MATH 202 Final, Summer 1996

FINAL EXAMINATION, MATH 202 September 2, 1996; 11:00 A.M.-1:00 P.M. Name: Signature: Student number: Section number (Encircle): 1 1. Instructions: e No calculators arc 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. e GIVE DETAILED SOLUTIONS FOR THE PROBLEMS OF PART 1 IN THE PROVIDED SPACE AND CIRCLE THE APPROPRIATE AN- SWERS FOR THE PROBLEMS OF PART II. 2. Grading policy: e 9 points for each of the problems 1, 3, 4, 5 and 7 PART I. 10 points for each of the problems 2 and 6 of PART I. e 5 points for each correct answer of PART II. « -1 point (penalty) for each wrong answer of PART II. e¢ 0 point for no answer or more than one answer of PART II. GRADE OF PART I: GRADE OF PART II: TOTAL GRADE: Not To Resarve Be Taken Out Resding Room Part I(1). Find the general solution of the differential equation (3x + 2)?y" + 3(32 + 2)y’ — 36y = 0, by using a change of variable of the form u = 3z + 2. Part I(2). Use Laplace transform to find the solution of the initial value problem ty” -y' =0, y(0) =0. Part 1(3). One solution of the differential equations (1 —2?)y” — 2ry' + 2y = 0 is yi(x) = 2,0 € (-1,1). Find a second solution that is linearly independent with y,in the interval (—1, 1). Part 1(4). Find 8 s? s+2 Part 1(5). Find the general solution of the differential equation xy" + cy! + (16x? — 1/4)y = 0. (x > 0) Part I(6). Find the general solution of the differential equation © y" — 3y’ + 2y = cose™*. Part I(7). Find the differential equation di 1 = =cosz — 3 sinztanz+ glsecz)y*. knowing that y = sina is a solution. Part II: Multiple Choice Problems. Part II(1). If t F(t) =cost +f f(r) cos(t — r)dr, f(0) =1 then the value of f(2) is (d) -1. (e) None of the above. Part I1(2). The solution of the initial value problem dx wy +2¢ ina = ze’; y(1) =1 passes through the point (a) c= ef /4 y= 2. (b) cae? 4 y=d. Ny = 3. (c)z=e (a) a =e? y = 3. (e) None of the above. Part I1(3). If y: = 2° is a solution of the differential equation xy" — Say’ + 9y = 0, and yo = v(x)yi(x), with v(e) = 1, is another solution that is linearly inde- pendent with y:, then the value of y2(e~2) equals (a) e/3. (d) 2e?/3. (e) None of the above. Part II(4). If cy" is an integrating factor of the differential equation a(x + y)dy — yda = 0 (a) m=2 andn=1. (b) m=2andn=~—1. (c) m=—2 and n=1. (d) m= —2 and n= —l. (ec) None of the above.