EXAM-3 MATH 261: Elementary Differential Equations MATH 261 FALL 2006 EXAMINATION COVER PAGE Professor Moseley

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1 EXAM-3 MATH 261: Elemetary Differetial Equatios MATH 261 FALL 2006 EXAMINATION COVER PAGE Professor Moseley PRINT NAME ( ) Last Name, First Name MI (What you wish to be called) ID # EXAM DATE Friday Ocober 27, 2006 I swear ad/or affirm that all of the work preseted o this exam is my ow ad that I have either give or received ay help durig the exam. SIGNATURE DATE INSTRUCTIONS 1. Besides this cover page, there are 10 pages of questios ad problems o this exam. MAKE SURE YOU HAVE ALL THE PAGES. If a page is missig, you will receive a grade of zero for that page. Read through the etire exam. If you caot read aythig, raise your had ad I will come to you. 2. Place your I.D. o your desk durig the exam. Your I.D., this exam, ad a straight edge are all that you may have o your desk durig the exam. NO CALCULATORS! NO SCRATCH PAPER! Use the back of the exam sheets if ecessary. You may remove the staple if you wish. Prit your ame o all sheets. 3. Pages 1-10 are multiple choice. Expect o part credit o these pages. There are o free respose pages. However, to isure credit, you should explai your solutios fully ad carefully. Your etire solutio may be graded, ot just your fial aswer. SHOW YOUR WORK! Every thought you have should be expressed i your best mathematics. Partial credit will be give as deemed appropriate. Proofread your solutios ad check your computatios as time allows. GOOD LUCK!!!!!!!!!!! REQUEST FOR REGRADE Please regard the followig problems for the reasos I have idicated: (e.g., I do ot uderstad what I did wrog o page.) Scores page poits score (Regrades should be requested withi a week of the date the exam is retured. Attach additioal sheets as ecessary to explai your reasos.) I swear ad/or affirm that upo the retur of this exam I have writte othig o this exam except o this REGRADE FORM. (Writig or chagig aythig is cosidered to be cheatig.) Date Sigature Total 100

2 MATH 261 EXAM III Fall 2006 Professor Moseley Page 1 The dimesio of the ull space of the liear operator L[y] = y + y that maps A(R,R) to A(R,R) is 2. Sice the operator L[y] = y + y has costat coefficiets, we assume a solutio of the homogeeous equatio L[y] = 0 of the form y = e rx. This leads to the two liearly idepedet solutios y 1 = cos(x) ad y 2 = si(x). Hece we ca deduce that y h = c 1 cos(x) + c 2 si(x) is the geeral solutio of y + y = 0. To use the liear theory to obtai the geeral solutio of the ohomogeeous equatio L[y] = g(x), we eed a particular solutio, y p, to y + y = g(x). We have studied two techiques for this purpose (attedace is required): i) Udetermied Coefficiets (also called judicious guessig) ii) Variatio of Parameters (also called variatio of costats) For each of the fuctios g(x) give below, circle the correct aswer that describes which of these techiques ca be used to fid y p for the ohomogeeous equatio y + y = g(x): 1. (2 pts.) g(x) = x 1 e x A B C D E 2. (2 pts.) g(x) = ta (x) A B C D E 3. (2 pts.) g(x) = si(x) A B C D E 4. (2 pts.) g(x) = e x A B C D E A) Neither techique works to fid y p. B) Oly Udetermied Coefficiets works to fid y p. C) Oly Variatio of Parameters works to fid y p. D) Either techique works to fid y p. ABCDE) Noe of the above statemets are true. Total poits this page = 8. TOTAL POINTS EARNED THIS PAGE

