ma366final10-2Soln

# ma366final10-2Soln - l l D> 9L4(7,k(1 W" L 2 0M C(0(1...

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Unformatted text preview: l l D > 9L4 / (7 ,k/ \(1 W" L) 2 0M C (0 (1) The following vectors X1 and Y1 are eigenvectors for a certain 3 X 3 matrix A corresponding to the eigenvalues 2 —i and —4 respectively. Find the general solution to the system X ’ = AX 5 pts. ‘ in real form. No complex numbers allowed! 7L + 1 1 x1 = ‘ i , 3/1 = 3 —2i ——1 Far mans/r/Msﬁﬂ 3-2, . t X ‘ e e- 00) , ' M, ' t {Hf ‘ ' d ) [ uL_ j vyv ’t ‘ mm Jre'hm 45 téwcvast , Lezisiwt! New .4 1» e‘i'cast. ' l-zltg'm-z; — lie“ (as—t I I (AWL 4 LL) 4L it (10% 1‘;th ‘3 i it 3 we {3 "t wag, t (Z {5 2%th “12w“. a CNN 4'59“" OWL " WW AW” 6! if?! £7! F. bl 2/ l 09 iv 5% z > pm 6’" ré +3” VLWX «1, we r: u» ﬁx . ‘ [ZN mama/elm“ if: a V; [Vt [§ 3’ 1 / a v ‘r M fhﬁgwmg FﬂjMf/féﬂ 9'3 are +1 s/cl ;» AW 2? Mt)’ (wit \$th 0 ML (Lgtifmtém «if C323 [3" J ‘3: M t ”9% , 1 Law . ‘1 Midi, J E? ii) I oi) (2) The characteristic polynomial for the following matrix 15 p(7")= —('r — 4)2 (7“ — 5) and the vector Yl is an eigenvector corre- sponding to 7" — —.5 Find the general solution to the system X’ : AX. 10 pts. FM Fri; : U‘LTL) his {A N/ mam ‘1": \$7M 43M ‘7sz :M‘rﬁ/ mozmemrg, by {vii/”.3. CA HYX o . , W é""1“”: 3 a a ' . [to -2 0 ~19 ,3 o , 92_[:3:1 ”ﬁg? w; 3“; 3 ﬂ =- 5* Ni: «1 al: [LE 3::\ (A'Jﬂyxiﬁw » ,1 5‘5 :3 L: (L: 91\$; “ 3 ”l4 £7 YEN/“1”] / New” Rm 1:: :3; l: , 111 4:1; "f , A :Q:% EDA—M; :l’l] 10> (219143“ 6 6 tier? 1 W111 : cit/(M’é‘l’l FM V 5’, 1‘ T / m hm”)? WM” 11-: that r 13‘” llifﬂ reﬁll \$553} atrial g: 2.4%, 1H3}; / U0 4 (3) Given that X1, X2 and X3 are eigenvectors for the following matrix, ﬁnd the general solution to X’ = AX. Hint: To ﬁnd 5 pts. the eigenvalue, compute AXi. 2 6 —3 1 ‘ —i —1 A: ——2 —-10 5 X1: -1 7 X2: 2 X3: 3 —6 ——12 5 —2 2 v 5 705 ﬁlm! «3'; “1 (A'r!1)Yl ‘ 0 [2W {an ”a? [, [:5 ,1 ”Ir “,9 c 5 . x _ l '5 ”’12.. {if n. "1"} 3:1 Xv}: €41 {.5 "L (9"?) i +4 : 0 ‘7 m H a” r .' A ‘ Wm”; a A”? l; ‘3 ”E 'l “Y a .3: “My? g§[u v e] «3;, 5;ij ”L 0 .. , o , '41 1! {Ml—knife: 0, _ 003(0)~€ [3;] f h “l girl" i Y 1 v (, “.7 - -1 «(w 5% [ 1;) 5 [if 4' "£1 34/ S" 0 l .4, .. «Hkﬂgdg» D g? MUH'e ['3] a“? v1 7 O ' // 04 f“! 75 mféfi .1353 lwﬂh’i’ 5% / (4) The following matrix A has characteristic polynomial p(r) : —(7~ — 5).3. 