solution for practice exam I

Solution for - Practice Exam 1 Math 244 Diff Eq.s Prof Tumulka 12345678910112 Name Given formula The general solution of the first order linear

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Unformatted text preview: Practice Exam 1, 2/15/2012, Math 244 Diff. Eq.s, Prof. Tumulka 12345678910112 Name: Given formula: The general solution of the first order linear ODE y’ + p(t)y : g(t) is y(t) = ~— (/ Mt) g(t) dt + c) , W) : 61pm dt Calculators, books and notes are not allowed. Good luck! 1) [worth 4 points out of 100] True or false? The global truncation error in the Runge— Kutta method has an upper bound proportional to (A05. False , it Is promfhonal +0 (45%)“ 2) [4 points] In which case does one say that the DE y” + p(t)y’ + q(t)y : g(t) is homogeneous? (A) A en 31 f ) 3’0 3) [7 points] a) What does the “superposition principle” say? an: gin/<4 flaws +0 0K (wear, hamymews eihqatigu gn’rpWaUEHLifig’ Mm my (meow rambmflcn Cigtfagle 11$ .5: fifiaimu b) To which of the following DES does it apply? (1) y” — 4y : sint El yes [X no Mi homoyuqmg : (ii) y” + (sin t)y’ — 4y 2 0 E yes El no (ill) y” — 4y’ + sing 2 0 El yes a no 7107" Ilfleaf 4) [5 points] Determine the longest interval in which the IVP (t— 1)y” — 3ty’+4y = sin t, y(—2) = 2, y’ (—2) = 1 is certain to have a unique solution. Do not attempt to find the solution. Q9 1'} ~ w! e efiucp‘fmn (713 a H - {if 'k i 3 SIM“ if.) Si LL [I . 1L" f?! '3’! M ’ [—7 a/ Org/5‘1 ML (“1"”“9‘42 at 15:). 0312pr9 micrvq 3;; (' 0", D a“! (i; M?) +0 2 _ i 2 Is 17! (40,1), go up 1 tidal/Q, 01101423,“! Eli/«iron an {-wl”) 91:50. For the following problems, your answer must also show your work. 5) [12 points] (a) Find the general solution of the DE 3/ + (1 /t)y = 300s 2t, t > 0. (b) Find the solution of (a) with initial condition y(1) : 1. Am ab; 1 1L 3 a < u w J; l, \- m— ‘éd‘i W all 5e :nieifoflllg- 0 ’5 t M ’6 “ 5’ : HI 55' '6?0 (.6: alt/01.5: z i figpnl‘l" J’s‘mlfd} “9f alvfiCofiz/éé. V‘l ‘5 co > ‘5 CM 16(4 I? ) elm/‘3 vagina .1 ifmz‘b 4'3“ Z-tr-gi- 2' 9% as f 7 E: Shah) +' 3L~/(03(2) ;3C : a C; /' ESInlzé- 3. (03(2) r34»; writ {2‘} g: Emu +3 Cog21‘ +— _’_.._.“H m “’5 “H” 4: . 6) [12 points] Solve the DE (1 + $3)yy’ — $2 = 0. 3 l: 2 . ’; x‘?’ gigs, fi (Mdaél x a is; j) 901x / s 2 € 2.. , ‘ l3 ’ jltxg 0’” € i’7éjf0’“’> %‘;SL(A/U/*C’ éfln/W/fl. H= l+x3 Aw szo'x ’ _ 21 2 , w) é’ ~§~Qn/’+X3/+C imp/my; gt 3 i“ magma/i 7) [12 points] Consider the DE y’ 2 fly) with f(y) = ay — bfl, a > 0, b > 0, Sketch the graph of f versus y, determine the critical (equilibrium) points, an classify each one as asymptotically stable, unstable or semi-stable. Draw the phase line and sketch several graphs of solutions in the ty—plane. 61/156“ magi Rng/Lfia ‘ Ty l/‘L/a; Vzvfijzc‘? :9 phase lime m 912:2” 31?