3 MATH 261 EXAM III Fall 2006 Professor Moseley Page 2 The dimesio of the ull space of the liear operator L[y] = y y that maps A(R,R) to A(R,R) is 2. Assumig a solutio of the homogeeous equatio L[y] = 0 of the form y = e rx leads to the two liearly idepedet solutios y 1 = 1 ad y 2 = e x. Hece we ca deduce that y h = c 1 + c 2 e x is the geeral solutio of the homogeeous equatio y y = 0. Use the method discussed i class (attedace is madatory) to determie the proper (most efficiet) form of the judicious guess for a particular solutio y p of the followig ode's. Circle the correct (most efficiet) fial form of the judicious guess for a particular solutio y p of the followig ode's 5. (3 pts.) y y = 2e x First guess: y p = Secod guess (if eeded): y p = Third guess (if eeded): y p = Fial guess. A B C D E 6. (3 pts.) y y = 3 si x First guess: y p = Secod guess (if eeded): y p = Third guess (if eeded): y p = Fial guess. A B C D E 7. (3 pts.) y y = 4xe -x First guess: y p = Secod guess (if eeded): y p = Third guess (if eeded): y p = Fial guess. A B C D E Possible Aswers for Fial Guesses. A) A B) Ax + B C) Ax 2 + Bx + C D) Ax 2 E) Ax 2 + Bx AB) Ae x AC) Axe x AD) Ax 2 e x AE) Axe x + Be x BC) Ax 2 e x + Bxe x BD). Ae x BE) Axe x CD) Ax 2 e x CE) Axe x + Be x DE) Ax 2 e x + Bxe x ABC) A si x BE. cos x ABD) A x si x ABE) A x cos x ACD) Asi x + Bcos x ACE) Ax si x + Bx cos x ABCD) Noe of the above Total poits this page = 9. TOTAL POINTS EARNED THIS PAGE

4 MATH 261 EXAM III Fall 2006 Prof. Moseley Page 3 Solve y" + y = 3x + 2e x o the back of the previous sheet. 8. (3 pts.) The geeral solutio of y" + y = 0 is. A B C D E 9. (4 pts.) A particular solutio of y" + y = 3x is. A B C D E 10. (4 pts.) A particular solutio of y" + y = 2e x is. A B C D E 11. (1 pts.) A particular solutio of y" + y = 3x + 2e x is. A B C D E 12. (2 pts.) The geeral solutio of y" + y = 3x + 2e x is:. A B C D E A) c 1 x+ c 2 B) c 1 e x + c 2 xe x C) c 1 e x + c 2 xe x D) c 1 si(x)+ c 2 cos(x) E) c 1 x si(x)+ c 2 x cos(x) AB) c 1 e x + c 2 e x AC) 3x + 1 AD) 3e x AE) 3x BC) 3 si(x) + cos(x) BD) 3 xsi(x) + xcos(x) BE) 3e x + e x CD) x + 1 CE) e x DE) 3e x + e x ABC) 3x + e x ABD) 2e x ABE) 3x + 2e x ACD) 3x + e x ACE) 3x + e x + c 1 x+ c 2 ADE) 3x + e x + c 1 e x + c 2 xe x BCD) 3x + e x + c 1 e x + c 2 xe x BCE) 3x + e x + c 1 si(x)+ c 2 cos(x) BDE) 3x + e x + c 1 xsi(x)+ c 2 xcos(x) CDE) 3x + e x + c 1 e x + c 2 e x ABCD) e x + c 1 x+ c 2, ABCE) e x + c 1 e x + c 2 xe x ABDE) e x + c 1 e x + c 2 xe x ACDE) e x + c 1 si(x)+ c 2 cos(x) BCDE) Noe of the above. Total poits this page = 14. TOTAL POINTS EARNED THIS PAGE