10 pts. (a) Find e“. (b) Find the general solution to X’ = AX. 5 2 1 0 O 0 5 1—401 (DU (5) A certain 6 X 6 matrix A has characteristic polynomial p(r) : —(7"——2)2(r~5)4. Let X be a generalized eigenvector for A corresponding to 7" = 5. Give a formula for etAX that does not require summing an inﬁnite series. Your formula should use as few matrix products as possible relative to the given 5 pts. information. LQ7‘ (/9 51)X:o» 15= Ill ‘31 :W 7,7 8+5]: €777) €(6’+5.Z)f @157 ‘ @777} (6) Find all singular points for the following differential equation and state which are regular. Don’t forget to justify your 6 pts. answers! \$2(2:1: — 5)3y” + 233(2x ~ 5)y’ + (a: _ 1);; = 0 (7’40" «Nu Pew}?! pumﬂi a}! my)? :9 a: 3L 0 W; WE can Ll) My: )5 3?? ﬁlm? 4574 f/sA-waﬂ‘f jjwiﬂfJ. MHCD Mitw ‘l’hH X I) }5 a maul/w qéwvli’eg PM“, ~ >4) (w mix) 4, rm (xx 9).? + (.4 {Big 9H) 70(4): ova/)7 .5704): X (2x 57°) 7 pm) . PM). i “ as] I a (Mr/Le WK), 7‘70 aw? WV) any ﬁmégy‘w [927, “9553?- K¢ ; 420 K b ,5 a [9574,1041 gm7m/a? jaw-m 3}“ (ﬂwél) ﬁﬂnw MM my 7% _ ”ﬂaming 71:79: Mir/huff / 5 l) a; :‘sz a3] £145.34 .7 yr {2354;7in «({K PM) \( (by g? 57“): x; , Vf‘éj-‘ff‘iryjﬁ / PLEQ‘O 7/77 ' 7 / If)”, M 7W; .‘i 9,1qu 9/4 {WMU‘V élbﬁmfé’y fﬁi'h”; ~ 71? Ere , - v €\V\LQ l §> 100 l I / if; (7) Substitute y 2 271—0 anm” into the differential equation 49” +. (m2 +/l.l)y ﬁlmy :.g and simplifyoo until youoo obtain an 0C)expression of the form 2%” +;?x” +21%” =0 71:? 71:? Where the exponent of cu in each sum is n and the question marks are explicit expressions. (You do not need to use exactly 3 summation signs.) Do not simplify further! 10 pts via “3 em; I], m , V): I?“ 19!?an I: £02 Viﬂmx / V‘ goi’llngﬁﬂxh 91’ 7" =l> H17”'génm DIM/9"? w M f? .2 ,m m” if” I, : §[email protected]ﬂ¥})(m’fi) 47?»;er . m:—). we Liv/I”: 5:1 [I‘M-fgﬂwa \$1.an aw: Br ”I :1) at)!" giwnXM’ . _ 91:17 W Mzm‘l MY”?! I ”g m 1 9L4 ’£,\_Mv1)amx 1 1 49 XV] " z UPOQWDEW n2; . 00 4(99 : €41) ﬂ?) ){P‘H n50 / (1“? Mt?! ~ A3VVH . // I m l I f ‘ // 4m / ﬂ Hwy/Imam / wt , if I :- ‘0 ,1” f? « 21 40179 at)“ // ’l m 14513 ”I 4;; /ﬂe€r ‘3 . ﬂ , D: 2:7 11'- (III’WI’PI‘Uﬂmin ~r Z m WM y” ”f a; “Will/is ii if: 1/;in LO / (7 (:E (8) In attempting to solve a certain differential equation, we sub— stituted y 2 22:0 anx" into the differential equation and sim— 10 pts. pliﬁed, obtaining ' . 2371(71— 1 )anx” 2+Z— —3n(n— 1) )anx “+2 —3nan:r” +2 —ana:“ — *.0 17,20 (a) Continue the solution process to obtain the recursion re— lation. E :3: in“) I) Mr) 1% W M: M . “’m‘” 00 00 M a 5mm) and“ 7, ﬂ 77WM-H)( M“) arm»): “:0 in”): rm} 9‘1: '2 : ﬁg 73 Um) (n+2) Mm)! %Em+om+l)0mx "’ 50 (””0414 when '1” 6454:1364; D ’7 1 ‘- Brash; {5 “m (“WWW “W 3) 5nUI-P) an ﬁner/2-5m :0 E (@007 I) + 50 +1 :lﬂwx’ an“ : iii/H!) (1H2) : (ran +2) 3w 0 (w) ' i ' \$0“me MM? 0:0,“)Ie‘” (b) Find the ﬁrst three non-zero terms of the power series expansion for the solution yl satisfying y1(0)— — 0, 111(0) : V6; 0% A”)! L 010-? @X 5; mic 4r: ‘ M it”): D / W”) E f VQIMDIQ / vi L7 5:79 a £2 ﬂint? ﬂaw? / 4,2? 