‘ 19%): (“9% V‘Xa-ibflzl ’ g’éfafzvb)(a%gffi) : /aa'/z—5)(af-£: ‘1”;ieihom pi; whm 46:; Va 4; i Q of“ aging]: A: g *7" 5.21 0' r g : $3. Vat a r a yé'béo a2 ° m, W’i a kayaz Unikkifi figusfilbflum - (Adi: CoACovwi ,’ ""‘WL ' a z z y? W «fr; 2%? 'é—a)(a{:%7‘é-w=Maxis” >0 53L: (EL/iv t‘ P” z Conan/t “49' 1‘». a f a,» 0' » Mix” a' o“ 6’ a” x34 a?)‘*A>(«(§)’*v:9:(v%—b)(e%) <0 O 4 CE. {V g AL! ‘ Concave (yield/Itfl a a? an: és)(«{§)"’%é):(§~—b>(2~g zo Cquv‘g (4’). a: ‘2‘ 8) [12 points] For each of the following DES, determine whether it is exact. If it is, find the solution. a) (xlny +my)dx + (yln$+ xy)dy = O, 33' > 0, y > 0. b) cosa; + e2yy’ = 0 c) + 2x)da¢ + (Ina: — 1)dy : 0, :17 > 0. M: k a r a) N 3)”ng K00. M; f- lx n04” exam” 3. fix fix? N; ; "CL 1*; /l/: 92} Z J 2 %(x,é’)=~3mx + :92] Mu “(33:3 # ’9 Mi)" “2" a; é“ ‘2 J a, C'fnem’ Solufion 2W” § S’nx’Léé'zJ: c) I145 3%;er M 2 i- x 3 X #/ N‘an~/ ,vx ;.fiV/€ch N; 243:“! » k/Xg):ggnx+z»¥ @fi_ , ft? )=~I aux}- 3a A Rn): +hfg) j c; C? 9) [12 points] Find the solution of the IVP y” + 53/ + 3y 2 0, y(0) = 4, y’(0) = 0. If r (4' (I; 2 ff. 5 I; c I g If e I ’ have‘ P/ . m ’w’ +3220, We (777/ y ; Gum/«aim Jenna/6‘ Vic“+ g 4 1+ , ,- 1L re *3: /0 ¢ 1 (r k>§r*3)c/+fy©- Z Jlk/x—c/(sm ’§:}:JB 7 r+fr+g;@ —/>{‘; W: -. Z Z, ,S'I’QB ’ W 5” C16 2’ é “5162” - V In " "m y ' {gig—we L +4“(;~§-\@ “we x ’ 1 Cl)@ L _ mm, «a g 7 a 6’5ng (gwrmwswm ’ 279/37 ’ ' 90M. 31(6) . TLC, +£75.53 (L30. a {WEE *{"5'\/§?)Cz=‘o- J [j 2819:- Vf§+fl37j -9C/= igjgé I , 10) [8 points] A mouse population satisfies the DE dp/dt = 0.5p — 450. Find the time at which the population becomes extinct if 19(0) : 850. l A: fig i 0L0 if” 3‘" _. , ’ , .L \ 1+7" Hp goo) 5?ij “flat ‘3' “PM” 154% =9 p-qoo=4ez* '5) P: Ae%t+q00 /)(o):8§O ; PG): -foeéé k P“): 4 4400 :'§0 ( r v H, ’ p e) §0e2 +900 v0: =2 6700;5‘042“ ,5) 18:67“? i+e1n095=9f= MM?) 11) [12 points] A body of mass m is projected vertically upward with an initial velocity v0 in a medium offering a resistance klvl, Where k is a constant. Assume that the gravitational attraction of the earth is constant. Set up a DE for v(t) and solve it for v(0):'uo. M, d7 g (Asgumg, fo3Jnv¢ JIVfCI'WFK lS downwh’U/S) i/ m - my “g'kv 3 34-8st- V/o)= vo_ Vlokv dv )4 k ° ; ~ .717— —v w- "“ ~— :: ,. m ( = ’4 ' air 9 M" Maw). V”) ‘3- -A “5‘ 0 ah/ :* k : = W1 -L‘: 21 ’V {Edi )VH') I r -63 M- L} j M k m 'un /r3'vl= e "‘“C V: £1 ~ '53 Kg, Ac ...
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This note was uploaded on 02/27/2012 for the course MATH 640:244 taught by Professor Ming during the Spring '09 term at Rutgers.

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Solution for - Practice Exam 1 Math 244 Diff Eq.s Prof Tumulka 12345678910112 Name Given formula The general solution of the first order linear

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