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240-2008&2009-3-M20-July2009 - Kuwait...

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Unformatted text preview: Kuwait University" Math. 240 July 30, 2009 Dept. of Math. & Comp. Sc. Second Midterm Exam Duration 90 Minutes 4 6 Calculators and mobile phones are not allowed Answer the following questions. . (3 Points) Let A = D - 2:1: and B : 93D + 3. Show that AB # BA. . (3 Points) Discuss the linear dependence of the functions fiil=23in$+cossm g(w)=sin:r—2cos;r. . (4 Points) Solve the initial value problem "3'” + 29” + y’ + 2y = 0; y (0) = '1, WC): 5, y” (0) : -4- . (5 Points) Solve the equation (D4 — 16) (D4 — 203 +16D)y : 0. . (5 Points) Use the method of inverse differential operators to find the general solution of the equation (D +1) (D2 + 4) y : 25re'"x + 20 cos 2a: . (5 Points) Find the general solution of the equation '2 If my +:cy’—y=3;r2, given that y 2 an is a solution of the corresponding homogeneous equation. Some Useful Formulas For Inverse Differential Operators OO 99 Math. 240 Second Midterm Solution July 30, 2009 ll'ABy : (D — 22:) (ID + 3) y : (D # 22:) (:ch + 3y) 2 Dy + $021] + 30y A 212Dy ~— 6mg AB : x02 + (4 — Zzz) D — 6m. BAy : (3D + 3) (D — 220) y : (mD + 3) (Dy A 22:11) 2 $D2y — 2:51; 7 21'2Dy + 3Dy — 6mg BAsz2+ (372x2)D—8m. 2sinx+cosx slur—2mm: 2- W (f, 9) = 2 5 f 0 d f (I) and 9(1) are linearly independent. Zeosxisinz cosz+25in$ 3. f(m):(m+2)(m2+1):0: m=~2, —i,i y =cle_21+02cosx+63sinx y(0) :71 :¢c1+C2:—1 c1:#1 y’ : —2cle_21 — 02 sinz + C3 COSI ::> y’ (0) : 5 :> 7261 + 63 : 5 => C2 : ” 2 4616—21 -— 62 c051: # 6351112? y” (0) : ~4 => 461 — C2 : —4 C3 = 3 y : #6'21 + 3sinx. 4. f(m)=m(m#2)(m+2)2(m2+4) [(m‘2)2+4] :0=>m:[), 2, 72, —2, 5:21} 2:21' yc : c1+ c232“ + (C3 + (14:6) e'zr + (C5 + Cash) cos 2x + (C7 + 68821) sin 217.1 5. 31:.1;c+11p,+yp2 yc -, ale“ + C2 cos 2x + as sin 210.] f(m)=(m+1)(m2+4)_0: "(n—#1, 2i, 2i——s 1 258': Se—I :———— 2 *r , , :2 2 l 5 '1‘ gm (D+1)(D2+4)(5$e ) D(D2_2D+5)(I) D (1 5D)(ar) 212(ac+4)e 1 (20cos2) ‘4 ( 2 'n29: 2) (n2 Zoo 2) =————————— x:———-51 —cosx::551:c— 32:. 9’” (D+1)(D2+4) (WM) 6. y=xu, ’:xu’+u, y”:xu”+2u' n 3 I 3 3 r 3 a —‘2 u+ — u=—, mu:x +01, uzmw—ZJ-x +02 35 z yzz2+kx_1+62x; k:—-C-2L. O. CO ...
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