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240-2007&amp;2008-3-F10-August2008

# 240-2007&amp;2008-3-F10-August2008 - 8 MATHEMATICS&...

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Unformatted text preview: 8. MATHEMATICS & COMPUTER SCIENCE DEPARTMENT ORDINARY DIFFERENTIAL EQUATIONS (MATH—240) KUWAIT UNIVERSITY August 4, 2008, 2:00 RM. Final Examination Duration : 2 Hours Answer all questions Calculators are not allowed Each question 5 marks . Find the general solution of the equation, (yeI +ylny) dx+ (x +ycosy) dy = 0. Show that the second order differential equation with variable coefﬁcients, xzy” —xy' +y = lnx, can be transformed to an equation with constants coefficients by means of the substitution (x=e'/r=lnx) Solve the equation. If y I x is one of the solution of linear homogeneous equation, {l—xcotx)y”—xy'+y=0, O<x<%. Find the second solution. Use the inverse operators to obtain the general solution, y” + 3y’ — 4y = 612'” + 10cos(2x). Evaluate a) 3H4 sin2(kt)003(kf)}- M ii"? { 325:1) } 322’, I Z 7:. <r _Let F(t)={ 0’ 0— (”5’ 3. Express F (r) in terms of a function. h. Use Laplace transform method to solve the initial value problem, WU) +2y’(f) = FCI); yl0) = 0, 3/0)} = 0- . Find the ﬁrst six terms of the power series solution valid near origin of the equation, (x2 — 4))!“ + 3xy' +y = 0. Write the interval of convergence of the series. a. Find £{jcoshkU—mekﬁdﬁ}. o b. The differential equation, % = Ptx)+Q(x)y+R(x)y2, ----- (1) is known as Ricatti equation. Given that yl is a particular solution of the equation (1). Show that the substitution y = y1 + u reduces Ricatti equation to a Bernoulli equation with n = 2. SEE THE BACK OF THE PAGE FOR FORMULAS :9 :9 :> CO CO Final Examination Math-240 Summer of 2007/2008 Q1. Solve the equation (yeJr +ylny)dr+(x+ycosy)dy=0 (l) M:ylny+ye", N:x+ycosy M : x M = 6M _a_N = I 5‘: 6y 8 +1+lny, ax 1: : 6y ax 6 +111)“ C- 1 ; a_M a_N e’+1ny i1 , I' ﬂy- -1“ _L x.) M(6y 53:) ’u eyie y_yC _ ylny-l-ye" J” Multiplied (1) by § (9" + lny)dx + (% + cosy)dy = 0 (exact equation ) - - - (2) §§:81+lny (3) i : i .. . 6y y +cosy (4) Integrate (3) F(x,y) = 9* + xlny+ 30;) . . . (5) 6F — = %+B’(y) = i +cosy, :> 3’02) : cosy :3 .802) = siny 6y _ y Sothat F(x,y):e’+xlny+siny:C 67] Q2. xzy"—xy'+y=lnx (l) x:e’, or t=lnx ﬂ_ﬂﬁ=iﬂ (pd: dtdx xdr’ _y=i[;_y)=ii[d_y)_;£ dx? dx d1 xcixdr 2dr 1 dzy dy , d3? g2E+y_ ..... (2) yc = C1e'+C2 te' ..... (3) 1 : r: i 2Dt=f 2 My D2_2D+1(} (+ } + yg=y5+yp=Cle'+Cgte’+t+2 l/ = CI 6"” +C2 lnxelmr +lnx+ 2. 7/ Q3. lfy = at one solution of linear homogeneous equation (l—xcotx)y”—xy'+y:0. 0<x<-;— (1) Find the general solution. Sol". let y = vx, yr = xv} + v, y” = xv” + 2v' , substitute in (I) G} (1 —xcotx)(xv” + 2v') —x(xv' + v) + xv = 0 (1 —xcotx)(xv” + 21") —x2v’ = 0 x(I —xc0tx)v” + (2(1 — xcotx) —x2)v’ = O i,dv' + e _ 4 d1 = 0 v (1 — xcotx) ijdv'wt 32€_ﬁ _xsmx dx=0 (of: v (smxAxcosx) lnv' +21nx — ln(sinx—xcosx) : Inc In x212“ = Inc 2 2812’ = c v’ : cl: sinx _ cosx ] (sinx *xcosx) ’ (sinxixeosx) ’ x2 x _ sinx _ cosx _ _ sinx cosx _ cosx v—c|:_[ x2 dx I x dx]—c[ x +J—x dx J—x dx]+c1 v = ---5‘SlxJ +61, 2) yg = xv = C2 sinx+ clx TL". Q4. Use the inverse operator to solve the following differential equation y + 3y _ 4}) = 6xe-2x + lOcos(2.x) (1% y( = ('18 + 026—4 U _. 