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6 Pages

2233-l08

Course: MATH 2233, Fall 2008
School: Oklahoma State
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
! &quot;#$ ( $ ( ( + . % &amp; ' ) % ' # $ % ' / # 0 ' * + + * + + * + + , , , + + $ $ + + '( * + % )( 2 $ ) # )))% % 13+ ) %# $ !# )( 4 5 6 7+ % ) + ) &amp;) &amp;$ $ + %# % ! % ( ) 5 + ) 1 5+%%# + 8 + $' * + 5 ($ ) # $ 7+ # 7+ ##
Oklahoma State - MATH - 2233
! &amp; ' !&quot; #$ &quot;%#* + ! % 0 % 1 2 % $ % 3 1 4 + 5 &quot; 6 % # -' 3' 4 # # ! 0 $ % &quot; # &quot; $ % / ' &quot; ! , &quot; ! %() %% $ ! &amp; % '&quot;.%'$&quot; ,% % .#!+6+%6!!7 8 % *6$%$ &quot;%%%% / 0 # + ' 2 % 3 % 4 ! 5 ! &quot; &lt; ! &quot;
Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
! &quot;#&amp; &amp; '( ! * ) . , $ &amp; !2' ) % %$ #) ) $&amp; '( %% % '% ! + , $ #% % ! $ ( /0 &amp; ' ( ( % 5 &amp; '3 &amp; ' &amp; '( 6 ) 6 ) , ! !2 7 3 5 3 ( / &amp;/' ( 2 &amp;/' ( / (/ &amp;/' 1 3 &amp;/' 1 $ 3 &amp;/' 81 3 $ (/ , ! ! 3 # 3 3 &amp;/' 1 &amp;/' 3 2 &amp;/' 3 2 4 &amp;/' 3 !5 &amp; '
Oklahoma State - MATH - 2233
$%! &amp; $% ) * ! ! !! $! &quot;# ' &quot; # #! &quot; % #! &quot; ( $( ( $ ! ! &quot; &quot; &amp; . #! &quot; % #! &quot; &amp; $ / $0! * $ ! ' #! &amp; ' !&quot; $1! &amp; $,! * $1! &quot;# !&quot; ' #! #! #!$ &amp; !&quot; ! ! $1! ! #! #! #! &quot; &quot; &quot; * $ ! * ! $0! &quot;# #! &quot; ! !&quot; ! * #! ! ' + #! % #! % , &quot;# #!! '&quot; #$#!%
Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
! &quot; ' ) + . &quot; + $&quot; &quot; + &amp; , -/ &amp;, &amp; , % $ &quot; $ %&quot; $# &amp; &quot; &quot; # ' &quot; # # # ! ,! ' % ( &quot;&quot;*+* &amp; ,01 &amp; ,0 1&amp; , *&amp; ,$2 $&amp;,$ ' $%&amp; ,*/ 0 ,3 %!.&quot; 4*$ * /!&amp;,&amp; ,/**/5$$0 1 * * * * * * &amp; ,* 0 &amp; ,1 + +&amp; , &amp;
Oklahoma State - MATH - 2233
! &quot;# $ % &amp; ' $ $ # $ $ $ * * ) &quot; &quot; &quot; + * ( $ ) $ $ $# # &quot; + * ' $ ) # $ * $# $ $ $ * &quot;, / 0 ('!&quot;* ) &quot; # $.! $1$ $&quot; &quot; $ # &quot;,$ $&quot;* # $,&quot; &quot; 1 ! -$ $ $&quot;,&quot;# &quot; &quot;, ## $# '2 ( &quot; , ) $.$ $ $$ , $,. '' '4'2(/
Oklahoma State - MATH - 2233
Math 2233.005 FIRST EXAM 12:30 1:45 am, Sept. 28, 2000 Name: 1. Consider the plot below of the direction field for the differential equation y = (y - 1)2 (2 - y).(a) (7 pts) Sketch the solution curve satisfying y(0) = 3. See graph above(b) (8 p
Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
! ($) (+ ) , ! &amp; !&amp; $&quot; . / ( )'#$%&amp;$'()( * )( ) (+ ) &amp; ( )$1( ) $ 2 $&amp;$. / ( )'0 $ !$ . &amp; $. $&amp;$.!$--$ /$ $ &amp;! $ ( ' ' ) $ . ! $ /$ ! &amp; ' ( )( * ) ' &amp; $ $! 0 ! !&amp; $ ( )'&amp; &amp; ! $ &amp; ' 3&amp; ! $ &amp; ! ' $&amp;/ ' ( )$$!
