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240-2008&amp;2009-1-M20-December2008

# 240-2008&amp;2009-1-M20-December2008 - Kuwait...

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Unformatted text preview: Kuwait University Math 240 December 18, 2008 Dept. of Math. and ‘ I Computer Science ' S-econthidterm Duration: 90 minutes Calculators and mobile phones are not allowed. Answer all of the following questions. 1. (3+2 pts) (a) Consider the diﬂerential equation y” + ay 2 0, a. is constant. Let y; and .912 be two linearly independent solutions. Show that the Wronskian W(a:) of y1 and 3;; is (:Onstant. f 1 i . a; eos(a:r:) D2 + a2 sm(aa:) V 2a . (b) Prove that: 2. (2+3-pts) (3.) Determine whether the fol'lOWing set of functions are linearly independent 1 - 1 1' £2 _ 11 f2(\$) Z ;——1’ 131(3): x—H' Verlfy your or linearly dependent. f1(x) = answer. (b) Find the solution of the equation. (172 — '51) + 6)y : 0, y(0) _= 3, y’(-0) : s. 3. (5 pts) Use the method of variation of parameters to ﬁndV'the general solution of m 2 y +..y"-=.esc m. 4. (5 pts) GiNen: that gy-l = :52- -is a solution. of - Igyi” — 3563/ + 4y : 0, use the method of reduction of-or-der to ﬁnd thegenerailssolntien of 1523;" — 3333;" + 43; = a: . 5. (5 pts) Use the method of inverse eliﬂierential operators to ﬁnd the general solution of the differential. equation: . ' (D_+ my2 +621)- +'8)y : 6.—23(\$2 + sin-(23):). Some Formulas for Inverse Differential Operators 1. I cu: __ a2; 1' . 801\$- _ eammk' fCDl-(e Mm» — e ——_f(D + a) 11(50): W a k!g(a)’ g(-a) yé U 1. . . . m SINCE“) = ‘Wa D52 i 02 cos(a\$) = __x Sligiam) O. CO Math 240 ' Solutions . December 18, 2008 (a) W: (15" [:1 33— ~ ylyz — yiyz W' = 1111912, ’ yi'yz = 311(‘0-92) — Lieifayl) = 0 => Wis constant. (b) Direct computation: (D2 + a2){ — M) = 2a .. : sin(o:c). 1 x2—1'zs1 3+1' 5:11- 2. (a) The functions are linearly dependent since: (b) 0 = (D2 - so + 5) = (D — 2)(D — 3). Hence y : £1621 + age“. 1,1(0)=3I y’(0) :8 :> C1+62 =3 21:1 +3cg =8. This yields 61-: 1, 62 = 2, hence y = a” + 2e“. 3. Since the constant is a solution of the homogeneous equation by integratingwe can seem there is a solution y that satisﬁes the equation 31;” + y— — fcsc2 zdx_ e — cote. We set 1;" — Acosa: + B sins: subject to the conditions: ' A'cosx+B'sinz=0 and .—A's1nx+Bcosz:—cotz. Solving this we obtain: COS IE and A' i cos :5. t9 1! sin :1: By integration we get —51n :1: B=—/-i—.—dx:—ln|cscmicotml—cosz and Azsinz. 5111 I Hence 3,1,, = L sins: In | cscz i cot ml and for the general solution we have yg =yp+clsinaz+czcosm+C3. 4. The second linearly independent solution of the homogeneous equation is ‘ 1‘13? iven b y - \$2 —11(‘ef=z)dz] : z ln(z ). .-_ iising ridifction drfder method iet y— = :1: 2v be a solution of the non-homogeneous 4 rr equation. Then after substltution of y"a.r_1d 3; into the equation we get :1: v + M3 ' — 3:. Put in —— —',n and soive for w, to get 111 : .So,u = E- enee, 2 ya- — 0112+ Cass 1n(z) + as. . _ _ - w 5 The Aux. Equ. is ('r1r_H-2)(rn2 +6m+8)- — 0 With soiut1ons m _ —'2 2 4. So yc— - (01 + Cam)e 25 + 036 4‘5, and a particular solution Eypl— _ 3,11,1 + 9m: _ 2 4_ 3:4 : IfWith4Up1—(PW—TD+2 D +SD+B {8 2: 33:2} 530+?) ){93 1" 512 0+2{ } ' 1 _. And 63 51(sc4 J2: +3z2 3x + g) 422: 1 - - — :l “3: sin 2:1: = 11m: t'D+2i5(D+17{e_2: 311103)}: e D (0+2) )[31n(‘21:)} _ 4 6 9+5{ ( )1, _T16-2z D T71_{Sin(2x)}_2= Tﬁgf'zﬂcosﬂr) - siﬂiglii- 8—2:: OO 99 ...
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