MAP 2302 Exam #2 Review

Instructions: Exam #1 will consist of 4 questions, although some questions may consist of multiple parts. Be sure to show as much work as possible in order to demonstrate that you know what you are doing. The point value for each question is listed after each question. A scientific calculator may be used but no graphing calculators or calculators on any device (cell phone, iPod, etc.) which can be used for any other purpose. The exam will be similar to this review, although the numbers and functions may be different so the steps and details (and hence the answers) may work out different . But the ideas and concepts will be the same. There is a formula sheet after the last problem , before the scratch paper.

(1) Determine whether the given set of functi ons is linearly independent on the in­terval (-00, (0). (8 points each)

(a) h (x) = 1 + .r, 12 (x) = x, h (x ) = x2

\-he Y. Xl. Wl\t'£~ 't/'(Z)-=­

" (I~) \ '0 z: \_X \ ~ : \ ~ X~ \ 'v : \ 1..~

0 0 "l­

(b) JI(x) = 5, 12(x) = cos2 x, ./3(X) = sin2 x

tO~l X

o

o

- 5 o

o

o

(2) Find the general solution of each differential equation. (6 points each)

(a) y" - y' - 6y = 0

w,.z.. - '" - ~ =-0

(b) y(4) - 7y" - 18y = 0

'1. '2.. Cf\'I - l-vn - (0 =: 0

(1M +~) (11'\ - J) (W\'l- t 1') =-0

I ~ \,VtIL .} 2. ::'0

(Il~ :"- L

i1\ ~ t M -=- ±l JZ. =)> J..-=-D, P=~

- 3 ~ 3~ ~)<, ( - \r-(,e -1-(1. e -+ e C3 COS J2 ~ +(~ t)1'" rz:~)

r e- )); + C e1 ).. + L ~ rz: X -\- C't ~\~ Six - '--I 1.. 3

(c) y" - lOy' + 25y = 0

CIr\ -S) (M -5") -=-0

(d) y" - 4y' + 5y = 0

- ~ t J-c[-~(I) -

(e) 16y(4) + 24y" + 9y = 0

If.o~ ~ t 2"( ~~ +~ :::'-t)

'-2 ~ ~ J2'('2 - l{ (r~) (~)

J(lIt,)

(3) Find the general solution of each differential equation. (15 points each)

(a) y" + y = 2x sin x

GVl~~~ ~r:: (A'kt~""x + (Ck:+O) CDS)(

'J,."J. ~~~ : 1p:;: X(A-~tB)~)c of )c (Ck+D) CoSx

~ V-t xl. +~)()Si(\X-f (CkL ·Vi> k) (os X

Yr::: (cIAIt: .H~)~lV\~ +(t\-l+8lt) Co~~ oJ-- (2C ~t"b)COO't - (etl tDx)Sil\)c

-=- (-C-l -I- (QA-1») ~ -1'13) '5\1\ H (Pr<2 -I- (6~,1C) ~ vi)) CO~ l(

~f" =- (-Jb. + (;uI-D))Si"" +reb (2A-i»).+~ (os!\: .I (':lAy ~ (6~lL))Cu~X - (lh~ +(~1(.)A~ ~~)(

~ (-A-'? - CfJ-t~t)( _:rO)~It\)( +(- l'l!? -r(t(A- -1))~ -I (2&Jc)} Cos X

~~' 1f -= (-Itl-( I'l!-\{t) ). - 11» Si'o.'< ~ f- C."+('1.4 -1)1 ~ -I-- ( :l~.:It\~x -\-(kh6xh"" +(&,ZtPx)to.s~

::= ;Lx ~\'r--. '( ~ ~~'v\vJ Ot'\ Sc.rli\~ f¥Y '*

(b) 4y" - 4y' - 3y = 2 cos 2x

l{ \4 " ,J f , - '1,y - 1, =-0

~ W\1. -ltm -l> -:::- 0

(2m -3) (ltr\ +I}-=: 0

w'\A - ,}. - 1 "\ - 2.' 'Z

6v-t~s : ~r -= Aeft':, 2)( of- Bsi" 2~

ry :: - 2A "O"IV'\ 2~ .} 2 B c.o~ 2 Xf

y~1 :: -'fA-to) Z" -l{& s\V\ 2.)c

~ ~/ -41( - "3'1r ~(- I{ Jl.LoS h -"lB\i~2x)-Q (-2~'h+U CoS :Ix) - ~(A-tos 2~ +6~" 2-~)

=- - ((, A(os 2)( --( ~BS\t\ t.( +<0 ~1" 2. t -tBCo:; 2)( - &A-cos 2;\ - 38Q" '2.)( ~---~~ ~

~ (-IC,A -fB)C05 2>. + (~A -l~S)s·,,, 2~ ~ L~2)(

q(-(,,4 -8B ~J) ~ - r-SfiA -lD l(~ :: Il, It'1 (--('A - f (3 -::. .2) -~i-It\ - IX G-:;.~1 ~

\0, ( <64- \'\5::: 0 ) l$..A - "3~' B =: 0 I _~ (~A - f~B ~o) ..:.(,l{A r-' -lB~ - tt2~A- -=- :3 ~-Y2S~-= fL,

1><t "" = - YlS'

(c) y" + y = sec x tan x

Yp= ~(c>s~- <;'I\.)(-~I(\X· (t>\ ILoS ~d

Y-= CI Co<, '< .). (1 t x f'XtoS ~ - j.i\~ -~I\l( ol·k~. 1

(C2-l)S.\t\~ l:C 3<)I'''X

-=- X - -\-tw-'X

U. c ~ CO':> 'Y:. S{c..'K ~If\ ~ ~ ~ X

Z

lAL -:: J~~ ~~ JS~~~ ~

U:::.. to ~ )(

clu -:. - ~lV\~ AA

- J ~ Ju -::- \~ \\A \ ~ - \~ \ C()S)( \

~f= (l(-+<m ~) ton -l" \Co~l< [- S\"l< ::; 'I( tosx-~- 9>'\)< o(., Ito~ 1< I ~ .c,~ '« -::. Cb\" ')cCO~lI

(4) Set up the appropriate form of a particular solution YP ) but do not determine the values of the coefficients. (9 points)

y(4) - y" = 4x + 2xe-X

l"t}

Y

tn4 _ in1.- =- Q

t"f\'2. (m1._\) ::: I.)

",,1. (M+~ (m-i\=O

'~Gvf5s" 1r ='dA1<t-B) .!-l< (C\ +-1»)t-~ -:: (A-~~+B);) -\ (C):z+lh:) e-)(_\

Extra Formulas

Variation of Parameters:

Scratch Paper

-l{c. =J ;:, c=- { - JD =-0 ~ 1)~o

~A- -=- 0 ~A-::..v

.2.8\;).(. ~a ~ d-5+.J (--\.)= :2B-I~ ==!> a-= i

Scratch Paper