MATH 202 Exam 2, Spring 2009
American University of Beirut · Differential Equations · 7 pages
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Math 202 — Spring 2008
Differential Equations, sections 17-20
Quiz 2, April 11 — Duration: 1 hour
Reminder of instructions. READ AT LEAST NUMBER 5.
1. Write your name, AUB ID number, and section number ON THE FRONT COVER
OF YOUR AUB EXAMINATION BOOKLET.
To remind you, the sections are as follows:
Section 17 Section 18 Section 19 Section 20
Recitation W 8 Recitation Tu 2 Recitation Tu 3:30 Recitation Tu 5
Prof. Makdisi Mrs. Karam Mrs. Karam Mrs. Karam
2. You may work on the problems in ANY ORDER in your exam booklet, but
please make it clear which problem you are solving on any given page. In particular,
PLEASE INDICATE IF THE SOLUTION TO A PROBLEM IS CONTINUED ON
A LATER PAGE.
3. Do as much of the exam as you can, and budget your time carefully. The problems
are arranged APPROXIMATELY in increasing order of time needed. Take a minute
at the start of the exam to decide which problems to work on first and to plan strategy.
4. Explain your steps precisely and clearly to ensure full credit. Partial solutions will
receive partial credit. Each problem is worth 12 points, and there are 6 problems for
a TOTAL 72 points.
5. If you cannot do a certain integral, just leave it as an integral in your solution for
partial credit on the rest of the problem. YOU MAY LEAVE THE SOLUTION TO
A DIFFERENTIAL EQUATION IN IMPLICIT FORM.
6. No calculators, books, or notes allowed. Turn off and put away any cell phones.
GOOD LUCK!
(Remember, each problem is worth 12 points, for a total of 72)
cos ©
ys
1. Find the general solution of 4zy’ + 2y =
44 9242 _ ont
2. Find the general solution of y’ = yer
xy’
3. Find the general solution of (32? + y) dz + (2° + ay + x) dy =0.
4. Find the general solution of y” — 6y’ + 25y = cos5a + xe*.
5. Solve the initial value problem y’” — 3y” + 4y = 0. y(0)=1, y'(0)=0, y”(0)=0.
6. We are given that y, = x~? is a solution of the homogeneous linear differential equation
yl + Baty! + 4a *y = 0.
a) Use the method of reduction of order to find a second solution y2 of the above differential
equation. Please assume that zx > 0.
b) Show that y and ye are linearly independent.
c) Use variation of parameters (NOT reduction of order!) to find ONE particular solution
(yp) of the inhomogeneous equation
yf tba ty! + 4e-?y = 27 ?,
(\ Msg +24 = oe Rerroslly wth ne 3
make subsk Qeyl%= 4
J
Se ve Steg
rarlk gly Me DE Ly “2 Ge ther vu! *PPpeare
4
“hey + ly. Usx e fhe 2. wx
x
Aris is a \in ie inkearak | bx
Mn ea~ DE LF, vy cad ara doe poe
Unx lL
=e — x,
mvlkpy ky BM vu + 2xu = XCosy
ork Lits Cu): = %@sx
as aA deriva Nive:
kegrate: sus \xcesx des Cs \xt@o9 tC
= Xsinx — (‘sins | x CF Xo ECO EC
Sm
Ur sna carrer C = Sax asx
xt x SET
\,0 Us 4"
yO SInx Cos C
Se [s = x a + << in is Sehukon
ev car also poke Lars exe Ae 4 Vee.
> (AS aly ro \'
<e
| a
Ae ge Ye RP HAH Spanair Lennie ore pt
: ca
--————- c a depee 4 1 $0 ik's a
a hemeqenesrs DE Cin Me norlinear rege »
4 FC)
; HUA 4 \ ut Lm
ye. vo Yee Ax “A “ :
meu 4 - 3 ° = = (4) (3) - 4
xy qU oO
(4)
shackle Oe Sv yrs So 42 Xu 0 - We qet
|. 4 ge Arndbions £ Ned We, Lub Ks ease- 4 ke d(i4\
ag & SUK Won
Hr Lee % 4(v-")
sy
~ lal + C
' ”
FL atlial = tab} Co Me oh
- a| = lable
> ~ Henert §s\ hon .
