MATH 202 Exam 1 with Solution

MATHEMATICS 202 (Ist Semester, 1994 - 95) Time: 60 mins. — QUIZ L (closed books) Prof. Haidar FACULTY OF ARTS & SCIENCE; AUB Nov. 30, 1994 1. Solve the following DE's a) (5x2y + 6x3y2 + 4xy2)dx + (2x3 + 3x4y + 3x2y)dy =0 b) y+ 2y" + Sy'- 26y =0 ec) -x(y'B- y(y')2 +1 =0 2. Solve the IVP (y2sinx)dx + (1/x - y/x) dy =0; y(x)=1 3. Given one solution y;(x) =x for y"-xy'+y=0 What is the other solution ? Over what interval we are assured that there is a unique solution to the IVP: y"' + x2y" + (Inx)y' - (cosx)y = (2-x)"! ; y(l)=3, y(l)=-2, y")=-6 Support your answer with an explanation. Lt ( 1) b) 2) Solution Quizl 1994-95 (Sx2y + 6x3y2 + 4xy2)dx + (2x3 + 3x4y + 3x2y)dy = 0 we can conclude that N = xy is an integrating factor . (Sx4y2 + 6x5y3 + 4x3y3)dx + (2x5y + 3x6y2 + 3x4y2)dy = 0 OM ON Lyx + 18xty? + 12xy? Oy Ox => exact equation f(x,y) = [Mdx = x6 y? + x°y +x%y’ + g(y) 8 fy) _ ay => g(y)=0 = aye f(xy yaxry+xy+xty +e=c, y'" + 2y" + Sy'- 26y =0 m +2m? +5m—26=0 => m-=2isa solution (m - 2)(m2 + 4m + 13) A'=4-13=-9 m=-2+3i & m= -2 -3i y=ce* +c,e* sin3x + c,e-* cos3x x(y')3 - y(y')}2 +1 =0 ~1 acer . => y=cyx4(1/c2) x= -2/3 $1 2 4 “pp (y2sinx)dx + (I/x-y/x)dy=0; . y(m)=1 xsinxdx = me S y) dy . y —xcosx +cosxtc=y' +Inly| (separable) Lee 3) 4) ol er y= Vi J =x Ja (yn x Note : This integration is to be solved with the help of series expansion from Math 201 . The interval is : x>0;x#2 1=)0, 2[ he ImLoneg) agai should contain1 the coefficient should be determined .