5 MATH 261 EXAM-3 Fall 2006 Professor Moseley Page 4 Solve the ODE y" + y = ta(x) I = (0, π/2) (i.e. 0 < x < π/2) o the back of the previous sheet. Also let L[y] = y" + y. 13. (2 pts.) The geeral solutio of L[y] = 0 is y c (x) =. A B C D E A) c 1 cos(x) + c 2 si(x) B) c 1 cos(2x) + c 2 si(2x), C) c 1 e x + c 2 e x D) c 1 x + c 2 E) r = ±i AB) r = ±1 AC) r = ±2i AD) Noe of the above. 14. (3 pts.) To fid a particular solutio y p (x) to L[y] = ta(x) usig the techique of variatio of parameters, we let y p (x) = u 1 (x) cos(x) + u 2 (x) si(x). Substitutig ito the ODE ad makig the appropriate assumptio we obtai:. A B C D E A) u 1 (x) cos(x) + u 2 (x) si(x) = 0, u 1 (x) si(x) + u 2 (x) cos(x) = 0 B) u 1 (x) cos (x) + u 2 (x) si(x) = 0, u 1 (x) si(x) + u 2 (x) cos(x) = ta(x) C) u 1 (x) cos (x) + u 2 (x) si(x) = ta(x), u 1 (x) si(x) + u 2 (x) cos(x) = 0 D) u 1 (x) cos (x) + u 2 (x) si(x) = 0, u 1 (x) si(x) + u 2 (x) cos(x) = si(x) E) u 1 (x) cos (x) + u 2 (x) si(x) = 0, u 1 (x) si(x) + u 2 (x) cos(x) = cos(x) AB) Noe of the above. 15. (3 pts.) Solvig we obtai. A B C D E A) u 1 (x) = si 2 (x)/cos(x), u 2 (x) = si(x) B) u 1 (x) = si(x), u 2 (x) = si 2 (x)/cos(x) C) u 1 (x) = 1, u 2 (x) = si(x) D) u 1 (x) = si 2 (x)/cos(x), u 2 (x) = 1 E) u 1 (x) = 0, u 2 (x) = si(x) AB) Noe of the above. 16. (4 pts.) Hece we may choose. A B C D E A) u 1 (x) = l(ta(x) +sec(x)) + si x, u 2 (x) = cos(x) B) u 1 (x) = cos(x), u 2 (x) = l(ta(x) +sec(x)) + si x, C) u 1 (x) = x, u 2 (x) = cos(x), D) u 1 (x) = l(ta(x) +sec(x)), u 2 (x) = x, E) u 1 (x) = 1, u 2 (x) = cos(x), AB)Noe of the above. 17. (2 pts.) Hece a particular solutio to L[y] = ta(x) is y p (x) =. A B C D E A)l(ta(x) +sec(x)) B) [cos(x)] l(ta(x) +sec(x)) C) [si(x)]l(ta(x) +sec(x)) D) [ta(x)] l(ta(x) +sec(x)) E) si(x) cos(x) AB) 2 si(x) cos(x) AC) [si(x) cos(x)] l(ta(x) +sec(x)) AD)Noe of the above 18. (2 pts.) Hece the geeral solutio of L[y] = ta(x) is y(x) =. A B C D E A) l(ta(x) +sec(x))+c 1 cos(x) + c 2 si(x) B) [cos(x)] l(ta(x) +sec(x)) + c 1 cos(x) + c 2 si(x) C) [si(x)]l(ta(x) +sec(x))+c 1 cos(x) + c 2 si(x) D) si(x) cos(x) + c 1 e x + c 2 e x E) l(ta(x) +sec(x)) + c 1 e x + c 2 e x AB) [cos(x)] l(ta(x) +sec(x)) + c 1 e x + c 2 e x AC) Noe of the above. Total poits this page = 16. TOTAL POINTS EARNED THIS PAGE