=\ fea
Bs 7 as writen, re 4 Nts Khe. Gren Mdand 6
oh W. ®ag)ey — Ways at
=! +4) | =F Sa ae) = bety ey],
JM (3844) dy be pro Cenryes) \, 2 O
aew NTO
we bal — % (neo) - Bnew)
“ We,
= Ay tay A) = Pe Come) spre)
Th ting ak Maal tag ph = AOD Lece not Corte, br pepty) dee werk
B= may ~ 4a on) _~ 9 +B )
a
So Fs = p> “Was easily leade b Choy rz"
Se Ane Cyack ne 15 “ =,2 i %
x 4 \L
ey % £e (ye) 42 O
tus 4 fy. uS ne
+x) rel * éke s e) (3 r ys x)
ee gv ne, So we O woe sO jyc~ t- Stn)
© Ae nel Sohn as Hayy eC -
. een Mec ao
Ve, | CeusSxl}—= (3° st
LOmsx) = —BWVoar Sx \ ‘Te, > Te oS |
YP, LCxe” | = ( Derry) ~( exe) + 2S xe% =
— ~Y4eX,-Dorex
xe
L{e% J > e* - 6 ex + 2S ek — Wer
L[owrbel = (end) e
Yon}
eX
So ~
~YC+2eD 3 o
ec = \ “> Ck, Ded
oo
Co) z0
YS) = ©
5.\ Yin Bye =O ts.
POM Be = Gani (ne nr) = Mryim-2)
M,~-S Ma zMge
Shes oh \\: hers eo + Ce * |
AWberer\iche ts ANre ( \atkfal Condibiers :
4> hes 4p Re™ + Cre
Yo Remy 2B OC BE)
Le
4"° Ae™ +4Re™ + Of He. dee)
whedly Ch +R ml
~A +B + Fo
t4HR r4C 2A
C= A +8 -_ A +8 = |
Be +C = { c=)
BR ES = |
SR + 4C 2-1
Bac F +2
<i“) |ea\ [5]
4 4
2.
asa srlka, & svdboshh {7 yve KOU ,
2\_! 2
a) v +(- uy Asay ty vy
chak oO. Nisha CAA
fe Oz yile Seyi Gy \= es
vi
- =)
apie See, fs aw WK
Se ean ee
Kore o* za eseslediy
; deo we rm _
Ske Wwe) se an —~ KW ibe de 2 tx Se
\,q wo = ~lex (we ony nook oye Wy earn ne we Corll eve \n ele -hy eC)
Se \ : nS x
Ly = x ~ Os <A Ss we take Pus tax} ( we ve oy teed
' Vhs. wet
So Any rae WA seule . qerer\ Vis
orAlaxr 8°
Check Ub.
Ye oh en eh, Cyr Cp.oce
" ; . _ 7"
L et: chal. 1 * mass Inx ° 3 se, so = x, CX lays o
——
x Slax te tn Ce
er crude, wei) . ee “an ik L ) Ne e)
P ki Wea Cy se alto,
_< x ~ rdeeall ere > __
= 2x Flap ee s 420° lay = , het Welker (% ek, revue 6
On HARA hone
trades Vrerly wg $4 Moves )
More. Vu, Lo, sah
bede Bre ype us Mate = xo ARV at
© 48° ra)
uv) \nx Oy) ee
—2 4
Vy © va “ix ,! = © =) ies
. 1 Ca (Oe . a
_ xu} +(x mer) U2! = x7 VS ares (© bo do) * lori) ve % &
Roo ree di,
(e——)) Uy + \qx 04! +0 oy = =~ Inx the, o> Lys
‘ ~ > ‘ L . ce Bx \ jal
‘Os = & ty = *
\
ute § xn dy = ~% 8 tage dG?) = Ae tay _\* 2): mA (ilo - 3 |
S fu, = ~ Wel & x
3 A