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Master MATH*2270 Test 1: Key Strategies and Solutions

School
University of Guelph**We aren't endorsed by this school
Course
MATH 2270
Subject
Mathematics
Date
Dec 12, 2024
Pages
11
Uploaded by ChancellorSeaLion4237
Test 1 MATH*2270, Fall 2024 October 19* 2024 from 3:00 — 4:15 PM Name g@]b{;@'/'oflS Student Number This assessment will be marked out of 49. A calculator and the provided HanDEbook can be used while writing this assessment, but no other printed or virtual material or electronic devices may be used. Keep your eyés on your own assessment and do not communicate with other students during the assessment. Academic misconduct will be treated seriously. Ensure that your answers are written dark enough for the scanner to pick up what you have written. Nobody can leave the room during the first 30 minutes or last 10 minutes of the assessment. This simplifies signing everybody in and the collection of tests. Unless it is in the last 10 minutes please raise your hand and a TA or myself will collect your assessment. If you need to use the washroom or need assistance during the assessment please raise your hand and a TA or myself will attend to you. You will be asked to leave your phone at your desk. There are two extra pages at the end of this assessment for rough work or extra space. If you use it for your solutions, please indicate this clearly in the question so we know where to look. Do not detach these pages or remove the staple from the test. This may result in errors during scanning. Take a deep breath and believe in yourself. Good luck :)
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MATH*2270 Test 1: 3:00 PM Version Part A: Quick Answer (8 Marks) Read these questions with care. Please place your final solution in the boxes provided on the right side of the page. Each question is worth 1 mark. Only the final solution will be graded for these questions. No partial credit will be awarded. Al Wha’cistheorderofi d2_y e dy 4+ 2 = cos (z)? ' dz \ dz? dz ) TY TR ol 2 5 A2. Ts cos(z)y” + 3 — zy = €® linear or non-linear? L/'flé’af "+ 2In(z)y = 2csc(In(z)) is a linear DE in standard form. [Fadse A3. True or Falseg% A4. True or False: 92%y + In(z)y fimgf,: 0 is a homogeneous DE. Falee A5, The equation y(73) = y™cos(72z) requires how many initial conditions to generate a unique solution, assuming one exists? +3 Page 1 of 10
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MATH*2270 Test 1: 3:00 PM Version A6. What substitution would you want to use to solve j—z — tan(In(z))y = v/ csc(x)y? rfl T ey = €sclOy e }—) Be yowllls AT7. Provide the equilibrium solution(s) for g—i =1In(y — 5). If none exist write “DNE”. m(W =0 So L{—-S_=j— O:\(\(G‘S—> %:‘ é. ‘-{ = 6 = D= In(.\\ O =0 d A8. True or False: every separable DE < f (y)% = F (a;)) is also exact. - fcx) + FL\OM:O / dx [ reee: e o W e M N M\\ =0 N K= O Page 2 of 10
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MATH*2270 Test 1: 3:00 PM Version Part B: Short Answer (12 Marks) Read these questions with care. Please place your final solution in the boxes provided on the right side of the page. Each question is worth 2 marks. If you are correct, only the final solution will be graded. If you are incorrect, you may receive one mark for a single minor error and/or demonstrating the correct process. \1“ ” X\,(‘ = 2x* ¥« B2. If y(z) = C cos(z) + Cyx is the general solution to a DE with initial conditions y(0) = 5 and 3'(0) = —2 what are the values of C; and Cy? Clearly label which is which for full marks. ({(0) _ 5 : C,(\) ‘O !C,\‘- 5‘ "{‘= - SN G (,('(,O):'Z': O +cCz gie 5 Cg = =2 s [Ez=-2} 1 B3. Given the IVP y" + In(z + 2)y' — Y= v/1 — 2 with initial conditions y(0) = 5 and x N y'(0) = 9 what is the largest interval for which a unique solution is guaranteed according to the existence and uniqueness theorem? If there isn’t one write “DNE”. SAon (g+2) —> X&(-2,%0) ' — xeR\xd5 XS - X — xelo, 1] xe (-2, 1) I e Al\ co,\,g::t‘«ows +he 2 Xy Fow? a%;?@JS Lheorfm onlv athets, bt Page 3 of 10 open br loyed bmkels il aceept clode for Nnere O3 narks.
