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240-2008&amp;2009-2-M10-April2009

# 240-2008&amp;2009-2-M10-April2009 - Kuwait University...

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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 ...
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