6 MATH 261 EXAM III Fall 2006 Prof. Moseley Page 5 Solve the ODE y IV 4 y + 4y= 0 o the back of the previous sheet. Let L:A(R,R) to A(R,R) be defied by L[y] = y IV 4 y + 4y. Be careful as oce you make a mistake, the rest is wrog. 19. (1 pt). The order of the ODE give above is. A B C D E A) 1 B) 2 C) 3 D) 4 E) 5 AB) 6 AC) 7 AD) Noe of the above. 20. (1 pt). The dimesio of the ull space of L is. A B C D E A) 1 B) 2 C) 3 D) 4 E) 5 AB) 6 AC) 7 AD) Noe of the above. 21. (1 pts). The auxiliary equatio for this ODE is. A B C D E A) r 2 4r + 4 = 0 B) r 4 4r = 0 C) r 4 4r 3 + 4r 2 = 0 D) r 6 + 4r 3 + 4r 2 = 0 E) r 6 4r 3 + 4r 2 = 0 AB) Noe of the above. 22. (2 pts). Listig repeated roots, the roots of the auxiliary equatio are. A B C D E A) r = 0, 2, B) r = 0, 0, 2, 2 C) r = 2, 2 D) r = 0,4 E) r = 0,2,4 AB) r = 0, 0,2, 2 AC) r = 0, 0 2i, 2i AD) Noe of the above. 23. (2 pts). A basis for the ull space of L is. A B C D E A) {1, x, e 2x, xe 2x } B) {1, x, e 2x, xe 2x } C) {1, x, x 2, e 2x } D) {1, e 2x } E) {1, x, x 2, x 3 } AB) {e 2x, xe 2x, e 2x, xe 2x } AC) {1, x, x 2, e 2x } AD) {1, e 2x } AE) Noe of the above. 24. (1 pt). The geeral solutio of this ODE is. A B C D E A) y(x) = c 1 + c 2 x + c 3 e 2x + c 4 xe 2x B) y(x) = c 1 + c 2 x +c 3 e 2x + c 4 xe 2x C) y)x) = c 1 + c 2 x + c 3 x 2 + c 4 e 2x D) y(x) = c 1 + c 2 e 2x E) y(x) = c 1 + c 2 x + c 3 x 2 + c 4 x 3 AB) y(x) = c 1 e 2x + c 2 xe 2x + c 3 e 2x + c 4 xe 2x AC) y(x) = c 1 + c 2 x + c 3 x 2 +c 4 e 2x AD) y(x) = c 1 + c 2 e 2x AE) Noe of the above. Poits this page = 8. TOTAL POINTS EARNED THIS PAGE

7 MATH 261 EXAM III Fall 2006 Professor Moseley Page 6 (9 pts.) The dimesio of the ull space of the liear operator L[y] = y + y that maps A(R,R) to A(R,R) is 3. Assumig a solutio of the homogeeous equatio L[y] = 0 of the form y = e rx leads to the three liearly idepedet solutios y 1 = 1 ad y 2 = cos(x) ad y 3 = si(x). Hece we ca deduce that y c = c 1 + c 2 cos(x) + c 3 si(x) is the geeral solutio of y + y = 0. Use the method discussed i class (attedace is madatory) to determie the proper (most efficiet) form of the judicious guess for a particular solutio y p of the followig ode's. Do ot give a secod or third guess if these are ot eeded. Put your fial guess i the box provided. 25. (3 pts.) y + y = si(x) First guess: y p = Secod guess (if eeded): y p = Third guess (if eeded): y p = Fial guess. A B C D E 26. (3 pts.) y + y = 4 x 2 First guess: y p = Secod guess (if eeded): y p = Third guess (if eeded): y p = Fial guess. A B C D E 27.(3 pts.) y + y = 4xe -x First guess: y p = Secod guess (if eeded): y p = Third guess (if eeded): y p = Fial guess. A B C D E Possible Aswers A) A B) Ax + B C) Ax 2 + Bx + C D) Ax 2 E) Ax 2 + Bx AB. Ae x AC) Axe x AD) Ax 2 e x AE) Axe x + Be x BC) Ax 2 e x + Bxe x BD) Ae x BE) Axe x CD) Ax 2 e x CE.) Axe x + Be x DE) Ax 2 e x + Bxe x ABC) A si x BE) cos x ABD) A x si x ABE) A x cos x ACD) A si x + B cos x ACE) A x si x + B x cos x ABCD) Noe of the above Total poits this page = 9. TOTAL POINTS EARNED THIS PAGE

