=
=, A=zif
=
rilt
[-i'l
*"
[-.'I
Lrl
:
z=
* "-*t
L
)
=
r - acl s.',r(a{) +
I
I
L 92
1e3
C.r'
Crr
lrt t + ca Siatat)
=
fi 1
(t-z:)t f-z"f
Li
-l
q't1
I
J "a.t lt"-tto.t
L 6'.\^ 12:t'r I
*
\
=]T=tTL cx' (z+'
I
:F
* -;f
-z;r Jl=' [L
- e
Czt)
(_\
I
itil =lil
is q co$^{'o-'*
.Ic't1 =e
t'i":'.tl
---. (1 )
{c+r
/
=o :-t>=
/
.;-]t;1{11
l^
l a?, _ .Ic.r
:
c_
(t)
o
o
g=o
d
-": lL "" \= L;I
o.=
-\
{
s-\
;t
I
6
.-_
1
-' r-tl
Sr'a(i
Probtem b Let
A: I o^ :0lI
L-3
5a (I0 points) Find an explir.it explession for the fundamental matrix crA o[ t]re
x'(t) : 4 *11;, i.e., compute the exponential of lA.
system
d*t--: 1. I :o -=7 rL+ 9 so :') r = t 3i
r:g=3i =t L:1'1,,1[il=[:\ =, -='i**31=o
t !r-Y'q
'--t
r\1
sat q=.
-r d " "[;\
rn=f =-3i: '
rI l 1
=l
ru-trA
I
:r
s{=
lS=L-:t\'
*-.'\
l-
=:
t+ \J=L;-il
"{
i!l
r
t rt
l--\ -t I - \ \
tJ-'I\
,,,-r
*J -- -l--rtL'
u - L".' jr.l ='A.
$
lft = u u), r'!r^t ='
+At
tr4
. .,
-'
I
trA - -L l- , t.l[ "
i
=) -''- = t L i -i\L " a-''J[ t
-iI
IIt
-!{-o"tt
"-3"t
= ZIiuart -,;t"JL, ;l
,
_ J- f
=.,L"'^.:j=.]t':
.art+c-.ril
"-t
'
J
. :it *rc
-rg ' --rtt
-1
I
i;=tt "ts''t * itt' J
{-c,ncat) * lti sr^tet'l I
= Lifz.s.\^(3t)l e'ocatl \
z
[ ;u"t-
1
oo
l-
clt I
s..^
cat I
5.\(sl)
c- cg tl
I
-;l
'rJI
5b( 15 points)
Use your response to Problem 5a to solve the
initial value problem:
t;lll:illl;:, ;.= Li'.-J
r1(r')
z. '=
--+
: 1, r2Qr): -1.
-[:i-,
\1,
l'F
\---1=-r'.{rt;
rti+,.o-ttA=
rnt-'
[s,l.
e ^'pc*1^\t*t,6ords
c+r=^*c+rc*)
fl
+
: .re d *" S .ct-s'r
! ,'r = [ !,1
'
ttl
=
[c.t<+]
:.:*l'l
t:{ 1 crr t:\tJ
A
,"
[:J
S--_-*---1,-*--'-*-
.
'1 c*r
t
C ICn 3<+-s'y sr\ 3tt-r, ][ " I Js
) L-=.r.e.t-r) cro 3c{-' IL-'l
si.rfacs-t>l Js
Ct Izc--[:cs-t1]t
1
-Cr
[sts-tr) J
c',[3ts-trt
f
\ :.*^f3ts-t,l
1 [t='
J r : T\
si"
f:cs-{}f
c-,
L3(s-tt
J
|
L+
\
t
=
f ,.
",t[srs-t,l -
c.r,
C{'(3q-=t' ll
(3n-"t)
2lsin
f
, f-r(an-s4 1
=bL-"
^ [-zc-r (3n-3tr . s'',
I
1--r - zs,.r(gt) - c^'r C3t) l
(I
s"^13tr JJ
tl + o'^Lo\-'
= 3L*z - zC-(3+)
_ -_nn I I I
_nrAa=\,,1 =, Z=.
L-,\
(a) i -1
o=)e
4t
d
*
I
.ct-+r tt.t = tn?X::-;,' Jil:i:.'ltJ[:,1
r-6ast -s'"3ttf lr
f-c-st{ss3t
ia
= l5rl" ct -c-, st lL-'I
'\
"v
:
'
=L
t'' st + c':
€
1
J
ir
-':'"'!{-
i:
'
-'
@ffiir,:.
T
<+r
f=L
r
-d
-)
c.-ngtl + <infrt) * { L-r - zsin tt*)
*,," (rt) + c., (t+) -- f z - 2- c.-, cG+)
d
\-
-t -t.s.\(it)-
L
-"
X.<+1 =
zr
c-ncet)
I
+{s,,rr1tt) -tc--,t3{r
t*
[
{
Xz.ll : }lr, t{
ca+)
s,;
<e+
'- Qc'4 (s tr
J
-rl
t -t Cnt?Lt - zl
cn rr+rl
s rn
I
ttrrl
J