5“ "’ 4 I’L/ , W ’“ mifTﬁ’ 5' Wk 1 ail , e ,, - , the M 133%? ij {WET}? M, 4 mew m ' it; t g (9) The following differential equation has a regular singularity at m : 0. 332(333 + 293 + 1)y"+ 20(4133 + x + 6)y' + (II? + 6)y = O. (a) Give the approximating Euler equation [900-)L5wyﬂ p40): 1 / 45’0” 4zl+x+é ,gzm ,. a Hg) ; ML ; r60)“é 96» H4? ﬂxﬁwaXéMﬁﬂhf éMLgy gnawi%‘i‘” MY! 8) x0 +9!” + 617“" (b) Give the indicialequation Jrlr , I WHO : t {N} X 2 (1+ gar? ,«4’ Nr)>< + emu +636 (F>—l’-+Z>r-HD)JC ' (_ r5+gr+é§ 5'0 30 HA" Mails/Ml ﬁmﬁwf‘rw lg Tzvférﬁr'é. =7?) (C) Use Theorem 5.6.1 on p. 289 of the text to describe the expected form of the solutions. Do not ﬁnd the coefﬁ— cients of the series expansions! F1+5frf6 ’0 (r—r35( n+9 :0 waQ WW4 H “'3: ”A: t SW hm mm“): x "5;: M A ,l em l“, — r3 : «was; a i ; «W WI?!“ W t , OM) anjﬂwl ﬁlxwﬂbw 00 4% 9 pts. 6 pts. 10 «K's (10) You are given that y(a:) = \$2 is a solution of the following differential equation. Use the method of reduction of order to ﬁnd a second independent solution. Other methods Will not receive credit! fly” - 3052/ + 4y = O.‘ U ewe WkW Vi? W3; mm team a”? a raga? , ﬁrsmxjg, 7* maxi" ‘, / WAN? Mggﬂtfﬂﬁéﬁﬁm‘gﬁ if” 6’0“" " : “V (Jig u’:14 M S f X“ y/x 1’ 1mm a" 41’” V] g" 11 (11) Use the deﬁnition of the Laplace transform to compute the Laplace transform £( f) of the function f (13) deﬁned below. Other methods Will not receive credit! 4 pts. 82t O<t<3 t: 7 _ ﬂ) {4, 3st. 6y {Ax M/é‘mH/‘vw A1715 16% gﬂﬁeg 7%AJ/120YM, I790 A Ugo) " J0 ﬁgjg’ﬁi' ”G: .5 N Wu i} €3£€'\$*J¢ ~16 / 4521““:- J g i «>0 % i5 e a; ’ t ‘ itaﬁybﬁfgf’édp’ﬁﬁj of if) #3 {M “" (24) J 3 ”'5 ‘ C249» 3 m * mi / + (:27 _) ﬁg) i0 \v ”5 3 (24M) __ Q L [+235 WW _ / 22's D“\$ “T “Ty/ 4 pts. 4 pts. 1; if)“ ( i (12) Find the inverse Laplace transform of 252+4 F(S)_s(52+4)' ' ’ " 3% MW?) MUM 2-: W5‘+H + (Bfwjg 9’9, 4&6? ) MIL/5%) (“=0 “3‘! 8:3»i Z/ f/g); .1.» + g ‘3 saw (13) Find the inverse Laplace transform of F(3) : W_ 3(82+4) M an); an“; a???) z ,L + 5 “g 53M :0 ”(5%) F ‘ I + (”049%, ;}(JC)» I“ M} Q'égﬂa) ﬂ ( Mgf-t‘) §c¢49 j ﬂy C(Ms;l{)( 450g596+§ ) .. i i fa H) : MxetH 1+ {9£{3~&»F03> (14) Find the inverse Laplace transform of €23 /F(S):(s+2)5' L64 3 6M ) Elf—W 3;) W (5+2); h6U-‘JLMT \$4 13 67 4 pts. ; ’2 I ’24; 99561“) :‘F(ff)):ﬂ:A 64g )6 42(1)) “/v {NE ”5 :)/ ’1’H2-t ‘f 7% 6 GM) LC (ha) 3mm) V ,L, ( VIM-L) Ca’umhﬁoy (7%)) ...
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