MAP 2302 Exam #1 Review

Instructions: Exam #1 will consist of 6 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.

(1) VerifY by substitution that the given function is a solution of the given differential equation. Primes denote derivatives with respect to x. (12 points)

x 2y" + xy' + y = 0; y = sin (In x)

- Sin (r" kJ - COS(/">;) X2..

(2) Find a function y = f(x) satisfying the given differential equation and the pre­scribed initial condition. (12 points)

~&~~ ~ \'Z..r -t.?&

~t4 = ~c"l-9-de ~(C. e=V'-rx~ ~ 1+)C'l.= sec. '2.g

=- - Co{ e - e

- -' k\" - I- X-~)(

y(x) ==- - ~ - ~.I X +C

Y(I\= - r:: - +--\-th-' (1)+ C. =";> -1; =-( - t +C. ..". 0 =-I+l­'-\ =) C ~(

\ ( ) =:: _ .L _ -Itm- 1 )c+ (\ 1:: NoIc'c \l-i!> ~ c.., lei '-"' ..{I. I.a... rJ.,...Y ~ K b, f'4,~J ~c.;f.j~s .

(3) Detennine whether or not the Existence and Uniqueness Theorem guarantees the existence of a solution of the given init ial value problem. If existence is guaranteed, determine whether Existence and Uniqueness Theorem does or does not guarantee uniqueness of that solution. (8 points each)

(a) ~ = cotx; y(O) = 2

(b) ..;y1J; = vx; y(l) = 1

~_ E..:: rT ~ - ~ ~j

, I

-f(~lj) == ~ ::- ~ Xii' -I-'('xl~) to'A-\i~"f16 ntM (L,I) '* ~ ~;'-Is c.. -ool.m...,

~ - _ ..1 X { ,, - t __..E.. ~ ~-\l~vt.\)~ ru, (I i) ~~ SoIJ"h~ Is vn6i~ ~- "l f -- ~

(c) ~ = y~; y(l) = 0 ~ -f{~(j) -;: J~ :: q

-fC>cj) Ulw\\" V~ Nt.<- (I, 0) ~~ f4,iS-ls '" 60 Iv,....

rd- (1,0) So ~1~I\iSS

~.

(4) Find a general explicit solution of the differential equation. (12 points)

dyx-=4y

dx

Sjc1J=~Jt~ l~ IJ\ -= ~\V"lt\ tC

\h 1,1":; I.. rd ~C I~ ():")ft

1,1 ":; e

J.< (~'4) C

I,l-=- ~ -e

(5) Find the general solution to the differential equation. Primes denote derivatives with respect to x. (12 points)

I XY +-y=2x

x2 - 9

x -p()c)~ t--~

JreXloW:O JXL ~ oVx: '" -i r{- d.u:: {l", lI. =' {: \" (i--'i')

l-t -;::. )('S.~

ctl\~2~~ ~ -t clu~/(~ {\" Cx~~) l{ (J'(lt-~ )

}A(~)=- e ~ ( =-~ ~2_~

.rM(~~ /~ 'f):: m (2)<.')

(fk'"-~ ·yY -= z\<S-x~-4

J(~ ' IYrh -= r2krX~-f ~ u-=.'J;,'l-i ctu:; ~)ctkx

Jz.k J.,.,-cjoIil: :: r~ dlA -= J'-I "2 ./,u -;. -\- u.'/'4. c.. ( \(1..

:::. ~ l-t-~) -tC

~~J1'1 = ~(t-,)*+-c =';i Cx2.-a,)'f'l:=' ~(k'-'j)~(''''(

~ ~~~/1= (~._,yI/2[}6:'-'i))/.+c] ,-'('l.. 1

=:;, 7(k):: -5 (x'-'i) -\- ( c'/-~ J

(6) Find the general solution of the differential equation. (14 points each)

(a) (3x2y + eY) dx + (x 3 + xeY - 2y) dy = 0

M(~,~') ~ 3,,-'lJ t(Y~ ~~ :: 31 -te '1 sc...... 50 '14 iwJi"" " ~Ii!.f ~ 'f i2tL - '2 2-.1J 1

N(x,J):: X +~e -2r ~ d)c - JX ''­

;~ :: fA ('<'j) ~ ~:: 3X2

J-te~ ~r(J;':J):: J(~l<.7.JH7)0/)(

::. X; ~ tX(Y + j ( Y )

~:: NC)('J)~X~+JI(J)= /t;t<?-2, ~jl(J) ~ -2, ~ j(Jl~ J-2, 1:;-1 l

(b) (x + 3y) dx - (3x + y) dy = 0

M(~,J')::' 'f.-f',/ ~ M ( h i -101) -=- -w 3-+;,/ ::. -I: 0.~ ~ 14 (~, ~) -is h"""3"'"lIv~ .+ -"Jru I ~(x,~)

N(l<,~h -~x -J ~tJ (hil-yl" -~~ - ty ~ -c (-3k­

Use 'flt ~Jos~~OV\. , -= ux -::;~ :::~~+ It d" ~SJos-\l1v-k 1~ onylrJ iwJt~

~ t ~x)oVx - C~)c f-(A~\ (u lk+'\:~lA\ ~O

x~ +-...LrJ.t,. -:~-"){q"-tA"'l<DllO -~~O -~ ~ ~

ex - l,t)c 1~ ~ ( ~A:1tt,{)c?)&'lA ~ o

ex -\,{'-~)~ :: (3)(2-~~)~ =9 'X Cl -~~} k =X2C~f4)otL\

x ~-MII J \ ~ - JF W ~ du.::: J t-\-1' 3 d-x>r4=- ~ct" ~ x - I- U'" / (It")r~u' ..

J {+ 1J _ A-(f"1.A\.f- \3 (I~ ') _ t<\~ 3 l~h:\ l~ ..{- I-u - (I~ ) (( - l..t\ - (ltl.\')(r\.\)

AC 1-\,\) l-13 (1+tA') ::. tA-t 5

l..\ -;. \: 2. ~:: \.{ .;, 6-==1...

\..\.:;.-\ : 2A"";.2.~A:;"l

j ~ct~ ~ 2 }I ~~ dM

1",1 \4t.\\ - 2\", \ \-"'\ 4-(,

\V\l~\ :::l",~ \-\-v.\-2l~\ \ -~\+L

\~ lxl :: \~ \ \+ ~\-L\v\\ \- ~ \-\c