Oklahoma State - MATH - 2233
!&quot;#$$ % &amp; # &quot;! ' !( $&quot; &amp;# &quot;!)! $ *+ !,-. '# # $ &quot; !&amp; # 0 -' 5 -$+ !'+ /! &amp;0! 1, &quot; ! ! $ &quot;!) # ! &quot; ! $ *+ ! !&amp; # 0 $! 2$!!&amp;0 # &amp; $ %! 1 &quot;!) ! $ *+ ! &quot;! 0 &quot;!) ! $ *+ ! , , ' ' &quot; &quot; &quot; &quot; $! $! 64 34' &quot;--, ,#! &quot;&quot; ( &quot;
Oklahoma State - MATH - 2233
1.2 = )0(y ta strats taht noitulos eht rof1 = )x(y mil ,eroferehT .1 = y enil eht sehcaorppa ti sa tuo nettafl lliw hparg sti tub ,1 &gt; y revenehw gnisaerced e b lliw )x(y noitulos a oS .1 = y nehw orez ot lauqe si dna ,1 &gt; y sa gnol os evitagen sy
Oklahoma State - MATH - 2233
12y 2x 1y 1x 0y 0x122.1 )2.1( y 122.1 = 121. + 1.1 = )1.0( )1.1( )1.1( + 1.1 = x ) 1y , 1x( F + 1y = 2.1 = x + 1x = 1.1 = )1.0( )1( )1( + 1 = x ) 0y , 0x( F + 0y = 1.1 = x + 0x = 1 = 1 =.)2.1(y etamitse ot 1.0 = x dna 3 = n htiw dohtem )rel
Oklahoma State - MATH - 2233
! &quot;# $ #%!%&amp;&quot;'( )&quot;* ' &quot; ! + &quot;%' )&quot;* ' &quot; ! + &quot;%' ,% % )&quot;' )&quot;* ' &quot; ! + % '%'.!&quot;' )&quot;* ' &quot; ! + &quot;%' / 0 %) 0 0 %) 2 0 %) ) 3 0 %) 0 0 !&quot;' 0 !&quot;' 1 / 0 /1 '%'!&quot;' 1 0 '%'!&quot;' 1 ,-&quot;'( $ &quot; % ) &quot;%' ')&quot;'( &amp;$ $ $ &quot;'( &amp;$ $ $&quot; ! !&quot;'1 &quot;'%)!&quot;'% % &quot;6 &quot; $
Oklahoma State - MATH - 2233
! &quot;## $ % !&amp;' # %!(# )*+,+ - ,+ , +. &quot; , / &amp;! ! + , +. &quot; , &quot; ! 0 % + , $! # 2 ! 0 * 1 $!# + ,&amp; 0 # 3 4 &amp; % 1 &amp;% %1 + ,* 1 + $ 1 &amp; 0 &amp; # + ,* 1 ,* 1 # , ! &amp; ! 2! ( &amp; # ) +5 ! ! / &quot; &amp;+,+ - , ! #$! $ #!% &amp;#+ , 5
Oklahoma State - MATH - 2233
1396.1 = )1.0( )34.1 + 2.1( + 34.1 = x ) 2y + 2x( + 2y = x ) 2y , 2x( m + 2y = 3y 34.1 = )1.0( )2.1 + 1.1( + 2.1 = x ) 1y + 1x( + 1y = x ) 1y , 1x( m + 1y = 2y 2.1 = )1.0( )1 + 1( + 1 = x ) 0y + 0x( + 0y = x ) 0y , 0x( m + 0y = 1y 1 = 0y.396 .1
Oklahoma State - MATH - 2233
1x- = )x(y mil edulcnoc eW .evitagen erom dna erom teg tsum noitulos eht fo epols eht )1 &gt; |y|( regral dna regral steg |y| sa ,deednI . x sa gnisaerced sniamer dna ,gnisaerced yllaitini si noitcnuf eht taht snaem sihT .1- &lt; y rof evitagen syaw
Oklahoma State - MATH - 2233
1396.1 = )1.0( )34.1 + 2.1( + 34.1 = x ) 2y + 2x( + 2y = x ) 2y , 2x( m + 2y = 34.1 = )1.0( )2.1 + 1.1( + 2.1 = x ) 1y + 1x( + 1y = x ) 1y , 1x( m + 1y = 2.1 = )1.0( )1 + 1( + 1 = x ) 0y + 0x( + 0y = x ) 0y , 0x( m + 0y = 1=3y 2y 1y3x 2x 1x 0x
