MATH 202 Final, Summer 1997

Wocy Cumcoy enesay Lo INO uarney aq OL ON FINAL EXAMINATION, MATH 202 August 25, 1997; 11:00 A.M.-1:00 P.M. Name: Signature: Student number: Instructor: Abdallah Lyzzaik 1. Instructions: e No calculators are allowed. e There are two types of questions: PART I consisting of seven subjective questions, and PART II consisting of seven multiple-choice questions of which each has exactly one correct answer. ¢ GIVE DETAILED SOLUTIONS FOR THE PROBLEMS OF PART I IN THE PROVIDED SPACE AND CIRCLE THE APPROPIATE AN- SWERS FOR THE PROBLEMS OF PART II. 2. Grading policy: e 11 points for each problem of PART I. e For PART II, 7 points for problem 6 and 6 points for other problems. e -1 point (penalty) for each wrong answer of PART II. © 0 point for no answer or more than one answer of PART IL. GRADE OF PART I: GRADE OF PART II: TOTAL GRADE/120: Not To Be Taken Out Reseve R- Part I(1). Find the general solution of the differential equation (5c + 1)?y” + 5(5a + Ly’ + 25y =0, by using the change of variable u = 52 + 1. Part I(2). Use Laplace transform to find the set of all solutions y(t) of the initial value problem ty” + 2ty’ + 2y = 0, y(0) = 0. Part I(3). One solution of the differential equation xy” —2(xt+1)y'+(c+2)y=0 is y:(x) = e”, x € (0,00). Find a second solution that is linearly independent with yin the interval (0, 00). Part I(5). Find the general solution of the differential equation ley) +(e ~ Sy =0. (2 > 0) dx Part I(6). Find the general solution of the differential equation y” — 2y! + 2y = e*cot x. Part I(7). Solve by the method of undetermined coefficients the difler- ential equation y” + 4y = sin’. Part II; Multiple Choice Problems. Part II(1). If t—2f(t) = [ f(t—r)(e — edt then the value of f(1) is (a) 0. (b) 1/4. (c) 1/3. (d) 5/12. (e) None of the above. Part II(2). The solution of the initial value problem passes through the point (a)c=n/4,y=r/4t1. (b) c= 7/4, y=7/4-1. (c)c=-7/4,y= 7/4 +1. (d) c= —71/4,y = 7/4-1. (e) None of the above. Part II(3). The solution of the initial value problem —zy' +y = (y+ 1)’, y(0) =0 satisfies (e) None of the above. Part II(4). The indicial roots of the differential equation 3x7y" + zy’ — (1+2)y=0 (a) -1, -1/3. (b) 1, 1/3. (c) 1, -1/3. (d) -1, 1/3. (e) None of the above.