240-2007&amp;2008-1-M20-Dec2007

# 240-2007&amp;2008-1-M20-Dec2007 - MATHEMATICS&...

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Unformatted text preview: MATHEMATICS & COMPUTER KUWAIT UNIVERSITY SCIENCE DEPARTMENT Differential Equations Date: November 26, 2007 Answer all questions Math-240 ' Each question 5 marks Calculators are not allowed Second Examination Duration : 90 Minutes 1. a. In the following question write TRUE or FALSE, and justify your answer. (D+ X2)(X2D- I) «1 (XZD- 1)(D+ X2) b. Determine whether the set of functions {sin 1:, cos x, sin 2x} are linearly dependent or independent for all x. 2. Find the particular solution ofthe equation, Xy"+(1— 2X2)}/ — 4xy= 4X, given y] = e"2 is a solution of the corresponding homogeneous equation. 3. Use method of variation of parameters to solve the equation, f’ +4}/ + 4y: fie-'2‘, x > 0. 4. Make use of inverse differential operators formula to solve the equation, _ I (D—2)(D2—3D+2)y= ez’(x+c053x). 5. a. Find the general solution of the equation, (D4 —1)(DS —2D‘ + 2173)y= 0. b. Let (my: y’ +p}/ + qy. Suppose that y; and y; are two functions such that 1(D)y| = ﬁx) and 1(0))»; = g(x). Show that there sum y= y: + y; satisﬁes the nonhomogeneous equation ITDLV = £1!) + 300‘ Inverse Differential operator formulas (2007) 1 a. m = 8,, 1 1' me ”‘ ) m: mm 80.1 I 2 1 I}... ﬂu) if M)?” —De : n as H ) Eerie} if M“) #0 where ND) = \$(D)(D—o)“ . I _ b 3 1 8111033) : 0‘ FE b I D2 + o2 n: cos (bx) = 225:1; a 9,5 b - W —xcos(e::) 4 1 511101111) _ T - D2 + a: a: sin (or) cos (ax) = 2a . _ 3111mm) 1 5111011) _ f (#12) 5. if f(—a2)7£(i . n92) WW Josue) 99 1. a) In the following questions write TRUE or FALSE, and justify your answer. (D + x2)(x2D —1)=(x2D —1)(D+ x2) (D+x2)(xzD— 1)y = (D+x2)(x2Dy~y) = x2D2y+ 2ny—Dy+x4Dy—x2y = (x213z +(Jr:4 +Zx—1)D—x2)y (xZD— l)(D+x2)y =(x2D~1)(Dy+x1y)= x2D2y+x4Dy+2x3yeDy—x2y = (1:202 + (x‘1 —1)D+ 2x3 —x2)y (x2D2 + (x4 + 21i:—1)D—x2)=it (152402 + (x4 — 1)]? + 2x3 exz) FALSE b) Determine whether the following set of functions {sinx, cosx, sian} are linearly dependent or independent for all x. C; sinx + C2 cosx + C3 sin(2x) = 0. sinx cosx sin(2x) W = cosx wsinx 2005(2x) = 3 coszxsinax) + 3sin2xsin(2x) = 3 sin(2x) t O esinx —cosx —4 sin(2x) for allx except at the points x 2 igir, n n 0, 1, 2,- - - - - ® But at these points (2on at 0 Therefore functions are linearly independent. 