8 MATH 261 EXAM 3 Fall 2006 Prof. Moseley Page 7 Solve y + y = 4e x + 20 cos(x) o the back of the previous sheet (3 pts.) The geeral solutio of y + y = 0 is y c (x) =. A B C D E A) c 1 + c 2 x + c 3 e x B) c 1 + c 2 x + c 3 e x C) c 1 + c 2 si(x) + c 3 cos(x) D) c 1 e x + c 2 si(x) + c 3 cos(x) AB) c 1 + c 2 e x + c 3 e x AC) c 1 e x + c 2 si(x) + c 3 cos(x) AD) c 1 + c 2 x + c 3 e 2x AE) c 1 + c 2 x +c 3 x 2 BC) Noe of the above. 29. (4 pts.) A particular solutio of y + y = 4 e x is y p1 (x) =. A B C D E A) ½ e x B) 2 e x C) ½ x e x D) ½ si(x) E) ½ x e x + e x AB) 2 x e x + e x AC) Noe of the above. 30. (4 pts.) A particular solutio of y + y = 20 cos(2x) is y p2 (x) =. A B C D E A) 20 cos(x) B)20 si (x) C) 3x D) 10 si(x) 10 cos(x) E) 10 si(2x) 10 x cos(x) AB) 10 si(x) + 10 cos(x) AC) Noe of the above. 31. (1 pts.) A particular solutio of y + y = 4 e x + 20 cos(2x) is y p (x) =. A B C D E A) ½ e x 10 si(x) 10 cos(x) B) 2 e x 10 si(x) 10 cos(x) C) 2 e x + 10 si(x) + 10 cos(x) D) 2 e x + 20 si(x) E) 3x 10 si(x) 10 cos(x) AB) 2 e x + 20 cos(x) AC) Noe of the above. 32. (2 pts.) The geeral solutio of y+ y = 4 e x + 20 cos(2x) is y(x) =. A B C D E A) ½ e x + 20 si(x) + c 1 + c 2 x + c 3 x 2 B) 2 e x 10 si(x) 10 cos(x) + c 1 + c 2 x + c 3 e x C) 2 e x + 10 si(2x) +10 cos(x) + c 1 + c 2 x + c 3 e x D) ½ e x 10 si(x) 10 cos(x) + c 1 + c 2 x + c 3 e x E) 3x + e x + c 1 xsi(x)+ c 2 xcos(x) + c 3 e x AB) e x + 3 si(2x) + cos(x) + c 1 e x + c 2 e x + c 3 x+ c 4 AC) e x + 2 si(2x) + cos(2x) + c 1 e x + c 2 xe x + c 3 AD) ½ e x + 2 si(2x) + cos(2x) + c 1 xe x + c 2 e x + c 3 xe x AE) e x + c 1 si(x)+ c 2 cos(x) BC) 3x + c 1 xsi(x)+ c 2 xcos(x) BD) 3x + c 1 e x + c 2 e x BE). 3x + c 1 x+ c 2 CD) 3x + c 1 e x + c 2 xe x CE) 3x + c 1 e x + c 2 xe x AB). 3x + c 1 si(x)+ c 2 cos(x) ABD) 3x + c 1 xsi(x)+ c 2 xcos(x) ABE) 3x + c 1 e x + c 2 e x ACD) Noe of the above. Total poits this page = 14. TOTAL POINTS EARNED THIS PAGE

9 MATH 261 EXAM III Fall 2006 Prof. Moseley Page 8 O the back of the previous sheet fid a recursio formula for fidig the coefficiets to a power series solutio about x = 0 to the ODE y + x y 2 y = (1 pts.) To fid the recursio formula for the power series solutio about x = 0 of this ODE we let:. A B C D E A) y = a x 1 N B) y = a x C) y = a x 1 D) y = E) y = 0 0 a 0 ( 1)x a x 0 2 AB) y = a 2 2 x AC) Noe of the above. 34. (2 pts) Substitutig ito this ODE we obtai. A B C D E 2 1 A) B) a ( 1)x a x 2 a x 0 a ( 1)x x a x 2 a x 0 C) a ( 1)x x a x 2 a x 0 D) 2 a 0 ( 1)x 2 a 0 x 0 E) a ( 1)x a x 2 a x 0 AC) Noe of the above. 35. (3 pts) By chagig the idex ad simplifyig, this equatio ca be chaged to obtai. A B C D E A) [a ( 2)( 3) ( 2)a ]x 0 B) [a ( 2)a ]x C) [a ( 2)( 1) ( 2)a ]x 0 D) [a 0 2( 2)( 1) a ]x 0 E) [a ( 2)( 1) ( 2)a ]x 0 AC) Noe of the above (5 pts.) The recursio formula for fidig the coefficiets i the power series is. A B C D E A) a 2 a B) a 2 a C) a 2 a ( 2)( 3) ( 2)( 1) ( 2)( 1) 2 1 D) a 2 a E) a 2 a AB) a 2 a ( 2)( 1) ( 2)( 1) ( 2)( 1) 2 AC) a2 a AD) Noe of these is a possible recursio formula. ( 2)( 1) Total poits this page = 11. TOTAL POINTS EARNED THIS PAGE