6 = Ego—6.1m 2* nosing; _ 721‘ M? — MD: + 31) _ 4 (6“ )+ WWW yp : 69'2’%1——*—(x) + —-—10-——— 005(2):) {Dw2)2+3(D—2)-4 —4+BD-4 yp = 6e'h—1—(x) + 10 005(2):) D— 6 313—8 yp = 62‘h(?1+31—6D)x+10—3~D—+~8—cos(2x) 6 (9D D2 —64) yp = e'2x(—x+%)3+10m( 32%“ (133(2):) yp 2 (3'2” (—x+ 713-) — %(— 65in(2x)‘+ 8003910) (:12 @3 Q5. Find a) i{4sin2(kt) cos(k!)} £036 {m} a) of{£lsin2 (k!) 005061)}: £{25in(kt) sin(2kr)} — I{eos(kI—2kr) —eos(kt+2ki‘)}= £{cos{— —kt) —cos(3kr)} @‘El _ _ _ 851:2 “ ‘ 3 +18 5 Zs+9k2 (s +k2)(s +918) 9-25 i-l 3—25 — 6—23 blgﬂ ls{2(s-l)}:tf {(Se_2])_e .S' 32} «‘41 — HU— 2k“”7a(r—2)—a(:~2)(t—2)- Q: Q6. Use Laplace transform to solve the following initial value problem; y”(r) +2y’(r) = FU); y(0) = 0, ya» = 0, ---- (1) whereF(t)={0 031““? ....(2) 822' {3.1: Fm = Sa{£—n)ez“‘“)(ez") C9 13mg} = £{San— 7r)e2(“")(ez“)} = 8e2"i{a( 7 mew—M} = 892” Let Jab/(1‘)}: y{s), take Laplace transform of (1) iwm +2y (t)}= ion» s yo) ~ sy(0)— y (0) + 2(sy(5)— —y(0)) = 892” e 59:”; ....(3) Q} ft: m 3— 2 3(s+2)y(s) ~— 882“— 8 SR;- 2 ® Ms) = 892” —-—~—e " S(S+2)(s — 2) ‘ = 2“ 9771's _ 2w!“ 6—1:: : 2:: e—rm _ 2871“ 8—“ y“) 8‘3 (so—2) 4S+8(s+2)) 9 ((8-2) 5 +(S+2)) 2V“) : EZRGU -— n)[32('—"J _ 2 + 9-2043] = .9an0 —- :r)[2cosh2(1_ 7:) 7 2] (2—) CO Q7. (x2 —4)y” +3xy +y20 ----- (1) Solution IS valid for lxl < 2 Let y= Zoanx” , y - Znanx", ' y” = 2:1u(r1~1)a,1x”’2 . . .(2) ”‘0 n—D Substitute (2)0 into (1) we get 6:} (x , 4)Zn(n — l)a,,x" 2 + 3x2nanx’“ ] + Zanx”: ":0 211(11 — ])a,,x” 7 243701 —1)anx” 2 + 23.1mm}: + Zanx” : n=0 n=0 Z[n(n — l) + 311+ Manx” 7241701— l)ar,,x"“2 - 0 - - - (3) "=0 11-2 Replace 11 by n + 2 in the second sum in equation (3) 2M2 +2n+]] Manx —Z4{n+ 2)(n+ l)ﬂn+2X" - 0 11:0 Z[(n+ l)2 ar4(n+2)(n+ l)a,,+2]x” : 0 0 Equating the coefﬁcients of x” to zero (31+ ])2an—4(n+2)(n+ Uri-3H2 = 0, n = 0,1,2,- - - n+1 - n- = r1, :Oalazsn' L a 2 4(n+2)a J"! O ‘12: "31"") ’ “32%“! “4: 1—33“?me “Stew?“ y=ag+01x+a2x2+a3x3+a4x4+a5x5+- y=ao+mx+éaox2+%mx3+ —1—38ao4x + 310a a1x5+--- y:ao(l+ \$114+ l—ggx-4+----)+a1(x+%x3+%x5+uo) 7:, Q8. 3) £{jcoshka—me-kﬁdﬁ} a... = £{cosh(kf)}cf’{ 94”}: {523 ﬂ} 5 + k =2(s — kg)“ + I") Q}? Q8. b} The differential equation, d a}; = P(x) + (2(ny + Row, ----- (I) is known as Ricatti equation. Given that y] is a particular solution of the equation (1). Show that the substitution y = y1 + :1 reduces Ricatti equation to a Bernoulli equation with n = 2. dzylgu ..... (2) .l:_yL ﬂ ..... dx dx+dx (3) Substitute (2) i“and [3) into (I) %+ + =P(x) + Q0001 + u) + R(x}(y1 + u)2 ‘ 9L“ — —P(x) + Q(x)y + R(x)y. + Q(x)u + R(x)u (yI IS a solution) — x é’f-i-de = Q(x)u+R(x)u2 (This Is Bemouli’ sequation n = 2) d2. J on)» to ...
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