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240-2008&amp;2009-2-F10-June2009(4)

# 240-2008&amp;2009-2-F10-June2009(4) - Kuwait University...

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Unformatted text preview: Kuwait University Math. 240 June 21, 2009 Dept. of Math. & Comp. Sc. Final Exam Duration : 2 hours Calculators and mobile phones are not allowed Answer the following questions. . (4 Points) Find the general solution of the equation y(2X+ y)dX+ (4xy+ 3X2 — 4y)dy= 0. . (4 Points) Find the general solution of the equation (3X— 2cosy) cosydx+ X2 sinydy = 0. . (4 Points) Use the inverse operator method to ﬁnd the general solution of the equation (D+1)(D2 + 2D+ 5)y = 16e‘xcoszx. . (4 Points) Use the method of variation of parameters to solve the equation y' + y = sec XtanX. a. (3 Points) lfL{P(t)} = 1(5) and a > 0. Show that L{a‘F(t)} = 1(5— In a). b. (3 Points) Find L{3“+ 2t 22 + 26’ t3”). . (6 Points) Evaluate (a) Lﬁzéﬁ} (b) L4{1n(1+%)}. t, 0 S t<1 . (4 Points) Leth) = 2, r21. a. Express Hr) in terms of the a function (step function) and ﬁnd L{F( t)}. b. Use Laplace transform method to solve the initial value problem Y'U) +y(l‘) = Flt); y(0) = Y(0) = 0- . (4 Points) Use Laplace transform to solve the following equation for H(t) with the condition H(0) = 3 t H(t) + 2! e‘Z”H(t- u)du = 6‘2’. 0 . (4 Points) Find the ﬁrst ﬁve nonzero terms of the power series solution valid near the origin, of the equation (1+X2)y”+X}/—y= 0 and state the region of validity of the solution. See backside for formulas OO 99 Kuwait University Math. 240 June 21, 2009 Dept. of Math. & Comp. Sc. Final Exam Solution 11/13) —— N9: _ ~4\$ — 2y 2 2 . . ——'=——:H y =y.Thesolut10nisF 33,31 :0. M 2xy+y2 y () ( ) Fm =2\$y3+y4 85 Fy :4my3+3m292—4y3 => F(:c,y) =y3 (x2+:cy—y) =c. 2 2. Putu:c0sy::>u’—-§u=—2i (BE); n=2 w 1'2 2 tzu‘lzsecyzwﬁ’+gt=—. (LE); M=x3 1-2 3, yzyc+yp; yoze’\$(c1+c2cos2x+63sin2m). 1 _x 2 86‘3” 111p: — (156 COS 95) :m (1 + cos 2.1:) (D +1) [(0 + 1)? + 4] Se‘x : D2 + 4 (an + % sin 2x) : 86—“ (inc -— éim cos 2x) = 3738’m (2 — cos 2%). 4~ y=yc+ypi yc:clcosm+02sinm. A’cosm+B'sinax :0 ypzAcosx+Bsinx= A’(—sinw)+B’cos:1: =secwtancv A' :—tan2x:1_sec2\$ A =LE—tanm I ——_——> B =tanx B :lnsecx yp = xcoscc — sina: +sinxlnsecm. y:clcoszr+ksinm+mcosm+sinmlnsecrc; k: Cg — 1. 5. a. From L{e“tF (15)} = f (s — a) :> L{a,tF (t)} = L{e“naF(t)} : f (5 — Ina) mm 5.b. L{3-t+2tt2+2ett3/2}: 1 +_2§+_i52 s+1n3 (3&1112) 2(s*1)/ 2‘ 6.a. L_l{£—L}=L“l{l+ 1 — 3 2}—1+e“t—3tet‘ s ) s 8+1 (3+1) 1 S 6. b. Let f(s)=1n<1+ >=ln(s+1)—ln3 LHFW :f'“) = i ‘ E = ”W - 1} => L-1{f<s>} : Fa) : 1 ‘6‘. 7. a. F(t)=t—ta(t—1)+2a(t—1)= t+[1—(t—1)]a(t—1). 7. b. Let L{y (13)} 2 11(5) => L{y" (75)} = 3211, (3) My” (t) +y m} : L{F(t)} :> (s2 + 1) ”(8) 2 52 + (l _ i) 882 l 1 1 —8 M5) =m+ [m_32(82+1)]€ _1 1+1 5.1+1973 ws2 52+1 . I y(t)=L-1 u(s)}=t—sint+[1-cos(t—-l)—(t-—1)+sin(t—1)]a(t—-1).‘ 8. Let L{h(t)} =h(s) : L{HI (15)} = sh(s) —3 & L{/e-2uH(t—u)du} 2:13; 0 2h(5)_ 1 _ 33+? _3(5+1)+4 sh(5)-3+3+2—9+2:”‘(5)_32+2s+2_(3+1)2+1 11(15): L_1{h(s)} : e’tL‘1 {3s __ 4} = 7t (3 cost + 4sint). OO 00 00 00 DO 9. Lety: Z un\$”=> Zn(n—1)anx“+ Z n(n—1)an\$"_2+ Z nanxn— Z an\$"=0 n=0 n=2 n 2 n=1 n=0 :0 [(71 + 1) (71 + 2) (1724-2 + (71 ~ 1) (n + 1) an] \$71 = 0 valid in |m| < 1 and with (10 and a1 arbitrary. 2 Wm ...
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240-2008&amp;2009-2-F10-June2009(4) - Kuwait University...

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