Oklahoma State - MATH - 2233
Math 2233.005 SOLUTIONS TO SECOND EXAM You must explain your work to receive full credit. 1. Given that y1 (x) = x-1 and y2 (x) = x2 are solutions to x2 y - 2y = 0. (a) (5 pts) Show that the functions y1 (x) and y2 (x) are linearly independent. W [y
Oklahoma State - MATH - 2233
!&quot; #$ % &amp;&quot; ' ! !( $) ' #$ , , %* / ! . !&quot; &amp; ! !( $+ &amp;&quot; '00'#$1!&quot;&amp;( /!'0 0 00 0$ / ! !&quot;!&quot;&amp;#02&quot;3 &quot;'&amp;4$! ' !(4$ !4$ 0#4$ # !00!+ $ 5 &amp;6 &amp;7 0 , ) *!&quot;&amp; 80 / ! !&quot; &amp;+) +* + 00) + 0
Oklahoma State - MATH - 2233
Math 2233 SECOND EXAM 10:30 - 11:20 am, November 8, 2006 SOLUTIONS 1. Given that y1 (x) = x-1 and y2 (x) = x2 are solutions to x2 y - 2y = 0 (a) (5 pts) Show that the functions y1 (x) and y2 (x) are linearly independent. W [y1 , y2 ] = x-1 x2 - x-1
Oklahoma State - MATH - 2233
! &amp;,&quot; *&quot;#$%&amp; '# (# ) &quot; $*)'*+* #$+ * ( ('(-($% &quot; (# ()0 )' * * * *****!*++ &amp; &quot;( )# * * * )$ )$ (*+* , % + *+ * #+** $ (#) ./# ( $ &amp; ,&quot;( &quot;./# ( $ 0 * + + $ )/ ( $ &amp; *# # ) -(1 * +* )$+)$-6($!
Oklahoma State - MATH - 2233
1|x| nl 2-x 2c + 2-x 1c = )x( 2y 2c + )x( 1y 1c = )x(y |x| nl 2-x = =1xd 5-x12 2-x -x1xd2xd2x51x- pxe 4- 1pxe2) 1 x( 4 x ) 1 x(x=1xd 2xd 2x) (p1x-]) 1x( 1y[ 1x)x( 1y = )x( 2y.noitulos lareneg eht enimrete
Oklahoma State - MATH - 2233
Oklahoma State - MATH - 2233
1)x( 2y 2c + )x( 1y 1c + )x( py = )x(y noitulos lareneg eht nwod etirw ot 2 petS ni denimreted py eht dna ,1 petS ni denimreted 2y eht , 1y nevig eht esU .3 petS , , xd ] 2yg1y[ W 2y + xd ] 2yg1y[ W 1y- = py :noitauqe laitnereffid suoen 1y 2y -egom
Oklahoma State - MATH - 2233
Math 2233 SOLUTIONS TO SECOND EXAM March 27, 2006 1. Given that y1 (x) = x2 and y2 (x) = x2 ln |x| are solutions to x2 y - 3xy + 4y = 0 (a) (5 pts) Show that the functions y1 and y2 are linearly independent. W [y1 , y2 ] = y1 y2 - y1 y2 = x2 (2x ln
Oklahoma State - MATH - 2233
1)x( 2y 2c + )x( 1y 1c + )x( py = )x(y eb lliw noitulos lareneG12y + xd 2y 1y)alumroF sretemaraP fo noitairaV(xd2y 1y2 .) 1- xx3 2 = g dna evoba denimreted 2y gnisu( noitauqe suoenegomohni eht fo noitulos ralucitrap a etaluclaC ] 1y[ 1y =
Oklahoma State - MATH - 2233
13x 2c + 2x 1c = )x( 2y 2c + )x( 1y 1c = y3:.nloS l'neGx = )x( 2x = xd 4x 4-x xdp2x =xd )|x| nl 4( pxe 4-x2x = xdxdx4- -pxe4x12x = xd-pxe2] 1y[ 11y=2y.noitulos lareneg eht enimreted ot redrO fo
