This preview shows pages 1–10. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: NAME #1 Recitation Instructor _____.____——— Signature __________.. Recitation Time Elementary Differential Equations
Math 240, Spring 2007
Second Examination, March 13, 2007 Show all your work in the space under each question. Please write legibly and organize your solutions
in a logical and coherent form; answers which are illegible or confusing will not receive credit. Each problem is worth 10 points. 1. Find the general solution.
2y” — 719' + 33/ = 0 307170 law
@043 ivy—9:0
we 30:}.
\l A’ (ﬂdﬂk tail“ 2. Find the general solution to the system of equations. 9%Uid—V02g slﬂ '1 3+” {5% “7%
iii, $ij  ,
Aim?” 7’1? 9% “31%. 10¢
x N7
an?» ‘%% “67x10
O\7’7\ _ a __ \
Kg; 753% \07k «9
DL’ 3b “we
QQ—ZﬂKDALﬂ :b
arc
we: 98* Jr L16
, a?” — )C ,‘H' 1 6+
\) ’l’bﬁ 27" %C\€6 ,LLQ/ . ’Z/C\&
 ~"Ur
x)? \ﬁgéﬁ ~3L~Vﬂ 3. Find the general solution. 3;” + 4y = 3x2 ¥f%uvo
Evilk \%;<ﬂu»1%xralaM2¢\ hﬂ 9&4w3x6»xc
93>2$v\%ﬁ
8931“ LIX 4. UMMCL¥ {37x 4 a» 2378”
LHHF*HB#%1A4QL23¢— quB
ugco
7)\VALPLD a» (wax w Lamaze
AC*%
L‘% 4. A mass of 2 kg causes a spring to stretch 9.8m. This is attached to a damping mechanism with
damping constant 5 The mass is allowed to come to equilibrium and is then set into motion by displacing it 3m and releasing it. Find the function a:(t) which gives the position of the
mass at time t. ”‘ my ﬁAL
‘ ammo» \A<«$\
a 7” \4
158%70 iuio NOV?» who
71?” +emLz O i \3 Lo WA :0 are [L 5. Find the general solution. CD7” «1% WW \: 0
mm + AWN m
b 1 1J—nglx «11 “LY 
\)2L\@ 4 (do J< LgLosu JV “W14: 6. Find the general solution. 1+e
(fab nab
{0 +\\k\om:o
D—xx my;
\jh: (4 Qj\ AV Lia/17k W Ta}
1 2\%\1+a\— a“ ﬁrm w“ u
2 6* MgﬂﬂfMZ’ gw*/L (W
\AAA H—Ux
‘— 45* \0}5\’H03L\ “in m — \ogmhaﬁ
2 a? \%\ HEW "éb 93“ WagVH 6 \\ (1r 43 a?» aﬁef’k + (o"‘+eﬁ°”‘\ MMHM 7. A mass of 1 kg is attached to a spring with spring constant 1 5:%. After coming to equilibrium, the
mass is set into motion with a variable force of cost N, but is not given any initial displacement
or velocity. There is no damping. Find the function :c(t) which gives the position of the mass attimet.
K Pa $W¥icont sumac $Mm20
5 .
‘ (Ll DLHCO
A 0:)L‘k 74“: g Lost +LLQMY‘
74g): Log’l *l’ Xﬁck%&%MPMﬂlWW%l®UMf N9“ 3 ’Pxérpl/ '* Premier l M l—wstl Jr 8 L054:
Jr 1% cost + Rt L» $9608
: ~7/Jx QVWC J'le cost *le LBS/C "Rt mid 9U 7km 1 owe): 3% £19;th +50 gut
O”; MMZM
Wm) 7km? LLSVWE 4 at wt
Whipo: (“inset % item % Lﬁmgk
Q ~— 70(0ﬂ2w
MALWM) 3 iltl" Khaki 8. A mass of 6199 is attached to a spring with constant 8 There is no damping. The mass is
set into motion with an initial displacement of 2m and an initial velocity. It is observed that
the amplitude of the resulting motion is 5m. What was the initial veolcity? 9. Consider the differential equation describing a massspring—damping system. 1
x" + 013' + Ex = F0 cos wt, 33(0) = 1, x'(0) = —— (a) Suppose F0 = 0. For which values of c is this critically damped? >4" >v L70 % A; $10 U‘KMW W nevi/w LLALtlpmio
9”»‘vtlhl’l‘ro :2) C»\ (b) If F0 = 1 and c = 0 What values of a; cause resonance? 74“ “37k: nois
waigqu a ‘PLW MW w‘/%/‘ (c) If F0 = 0 and c = 0, the solution is m(t) = 008% + x/gsin (You may assume this is true.) Write x(t) in theaform m(t) ﬂosﬁut — ¢) for some w and qb. x _ g); 025%; W/” 9; (gaugiqu
so Loaf/L l" ggmg : (LAM $0 U”\
\m’ﬁa PM Ch QW‘ 1 “E
s “EEK
FEM“— 7’3
3; US} , A i”?
95 wsgtﬁéwée \22/ Wist/k \ ’ Mi} 3\ 10. Match the graphs to the equations. Each graph ranges from 0 to 20 on the horizontal scale
and from —10 to 10 on the vertical scale. (a) 2:” + 1033’ + 323 = 0 93(0) = 0, m’(0) = 100
b) as” + 16110 = 0 ar(0) = 0, x’(0) = 30 0) 5:5” + x’+10:c = 0 :r(0) = 0, x’(0) = 10 d) 1:” + 400:1: = 200 cos 18t 33(0) = 0, x’(0) = 0
e) x”+:r’+a:=cost 32(0) =0, 33’(0) =0 (
(
(
(
i I “\KXXTX‘KH‘E—h Nlrl rerl’VJl’lm'nlll I lia'lll
This is the graph of equation (A . This is the graph of equation Cl . This is the graph of equation Q1 . This is the graph of equation This is the graph of equation C“ . ...
View
Full
Document
This note was uploaded on 04/07/2008 for the course MATH 240 taught by Professor Yetter during the Spring '07 term at Kansas State University.
 Spring '07
 Yetter
 Differential Equations, Equations

Click to edit the document details