{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

240-2008&2009-2-M10-April2009 - Kuwait University...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Kuwait University Math. 240 April 13, 2009 Dept. of Math. & Comp. Sc. First Midterm Duration 90 Minutes Calculators and mobile phones are not allowed. Answer all questions. Each question is worth 5 points . Solve the following equation (1 + 3f) (1 — sin 2:) da: + 2(cosx)(tan“1y)dy = O. . Find the general solution of the equation (x—2y+5)dx+(2m+y)dy=0. . Solve the differential equation y(y—2x+cosm) (133+ (Iy+21ny)dy=0. A Find the orthogonal trajectories of the family of curves m2+cy2 : 2. . Solve the initial value problem 3323/ 2 xy — y3 In at; y (1) = 72. OO 99 Math. 240 First Midterm Solution April 13, 2009 2tan'1y 1 +y2 lnil + sinxl + (tan‘ly)2 = c. l. (secx — tanx)dx + dy = 0 => lnlsecx +I;anx|+lr1|cosx|+(tan‘ly)2 = c 2. x~—2y+5 =0and2x+y=0 =>P(—1,2) =>x=u—1andy=v+2 d 2dr tdt _ . _v y—2 “t4l+1—I?2‘+TI72’“°’ "7 ln|u| +2tan“t+ %ln[1+t2l= k :> ._1__Q_M_flV_ 2.1. :“Tzi 3' M(6y 6x) 2,10,) e y (y—2x+cosx)dx+ (x+ 21;”) )dy = 0 :> xy—-x2 +sinx+(lny)2 = c. 5. y’—%y = (—%)y3 (BE) => tl+32fl= Z—chlzli (LE); 1:)72 xly-Z = 2I1nxdx+c => xzy“2 = 2(xlnx—x) +c. x=1andy=-2:>- um AM 99 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern