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Unformatted text preview: Math 215/255 Final Exam (Dec 2005) Last Name : First name : Student # : Signature : Circle your section # : Burggraf=101, Peterson=102, Khadra=103, Burghelea=104, Li=105 I have read and understood the instructions below: Please sign: Instructions: 1. No notes or books are to be used in this exam. 2. You are allowed to bring a lettersized formula sheet and a smallscreen, nongraphic, non programmable calculator. 3. Justify every answer, and show your work. Unsupported answers will receive no credit. 4. You will be given 2.5 hrs to write this exam. Read over the exam before you begin. You are asked to stay in your seat during the last 5 minutes of the exam, until all exams are collected. 5. At the end of the hour you will be given the instruction “Put away all writing implements and remain seated.” Continuing to write after this instruction will be considered as cheating . 6. Academic dishonesty: Exposing your paper to another student, copying material from another student, or representing your work as that of another student constitutes academic dishonesty. Cases of academic dishonesty may lead to a zero grade in the exam, a zero grade in the course, and other measures, such as suspension from this university. Question grade value 1 12 2 12 3 14 4 14 5 12 6 16 7 20 Total 100 Math 215/255 Final Name: Question 1: [12 marks] Solve each one of the following firstorder initial value problems for a realvalued solution y ( t ) in explicit form. Also, determine the domain of definition for each solution. (a) y = y sin t + 2 te cos t , y (0) = 1. (b) 1 (1 t ) yy = 0, y (0) = 1. Solution: (a) Rewrite the eqn in the normal form y + ( sin t ) y = 2 te cos t . Thus the integrating factor is μ ( t ) = e R ( sin t ) dt = e cos t . Thus ( e cos t y ) = e cos t ( 2 te cos t ) = 2 t ⇒ e cos t y = Z (2 t ) dt = t 2 + C ⇒ y ( t ) = e cos t ( t 2 + C ) . y (0) = 1 ⇒ C = e ⇒ y ( t ) = e cos t ( t 2 + e ) . It is define on (∞ , ∞ ). (b) This eqn is not exact but is separable. yy = 1 1 t ⇒ 1 2 y 2 ( t ) = ln  t 1  + C ⇒ y 2 ( t ) = ln( t 1) 2 + C ⇒ y ( t ) = ± p C ln( t 1) 2 . y (0) = 1 ⇒ C = 1 ⇒ y ( t ) = p 1 ln( t 1) 2 . It is define on the interval (1 √ e, 1). Page 2 of 14 Math 215/255 Final Name: Question 2: [12 marks] Answer “True” or “False” to the statements below. Put your answers in the boxes. (20 points) (a) Suppose the Wronskian of two functions f ( t ) and g ( t ) is W ( f, g )( t ) = t ( t 1) which is zero at t = 0 , 1. Then, f ( t ) and g ( t ) must be linearly dependent functions. False. (b) The Laplace transform of the initial value problem y 000 + y 00 + y = 0 , y (0) = 1 , y (0) = 2 , y 00 (0) = 3 yields Y ( s ) = ( s 2 + 3 s + 6) / ( s 3 + s 2 + s )....
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This note was uploaded on 01/23/2011 for the course MATH 100,200,30 taught by Professor Dr.alejandrocortas during the Winter '10 term at UBC.
 Winter '10
 Dr.AlejandroCortas
 Math

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