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Unformatted text preview: KUWAIT UNIVERSITY Department of Mathematics and Computer Science
Math 240 May 2, 2006
Ordinary Diff. Eqns. Second Midterm Exam Time: 90 minutes 1. (a) Show that the functions ﬁx) = :22—1 and 9(33) = 33—2 are linearly independent in the
interval ac > 0. [2 pts] (b) Perform the multiplication (expand) D (322192 — 1). [2 pts] ‘2. Use variation of parameters to ﬁnd the general solution to the differential equation: sin 3
32y”+4$y'+2y=—, I>0
:5
given that y1 = 33—1 and y; z 33—2 are two solutions to the homogeneous equation [5 pts] 3. Use inverse operators to ﬁnd the general solution to the following differential equations: (a) [D‘L _ 8132 + 16) y = 83; u a”; + 8353 [4 pts]
(13) (D+1)[D2w—2D+2)y:e‘%032$ [4 pts]
4. Find the Laplace transform of the function F(t) = 5 + 25m? 3t + 6cosh 2r. [3 pts] 5. (a) If Fm is coritinuous for t 2 O and is also of exponential order as t —} DO. and if PM is of Class A‘ prove that I, {PM} = s E {FM} _ Fm). [2 pts[ ([3) If y(0) = 1, y’(0) : —1 and C{y(t)} : 10(8), find the Laplace transform of the expression y”(t) — 2t y’(t) + 4y(t). [3 pts]
.9 CO Solutions 1. (a) Take the Wronskisn: ,—1 —2
‘1 £4 :4 =—2x’4+r_d:wx"47é0 sincex>0 @
,3; , :r
(b) D (x213? —1)= D{I2D2) — D : 2:21.33 + 23132 , D @ 2. The solutions yl and yg are linearly independent (prob. la). We have yp = Ayl + By; :
A/w + B/I2 leading to the equations: 1 1 1 2 '
—A’+——B’:G and ——?A’iws'=smI G]
x 5:2 x :53 a: which reduce to :5 A’ + B’ = 0 and mA’ + 2 B’ = ~12 sin: [since at > 0). Subtract the ﬁrst from the second to get B’ : “3:2 sins and integrating by parts twice gives
8:352 cows—2;: sin$~2cosz
We also have 11’ = eff/s; = 3: 511131;, which leads to A : shut — as costs. The general solution
becomes @
1 . 1 2 .
y = (c; + A)y1+(::2 + BJyg : — (c; + smz — x coscr) + 7 (Cg + :t' cosr — 27. smz— 2cosa:)
:1: I
1 —2 [c2 ~— (:12: — 2: Sins: — 20085:) (D
:c 3.3, The operator f(D) = {D2 — c1)2 = (D — 2)2(D + 2? so that 1 3x : 1 63.1: 1 33 (D Efﬁe n3) :%5 For the second exponential e—2r we have u : —2,k = 2 and (MD) 2 (D — 2)2 so that til—2) : 16 and we get
2
1 8—21 = 3: 672:5 : LIQe—ZI Q Hi?) mast—2) 32 The third term involves a polynomial and o 2 0.. so that
1  8 1 1 l A 1 .
8 .5 _ 3 = 1 __ D4 __ 2 1.. 3 t _ 3 CD
f(D)z 161+(D48D9)/16x 2 15( 8D )* 3 2t”: +33)
The particular solution we are after becomes
7 1 3x 1 2 —21 . 1 3
yin—258 +3235 T§(m +3x) Since the polynomial ﬁne) 2 (m A 2)2(m + ‘2)2 has repeated roots or = i2, the complementary solution is
ya = (Ci + “235) 6—2: + {Ca + 6415} F321 6) and the general solution is y : yc + yp. OO 99 3.b Here we have f(D) : (D + 1) (D2 + 2D + 2)., a : —1, and 11(3) = cos 22. Use exponential shift to get W 1 1 _ 1 1 _ _1— 1 _ #1 1
ypﬁfmle cos2z—e moos‘Zx—e D(D2+1)C032$“e D_4+10052rn
' _,1 11 “1—1 _1111 9.
— 36 Dcos2z— 36 foostdx— 66 31112.15 O The polynomial f(m) : (m + l) (m2 + 2m + 2) has roots m = 1 and m = —1 i i, which yield
ya = 1:167:c + 026—” sins: + age—x coax = 8“: (c1 + c; Sims + C3 cos x) {'24 and the general solution follows as y : ye + yp. 4. Simplify to get F[t) = 5 + 2sin2 31+ Stash 2t 2 5 + 1 — cos 6: + 3e” + 3.9—2t so that £{F(t)}:£{ﬁ—cosﬁt+352t+3e_2‘}zEH 8 + 3 + 3 :EW '9 + 63 .9 33—4—36 5—2 3+2 3 52+36 52—4 @ 5.a Proof in the book, @ 5.b Using part a, we more 15 {y’} = 5111(3) —1 leading to £{y”} = 3211(5) — 5 +1. For the term
ty’ﬁ) we get I d , d
Lity}2—dsLiy}=“E(31U(Sl1)=—3w'(s)—w(s) (D
and it follows that
£{y"—21y’+4y}mam—3+1+25w'+2w+4w:2w*+(32+5)w—s+1 @ OO 99 ...
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