10 MATH 261 EXAM III Fall 2006 Prof. Moseley Page 9 MATHEMATICAL MODELING. As doe i class (attedace is madatory), you are to develop a geeral mathematical model for the mass/sprig problem. Take positive distace to be dow. Suppose a mass m due to its weight W = mg where g is the acceleratio due to gravity stretches a sprig of legth L a distace Δ. If the mass is stretched dowward a distace u 0 from its equilibrium positio ad give a iitial velocity v 0, develop a appropriate mathematical model to determie the subsequet motio (i.e. to fid the distace u(t) from the equilibrium positio as a fuctio of time). Assume that the air resistace is c i feet slugs per secod ad that the sprig costat is k i pouds per foot (or slugs per secod squared). Assume a exteral force g(t) i slug feet per secod squared. 37. (1 pt) The fudametal physical law eeded to develop the model is. A B C D E A) Ohm's law, B) Coservatio of mass C) Coservatio of eergy D) Kirchoff's law E) Newto's secod law AB) Noe of the above. 38. (1 pt.) The relatioship betwee Δ, k, m, ad g is. A B C D E A) k = Δ m g B) k Δ = m g C) k m = Δ g D) k g = m Δ E) m Δ = k g AB) Noe of the above. 39. (3 pts.)the mathematical model for the mass sprig system whose solutio yields the distace u(t) from the equilibrium positio as a fuctio of time is. A B C D E A) mu cu ku g(t) B) mu cu ku g(t) C) D) mu cu ku g(t), u(0) u0 u(0) v0 E) AB) mu ku g(t), u(0) u u(0) v AC) Noe of the above. mu ku g(t) mu cu ku g(t), u(0) u0 u(0) v (1 pt.) The uits for the ODE i the model are. A B C D E A) Feet, B) Secods C) feet per secod, D) feet per secod squared, E) Pouds, AB) Slugs, AC) Slug feet AD) Noe of the above. Total poits this page = 6. TOTAL POINTS EARNED THIS PAGE

11 MATH 261 EXAM III Fall 2006 Prof. Moseley Page 10 For each questio write your aswer i the blak provided. Next fid your aswer from the list of possible aswers listed below ad write the correspodig letter or letters for your aswer i the blak provided. The circle this letter or letters. Fially, circle your aswer. MATHEMATICAL MODELING. Cosider the followig problem (DO NOT SOLVE!): A mass weighig 4 lbs. stretches a sprig (which is 10 ft. log) 2 iches. If the mass is raised 3 iches above its equilibrium positio ad give a iitial velocity of 5 ft./sec. (upward), determie the subsequet motio (i.e. fid the distace from the equilibrium positio as a fuctio of time). Assume that the air resistace is egligible. Apply the data give above to the model you developed o the previous page to obtai the specific model for this problem. DO NOT SOLVE! 41. (2 pts.) The sprig costat k i pouds per foot (or slugs per secod squared) is. A B C D E A) k = 2 B) k = 24 C) k = 4 D) k = 2/5 E) k = 10 AB) k = 5/2 AC) Noe of the above. 42. (3 pts.) The specific mathematical model for the mass sprig system whose solutio yields the distace u(t) dow from the equilibrium positio as a fuctio of time is. A B C D E 1 u 2u 24u si(t) 8 A) 1 B) u 2u 24u 0 C) 1 D) u 2u 24u 0, u(0) 3 u(0) 5 E) u 24u u 24u si(t), u(0) u(0) u 24 0, u(0) 1 u(0) AB) AC) Noe of the above. Total poits this page = 5. TOTAL POINTS EARNED THIS PAGE

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