# exam2 - Exam II I could not ﬁnd my work solutions or a...

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Unformatted text preview: Exam II I could not ﬁnd my work solutions or a. blank exam, so this is a recreation of exam 1]. Enjoy. In. Find the general solution of the differential equation 1;” — 9y = sin L Lamagem ecu: JOIUHOnI \I —c(\l - 0 rqrtmulgy' goln ', VP: A—gin (:1- {3 (as t‘ 1?” l — \(P ‘E — ,0 {3(03 t 2?..n t 1). Which solutions stay bounded (that. is, y{!.) does not; go to inﬁnity) as I. ~> 00? "Aqu hot/Q C‘ZO, 2. Solve the initial value problem 1.23:” — 2ng’ + 2;!) = 0, y(1) = 0, y’(1) = 2. Hint: This is an Euler’s equation. We. looK 4‘0" SOMS O4 g‘ovm \(3rt Flujary in \$0 our ect‘laf'.Un-. 13 r- (r-\) €“~ 2+, NJ" +24J=o 50 £”( Mm) — 1’ +1): 0 So "(F~‘)*2r+1 ~c ~50 Fix-3,» +120 50 3. Find the general solution of the inhomogeneous equation y” + Gy' + 133} = 1. 0““ 9% F1+6rfi330 ‘ + F‘T—‘ﬁ So F: Qu- 36 LLB 1 uélele Z 2 = *3 1 2i, 4a. Using the fact that yl = :L' solves the equation may" — :c(.’1:+ 2)y" -|- + 2)y = 0, find the general solution of the equation. We look {30/ a soim 04 “‘6 PW“ Y 1‘ X v I Tken \l'ng'a-V/ 7'=2V'+Xvu So X2(2v'1-XV" —x(x+2)(xv'rv) +(KF2)X V = O so xz(2v‘+x\,") - x‘(x+2)v‘=0 b. For which 11:0, yo, yf, does the initial value problem y(:l:o) = yog y’(:1:0) = y{, for the equation in (a) have a unique solution. )1 ‘L'fl l a): _. 'i 7‘ "0+ Cani": @ ﬁro Un‘Twﬁ :Glm‘HOM £7 +‘ve axcsf-enfe onci umguemess 'i'keavww °"‘ “"7 Q8 x0960. So 5. l-‘ind the general solution of the equation y" — 21/ -|~ y = e' In 1. (Mint: The variation of 1 r 1 ( parameters formula is Y = —y./~‘ﬁ}idt +312 #011...) 0W“ 91'“ 0=r“—2r+ I = (r—I)‘ 5" Y», = Cnet + citet Le!- : t ‘L 9/ \zetet W {\{i’jzg 2 :1" \fz ) — a . I, 8 (“get 6 Tku') t k t8 8 ln’t t 1"‘l 4: \ _.— *- .____H_\ a e. e w I— eat Cit + {:9 j e24; dg =—ef swat + tetflnedt dv U‘ V: du 6. Consider the initial value problem (1 — \$2M" + 2y = 0, 31(0) = 2, y'(0) = 1. Write the ﬁrst; five terms of the solution y(:v) as a. power series about 0. CO 6.: O0 .. n __ It _. - "'1 ._ \f ._ g QRX a) Y .— nZ‘: (1,. n (n 0X — [ﬂoann (nr2)(m-|)X \J SM“. Casr rum ferms W‘s/ﬂ C), xiv" = E <2” n (n—I) xn “e” d° a “We 0‘- "of. So -0 a n 00 0: ('“X ) \/ +27 = 2 (0m (mummy -0n n (“t-I) +2an> “To 50 OU" r‘e Curran Ce he le- hbr’l 25 0r”: (n+l)(n+l) - an HUI-I) +2Qﬂ 2 0‘9—2 Q332—Il-o+1l=o =7 (lat—A3. On. H 3—-2.21+1.-2= => Iz'qqzo c3: =70“: 7. A small object of mass 1 kg is attached to a spring with spring constant k : 1 N / In. An external force ﬁt) = £15th N is acting on the mass. At time t : 0, the mass is pulled 1m to the right from its equilibrium position and released. Ignoring friction, find the position function u(t) of the mass. m:i" t:( U" + U “3‘— (‘1 5m '5 Lilo): l, M'io):o r‘+ i=0 Uktctcost {—625.41 t U? 1; Atgos‘t ‘l‘ Bt 3;” t t; Timer wt) +s(2m-t-tsmon COS‘E', A:“2 tCestt sat ~24 2L1 U '1 ’1tcos“: P H o t U301 +U\P:C[cost+ Cgsmt QtCs h :M(O)=Qt _ \$.12] a 2. -— 2 1 .4. asmt "Z‘ECOSt q Is», Li , ,_ M Ma + loud-- +116 +e/M rhea '1 5 \‘j/ final» as La <35) I M o sounds; mar». my? {m .i *9 a 1; q l1} udes ( Have We sow «VI ( ‘E‘ ) ...
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## This note was uploaded on 07/17/2008 for the course MATH 415 taught by Professor Costin during the Fall '07 term at Ohio State.

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exam2 - Exam II I could not ﬁnd my work solutions or a...

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