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MATH*2270 Test 1: 3:00 PM Version d B4. What are the constant solutions for —2 — zy® + 22y — 3527 Assume y = f(z). dz X (117‘+ 2\1 ‘35) X (u\+ U~ D ] Y \»\:~’:l- . \1:~5 : dy _ B5. Solve: P aeds 12 # = ‘ZXZ 4’36)( AxX ilt_d’x - g IZ;('LI-BQ)( A X AX _|2x% £ 36xT 4 C 2 = L‘)(S-I- llxz~ + C_ B6. Provide an exact DE that has the solution p(z,y) = 42% + 2y% = 2. ¢ , 9 o J x 37 d X Bx * "ty%(!co Xx”"‘\{%‘i&‘o Page 4 of 10
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MATH*2270 Test 1: 3:00 PM Version Part C: Long Answer (29 Marks) For each of the following, please write your solution in the space provided, showing all your work. Partial marks may be given in certain cases, so please attempt all questions! d —2 Cl. (8 marks) Provided the DE kA de y—4 Hint: this is extremely similar to a question from the FYTT. (a) (2 marks) Determine the equilibrium solution(s) to the DE. dy . p= Y-2 R = ri'leem 501,“}:0/\ - L/{:Z s an @W:Z/‘b s ,,md,e/.'ned ond us s nNot en . ' ortenn Sohutron wit | @7' Mot fég,w'nocl. (b) (3 marks) Determine the interval(s) of y on which the solutions are increasing or decreasing. < ' { > o Seludvons v nereoasd |'w.5 o ) Z\— ’!( y €& (_-w,zfl)t"t,w \{ o B ¢ " pL('Gf@MI‘Hfl 4 - " (c) (3 marks) Determine the interval(s) of y on which the solutions exhibit quickening or slowing growth /decline. Awass )7_ NQT’OHVQ - A DU e - g R Z— = - - 7 L ..'-}) C,,I.— dv 1 1] 1 ‘;\w Swe o’ BM' M@'(:\'Md 0/%' Page 5 of 10
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MATH*2270 Test 1: 3:00 PM Version dy —2zy Direct t""'&“fir“"on L o C2. (6 marks) Solve: i 5= 215 ,Sepé’fa'bée ¢ X A1 4+ 2x q = Lineor’ AX KT+ S Z X X — @ X¥s 4 /_7-—— g o M= ’ /Z,mx% <) s | XEF ST YTy Foetor 8° L - vop Abs. M” )(7'4'5 vop KOS _ \n\)(7-+5\\ D (xreepy 4 24 y = ol Faster Selubion: B ! é‘tf“ -5 = Z2xy Ax P © N (Gres)y) = Bx7+28 | pugiply oy XS @ T Séaé»fizécl& | (,)(%S‘)Q@fr - S(K*+5) < -2%\ Lesrpae = e ) 5(X*+S) G*+s)y = 2 x*r25x *C IQ(L*S)%!‘*ZW . \/ = / No“f—e Produd’ Rude & Tge Exrae 5 Page 6 of 10
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MATH*2270 b Test 1: 3:00 PM Version Asing Vo mHUO. C3. (7 marks) Solve: 4ymj—z — 1% = 49/° . _— Hint: there are a couple good approaches to solving this problem.'D\Hfd~ Tutheqrokion ¢ X -fiéi— = _l:,_y_?::_t—)s—?_—‘ ()F forw ?(Klfl\ Wh@(é dcwl« é@P@%L@ ( X i . "i\{X &(Xu‘) '\"4:‘2’/3 L inear 7X q L; andlor - Exouk T 5;)03 . Cor Alwort Eowr”? @ IS Subshtukion? VIXEN = L\LV‘XB’L*"L Alve) Ylvx) A Vikev s e 2 b v x* ViXev = l:\_\l—-—i:j;‘ WV V= Uil (v . e V'K = By sl (M) U\ EEVe ko= Bl T * W € ., - b g : > vy =% ‘»Z\v\\x\ L C @ Recoll =X~ \r:,iL i % rC (VKT - i"/f_l«\l)fl*c - X oR S ’ 2 l = 0 ( = X (2 @ LL\/_;—— = \ \)(\ ew- l(—-—- —-\MX\*—CX 2v-T = InIXIFC o vrzginiXl 4 % Page 7 of 10
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MATH*2270 Test 1: 3:00 PM Version C4. (8 marks) Solve: dy _ _—ap® Direct Ivd—c‘j(w(-fm 7 X X\15+ (X"L{"Hl)% =0 (,,'nem;?x)( A Exovuk : , " 4 " Pimo s+ Exoct! M\{ - SX\{ > Na-l-— Exact: Y AxXy Page 8 of 10
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MATH?*2270 Test 1: 3:00 PM Version This page 1s intentionally left blank. If you use this page to continue a solution, be sure to make it clear that your solution continues on this page so we know to look. Ts the DE Erxont o Mudit Ecuct 'n @D 2 = P S UL e o M: —& ,L( - -./I—;—- N btyfx L x B v— - Swo. Page 9 of 10 couli e Solued st = sF - 5,(»«/% ?We’ 7/; Were osked k2 &
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MATH*2270 Test 1: 3:00 PM Version This page is intentionally left blank. If you use this page to continue a solution, be sure to make it clear that your solution continues on this page so we know to look "{‘1)(4,)( - Yy ® t | AV)oH/mr solukion to C3! —ad/wde by Y- L{Ki‘(» ’,)fi— = ‘4\1 dx Y . s A Yx dy. "}\{ g D qx \f = diviae b\( yx. X’zl - — 4 - XA Ay = ,)-S-— - - B@\moul/l"s yi =1 ~ % X vo- VT ) ; ' stomcdorck [0+ ' 2‘\/‘:_2_-_&___)/6/}@“(1% V- ,x_ . < 3—,7(0\* thoo 5C ¢ 0. @ /as e ,Z\J\\X\k- R ,zy‘ D h U ¢ Soluhon ‘ Ux = \ S=ax SO , Y oy & \V\\x\ i ! Z, Page 10 of 10 X S AP e O
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