Oklahoma State - MATH - 2233
Math 2233.003 SECOND EXAM 1:30 2:20 am, April 10, 2001 Name: 1. Given that y1 (x) = x and y2 (x) = x are solutions to x y - 3xy + 3y = 0 (a) (5 pts) Show that the functions y1 (x) and y2 (x) are linearly independent. W [y1 , y2 ] = (x)(x3 ) - (x) (
Oklahoma State - MATH - 2233
1.2 = )0(y sefisitas taht noitulos eht hctekS )b(;)2.2-=y ,4.0=x,]y[,)5/4(^y=)x(y(ffid(tolpdleifd .2 ;)slootED(htiw .1 sdnammoc eht yb decudorp tolp elpaM eht si woleB .5/ 4y= y noitauqe laitnereffid eht rof dlefi noitcerid eht tolP )a( .11
Oklahoma State - MATH - 2233
Math 2233 Homework Set 21. Solve the following differential equation using Separation of Variables. dy = xey dx We can explicitly separate the x-dependence from the y-dependence in this equation by multiplying both sides by e-y dx : dy e-y dx = xey
Oklahoma State - MATH - 2233
.)x(nis2x1= )x(ysuht si melborp eulav laitini eht ot noitulos ehT .0 = C ekat tsum ew oS 2 = 2 + 0 = 2 + )(nis 2 = )(y = 0 C C C 1 . C xfi ot noitidnoc laitini eht esopmi won eW x 2x + )x(nis 2 = C 1 2x 2x + sd)s(soc = C 1 x x s x2 2 2
Oklahoma State - MATH - 2233
Math 2233 Homework Set 41. Find an integrating factor for each of the following differential equations and obtain the general solution. (a) y + (y - x)y = 0 Suppose (x, y) is an integrating factor for this equation. Then y + (y - x)y = 0 must be ex
Oklahoma State - MATH - 2233
!&quot;# ! $ $ % !&quot; % % % % $ &amp;'&quot; ( ) %$ $ %&quot;)%$ % ) % % % % ) % ) ) %)% % )$ % % %$ % )$ % ) ) $ % &quot; &quot; % $ % %$ $ $ %* ! + ,% % % %* $ )&quot;%$%!--% % ) . -%------* ! + ,&quot;&quot;$/ %%* $ -&quot;$
Oklahoma State - MATH - 2233
12 8 1= = 1 48861 - 8 8 = =r 84 - 46 8 ,alumrof citardauq eht gniylppa retfa ,ro20 = 3 + r8 - r4 evah tsum ew oSrx 3 + r8 - r4 = rx3 + rxr4 - rx)1 - r(r4 = 02sdleiy noitauqe siht otni rx = )x( y gnitutitsbuS 1 = )1( y ,0 = )1(y
Oklahoma State - MATH - 2233
1n + 1-nx nan )1 - x( )c( nx a .n fo smret ni x fo rewop htn eht fo tneicffieoc latot eht yfitnedi ot ysae eb dluow ti taht os tsuj niaga xedni noitammus eht detfihs ew pets tsal eht nI0 =n0= n1-nx na)1 - n2( 1 - n x ) n a n + n a) 1 - n (
Oklahoma State - MATH - 2233
1i2 = x 2x + 4 2x + 40 = 2x + 4 nehw seitiralugnis evah htob = = )x(q )x(p snoitcnuf tneicffieoc eht ,esac siht nI 0 = 0x , 0 = y + yx4 + y 2x + 4 )c(1x4lavretni eht ni7&lt;x&lt;1 taht hcus x lla rof ,yltnelaviuqe ,ro 3 &lt; | 4 - x| x lla rof e
Oklahoma State - MATH - 2233
