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240-2008&amp;2009-1-M10-November2008

# 240-2008&amp;2009-1-M10-November2008 - KUWAIT...

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Unformatted text preview: KUWAIT . UNIVERSITY Department of Mathematics & Computer Science Math 240 ' 11 November 2008 Intro. Diff. Equations MidTerml Time: 90 minutes m . Calculators and cellular phones are not allowed during the exam. 1. Find the general solution: [4 pts each] (a) (2x+y+3)dw+(\$+2y+12)dy=0 (b) tlntdu— (u lnt+t)dt= 0', where t > 1. (c) yy’ + y2 cota: 2 cscz. t yem_ely may —'ejx‘ (d) y (e) (2y+3\$y2) d¢+ (1+2m2y) dy = 0. What is the solution with y =2 when m = 1 ? 2. Given the family of urves 312 2 ca: — 32:2 {ﬁnd the orthogonal trajectory which passes through the point (E, l}. ' [5 pts] 00 CO SOLUTIONS 1. Find the general solution: (a) (23+y+3)d:t+(a:+2y+12)dy || c: The lines intersect at (2, 77); take 3 2 u + 2 and y : v — 7 so that (2u+v)du+(u+2v)dv20 whose coefﬁcients are homogeneous. Put 1} = st to get 2 1+2t 21 1? Fri 1 2t dt:0 —d —dt: (+ + )u+( + )u -> u u+1+t+t2 0 —> lnu2+ln(1+t+t2):lnc —) u2(1+t+t2):c —> u2+uv+vg=c Back to my and you’ll have [\$—2)2+(x —_2) (y+7)+(y+7)2=c. (b) t lntdn— (u. lnt+t)dt 2 0, where t > 1. Write this equation in the form for which the IF becomes 1/15 and l 11:15]. dz+ct=tlnllntl+ct tzlnz (c) y y’ + y2 cot .T = csccr. This fellow is a Bernoulli: y” + y cot I = 030 any—1. Put 1; = y2 , then ef+2v cota3=2 cscs: IF becomes sings: and the generai solution becomes 1) = y2 = CSCQLE fairin cscz.dz+ccsc2:c: —csc:r: cota:+c cscgsc Z yew—way d ’=—————. l) y xW—ﬂ This is (y 693 — 89) (1:17 — (a: ey — 6"") dy = U which is exact. ARE—ye” ey > Fuyei‘ weylhﬂt) > Fyze‘E—mey+h’(y) —> h'(y):0 The general solution becomes 3} e?" — may : c. 99 (e) [2y+3wy2) dw+(;r+2zt2y) dy = 0. What is the solution with y : 2 when :17 = 1 ? Mg, w Nx = 1 + 2mg. When divided by N this gives ar’l, so IF is p = a: and we multiply the equation by at to get (2275; + 33329?) day + (2:2 + 23333;) dy : Fw dx + Fy dy = 0 Then, F = 232;: + :83y2 + My) which ieads to Fy = 2:2 + 235%; + h'(y) and we have h’(y) = U. The general solution is 512% + may? : c] and the particular solution is 1321; +3533;2 = 6. 2. Given the family of curves 1;? = CI — 3x2 , ﬁnd the orthogonal trajectory which passes through the point {2, 1). The parameter c 2 § + 3:5. Differentiate to get 2y dy = (c— 63:) 05\$ : (3"; + 33: — 6.1:) dst so that (g 732:) d3: 7 2ydy= 0 —> (y2 — 39:2)(1733— 2\$ydy= 0 The OT’S satisfy the equation 2:1:yda: + (”92 — 3:132) dy = 0 which has homogeneous coefﬁcients. Take 3 : 'u y 22: y2 (edy + ydv) + (5/2 — 329312) (13; = O 221(edy +ydn) + (1 73n2)dy= 2oydvi (v2 w1)dy : 0 2‘ 1 v2:1dvi§dy=0 % n271=cy The OT’s are deﬁned by the curves 1‘2 — if 2 (:93 and the particular OT passing through (2.1) is given by 1‘2 —— y2 3 figs. 00 CO ...
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240-2008&amp;2009-1-M10-November2008 - KUWAIT...

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