# Exam 2 Key - Differential Equations Exam#2 Fall 2010 Name...

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Unformatted text preview: Differential Equations Exam #2 -. Fall 2010 Name Show all your work neatly and in numerical order on notebook paper. Please circle your answers. DO NOT CROWD YOUR WORK OR WRITE ON THE BACKS OF YOUR PAGES. 1. Find the form for a particular solution yJD to the equation y” + 2 y’ — 3y = g(x) where g(x) equals: a. 7cos3x b. Se'“ 0. x2 cos(7r x) (Do not actually ﬁnd yp) 2. Solve: y"+4y’+4y=0 3. Solve: y”+2y’+4y=0 4. Solve: 3x2y”+11xy’—v3y=0 5. a. Find a linear differential operator that annihilates 13x + 9x2 — sin 4x. b. Find a linear differential operator that annihilates x3 35" 6. Solve: y”+3y’+2y : e” 7. Solve: y” + y = tan x 8. Given that every solution to y” + y = O is of the form y = c] cosx + 6‘2 sin 9: , a. is there a unique solution that satisﬁes the conditions y(0) = 2 and z 0? b. is there a unique solution satisfying y(0) = 2 and y(7r) = 0 ? c. is there a unique solution satisfying y(0) = 2 and y(7r) = —2 ‘? d. Is this a violation of our existence and uniqueness theorem? Why or why not? 9. Solve: Zyl’ = 3312; 31(0) 2 1, y’(0) = 1 (Hint: Use your initial conditions after ﬁnding y’, before trying to ﬁnd y.) 10. Solve the initial value problem: % = 3x —— 3y ' d—y = 2x — 23/ a: x(0) = 0, y(0) = 1 913‘"??- E61 5W” I425 — F2010 I); MZ—LZM’3 T—O rsx (VHQCm-Uto go): (1.6 +0,sz E m:‘5 )Mi] apt ;@ (1h: Acm3x+ Bsm 3X we (9 W: Axe?“ H @ :- (A SLQ+E>L+CDC¢DCWX> Pt: 73p + CDX1 +éx+ F) SinCﬂ’X) 51) m2+4m+4 :o . . , (WW—350 g: c.c=;*'”+ 0,,er2K mz—z 13"L25'7L45 :0 m: “Lil—{:1 2-21- 2H? "‘ m1+2m+430 - MZdi/Cg x ; m: .‘2 : {MT—W4) 5 : c,c”‘mﬁx +CzC’sinf3X 1 2. q):3xzm(m—\3Xm’z Hmmxw' - mezo :3sz XML?) MCm-D -|— H m #5310 i x’“ ( 5m1+ gm x5) :0 (Sm-*lYmmB M713 )m 3-3 m « -2_ I J 01 my Iqch I 1013’“ : 65y + 2 - A p; A 63: 131°": 63X :33: 2A6? x Ac”: x u.’ Sim—A IX‘ 2 : 83%.. mi fogi-l Mamie” X+\$-\$QC, m x x} ' 9‘30» F 4}}? 1:4‘ —\> Semi-ion is 1001' umé ub' EIQ‘CO’JX. + ast'nx. Nb 3 Our mama, and uniﬁuUuSS é) WVQNK, ﬁo W ﬁx Waj'mg Wk 'zmchO can 1 was r This pwlomm hob boundary conqu‘msJ YWU‘F ‘IVYVHQD COMH‘IGHS. d1. '35::3wv3 : ~ dJo/di, = zx—iﬁ M0) 0 “3(0),, + CD ~33x +3 - 0 ~ - ~I> o’lCD—abw to Q34 +(szb)j:.o ‘2CD‘33XKD‘ aagj LCD—33mm H5] 5 :0 €9,151: 3 ax, “ 920/: “\$162.67: 01x: (19.6315 +ZC. +£02.65 9.x : 2.0,. + sczeﬁ X 3 C" +‘%_ CLEJt ; mom—0:0 CD2” D3 :0 8‘: 0/1 +C2€Jtl Aft tlx~lgb ...
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Exam 2 Key - Differential Equations Exam#2 Fall 2010 Name...

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