1i2 = x 2x + 4 2x + 40 = 2x + 4 nehw seitiralugnis evah htob = = )x(q )x(p snoitcnuf tneicffieoc eht ,esac siht nI 0 = 0x , 0 = y + yx4 + y 2x + 4 )c(1x4lavretni eht ni7&lt;x&lt;1 taht hcus x lla rof ,yltnelaviuqe ,ro 3 &lt; | 4 - x| x lla rof e
Oklahoma State - MATH - 2233
Math 2233 Homework Set 91. Compute the Laplace transform of the following functions. (a) f (t) = t Let f (t) = t.L[f ] =0te-t dtIntegrating by parts, with = e-st dt = - 1 e-st su du= t = dtdv vwe get te-t dt0=0vdu uv|0 -
Oklahoma State - MATH - 2233
! $ % ( )&quot;# &amp;' &quot; $ %* &quot;$ + # , # ! % / 0 $ % 2 $3% 4 $'% / %* 1 # . ( ! ! &quot; * * $ $ ( + # #$ % &quot; %$4/ 2 5 -$ + / 9 : $ 4 * &amp; &amp;78 &quot; % 6 %( #0 *78; $ 3%0/2 # &lt; &quot; 29 ! 2! % &amp;78 * 0 = $7 8 % &amp; 7 8 ( !$$7 8 &gt; 3?% &amp; 7 8 $7
Oklahoma State - MATH - 2233
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Oklahoma State - MATH - 2233
1.setanidrooc lacirdnilyc dna naisetraC ot lacirehps morf ) ,2/ ,1( tnio p eht trevnoC .7 .setanidrooc lacirehps dna naisetraC ot lacirdnilyc morf )1 ,4/ ,1( tnio p eht trevnoC .6 .setanidrooc lacirehps dna lacirdnilyc ot naisetraC morf )4,3,0( tni
Oklahoma State - MATH - 2233
2 + 2y + 2x )0,0( )y,x( mil1yx)b(3 + 2y + 2x) 0, 0(mil) y,x()a( .tsixe yeht fi stimil gniwollof eht etupmoC .4)x( nis22x0milx x*)d()x( 2nis) 1, 0(x0mily xemil)y,x(mil)c()b(y 3x.U) 1, 0( )y , x
Oklahoma State - MATH - 2233
1.&quot;rotcev ciarbegla&quot; gnidno pserroc eht esirpmoc taht srebmun fo tes deredro eht ot ylesicerp dnopserroc &quot;rotcev cirtemoeg&quot; a fo daeh eht fo setanidrooc eht ,lareneg nI .)2 ,1( tniop eht ta daeh sti dna nigiro eht ta liat sti sah taht enalp eht ni
Oklahoma State - MATH - 2233
1R t ,s,tv + su + 0p = p | 3R p = Pmrof eht sekat noitpircserp a hcus noitaton rotcev nI .enalp eht ni gniyl u ,v snoitcerid tcnitsid owt dna enalp eht ni gniyl 0p tniop elgnis a gniyficeps yb debircserp eb nac enalp a ;enil eht no tniop
Oklahoma State - MATH - 2233
1|v| + |u| |v + u||sdleiy sedis htob fo toor erauqs eht gnikaT2 22) v + u(= v + v u 2+ u| | | |2||||2|| 2|v| + v u2 + |u| = |v + u||v| |u| |v u| v u,suhTmeroeht gnidecerp eht yB : foorp|v| + |u| |v +
Michigan - EECS - 486
Homework 3 Assigned 08OC01 Due 19OC01 All work is completed individually. All diagrams must be drawn using Visio. Homework set is two pages. Each problem is worth 10 points. Point distribution is described for each problem.Object Diagram 1. A.