2. Find the particular solution of the equation, xy”+(1—2x2)yf—4xy=4x, (1) given )1] = e"2 is a solution of the corresponding homogeneous equation. Let yp = w“, y; e (v’ + 2xv)e‘2, )1}? = (v” + 4xv" + (4x2 + 2)v)e"2 6) Substitute in (1) W! + 4n! + (4x2 + 2)v]e"2 + (1 e 2x2xv’ + 2mex’ — mex’ = 4x v” + (% +2x)v" = Ate-xi Let v” = w, v” = w’ So w“ + Q + 2x)w = 48": This is ﬁrst order linear equation (’7) . . i 1rd: lntegrattng factor Is ,u : 9k” ) = em”2 = are"2 2 72 1 2 72 -g—x(xe’w):4e‘xe‘ =4x :xe"w=2x2 =>w=v’=2.xe" ‘128‘2 = —1 u: W2 2 v=J2xexdx=—e‘ ﬂyp=vexzwe 3. Use method of variation of parameters to solve the equation J’” +4yf + 4y = fie—2", x > 0. . . . . . . . .(1) The corresponding homogenous equation is y”+4y"+4y=0, - - ~ - - - yc = C1 «2‘?-"+C2xe’1’r - - - - -(3) Let yJTJ = A(x) e'l" +B(x) ice-7’-at . . . . .(4) 6) y; = A’(x) e'z’ — 2.210094" 4- B’(x) xe'z‘ + 30:) (1 — 2x)?h Set A’(x) 24" + B’(x) xe'Z‘ = . . . . . . .(5) y; = —2A(x) (2* +B(x) (1 —2x)e-2x . . . . .(6) y“ = —2A’(x) e-ZX + 4A(x) (2x + B’(x) (1 — bee-21 + 430:) (x — 1 )e-lx - -- - -(7) Substitute (4), (6)_and (7) into (1) ~2A"(x) 2'” + 4A(x) 2"" + B/(x) (1 — 2302-21 + 4B(x) (x — I)?” + 4 (—2A (x) e‘” + B(x) (] — Ewe—2") + 4 (A(x) 2’” + 30:) xe'b‘) = sic—29‘2" . E \ O O "24%) 9'23 + Bl(x)(1— 2309—2)! = x—ze—z: .....(g) a) A’(x) e'z"+B"(x) xe'z" : 0- - I Solve the two equations 3’05) = Jr‘1 and A’(x) : —x" A(Jc) = —lnx and B(x) = —x"] Substitute into (4) . xe’l‘ = — e‘lenx yp = — eklenx — x’1 4. Make use of inverse operators formula to solve the equation (D—2)(D2 —30+2)y = eHx-icosBx) - « - - -(1) Corresponding homogenous equation (D—2)(D2w3D-l-2)y=0 -----(2) y6 = Cu.”r +Cgez" + ngez" - - - o - -(3) <9 Tofind yp I 1 =—~—m—2‘ + 33 =——~—m—h + 3 ” (D—2)(Dl—3D+2)e "‘ °° ’0 (D—I)(D—2)(D~2)e 0‘ m”) = h_~——1—-—m—_ +cos3 y” e (D+2—1)(D+2—2)(D+2—2)(x x) =2l——1 +53:1‘+i3—i 3 yp e (D+1)D2(x co x) e (D+1)(6x 9cos x) (13-1) :eh|:(1—D+D1+D3)(éx3)_m(%cos3x) (Bil) =ele: %x3_;_x2_x+1)_ (_9_1) (goos3x):| = elxl: %x3i%xzﬁx+l) —\$(3sin3x+cos3x):| yg=yc+yp @ 5. a) Find the general solution of the equation (1)“ —1)(D5 —2D4 +2D3)y e o, ..... (1) Auxiliary equation (m4 —1)(m5 ~2m4 -l- 27713) = ;m3(1lri2 w I)(m2 +1}(m2 —2m +2) = 0 Roots are {0, 0, 0, —i, 1, ii, Iii} yg = C1+ C2 x+ C3 Jr2 + C; e" + C52" + (C6 cosx + C7 sinx) + e"(C cosx + C9 sinx) @ 5. b) Let ﬂD)y = y“ +py’ + qy. Suppose that 321 and y; are two functions such that ﬂD)y1 = ﬁx) and 1(0))»; = g(x). Show that there sum y = yr +y2 satisfied the nonhomogeneous equation ﬂD)y = ﬁx) + g(x). Wm O O ﬂD)y =ﬂDXy] +y2) 2 5y: +Jlt'2)" +1901] +Hy2)! d,- (yi +y2} = 1+py1+qyzi+ (y: +py2 +qy2) =ftDJy1 +1113»: =ﬂx) +grx). (‘29 ...
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240-2007&amp;2008-1-M20-Dec2007 - MATHEMATICS&...

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