×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

## Week 1 notes : 8/24-8/28

by: Samuel J Richards

145

0

18

# Week 1 notes : 8/24-8/28 4100

Samuel J Richards
Mizzou
GPA 3.72
Differential Equations
Dr. Jose Martell

These notes were just uploaded, and will be ready to view shortly.

Either way, we'll remind you when they're ready :)

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

These are the notes from Dr. Martell's class including 8-24 and 8-26 and 8-28.
COURSE
Differential Equations
PROF.
Dr. Jose Martell
TYPE
Class Notes
PAGES
18
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 18 page Class Notes was uploaded by Samuel J Richards on Wednesday August 26, 2015. The Class Notes belongs to 4100 at University of Missouri - Columbia taught by Dr. Jose Martell in Summer 2015. Since its upload, it has received 145 views. For similar materials see Differential Equations in Mathematics (M) at University of Missouri - Columbia.

×

## Reviews for Week 1 notes : 8/24-8/28

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 08/26/15
Differential Equations Guilliams 82415 21 I Introduction a Differential EquationsWhat are they for i Usage modeling and real world behavior 1 Involve rates of changevariation a DERIVATIVES 2 Examples a Fluid motion b Electric current ow c Heatenergy ow b Forms i Example 1 falling object using LINEAR EQUATIONS 1 Variables ttime seconds vvtveoctity Air resistance drag T v Newton s 2nOI Law Fma FForce Newtons N vtvelocity Down is Positive mmass Kg aacceeration msZ dv 2 Acceleration is 1St derivative of velocity a E v39t dv thus F ma m E 3 Forces involved a Gravity g 98ms2 b Air resistance drag Tv where T is the drag coef cient and v velocity ms dv 4 NowF ma m E mgTV where drag inhibits downward motion of object and is thus subtracted ii Example 2 Falling object but now with numbers 1 m 10kg T 2 kgs v0 0 mswhat is vveocity in this situation 2 Consider the general solution a vt491e39 392is this a solution Differential Equations Guilliams 82415 21 dv i E 98e39 392t is the rst derivative of vt which is also acceleration dv 3 Fmam E mgTv dv a LHS left hand side m E i LHS 1098e39 392 b RHS Right hand side mgTv i RHS 1098 2491e39 392 98e39 392t c LHSRHS thus vt491e39 392t can be considered a solution II First Order Differential Equations dy a E fyt where y is a function of t b Baby Model dy i dt ft ii y fftdtc Section 21Linear Differential Equations with variable coefficients a Method 1 Integration factors dy i Ex 5 ay b where ab are constants dy dv 1 E ay b reminds one of the m E mgTv ii General Form dy 1 Pt E Qty Gt where PQG are given Differential Equations Guilliams 824159 21 I Introduction a Differential EquationsWhat are they for i Usage modeling and real world behavior 1 Involve rates of changevariation a DERIVATIVES 2 Examples a Fluid motion b Electric current flow c Heatenergy flow b Forms i Example 1 falling object using LINEAR EQUATIONS 1 Variables ttime seconds vvtveloctity Air resistance drag T v Newton s 2nd Law Fma FForce Newtons N vtvelocity Down is Positive mmass Kg aacceleration msz 39 1 4 d 2 Acceleration is 1st derivative of veIOCIty a 61 v tthus dv F ma m dt 3 Forces involved a Gravity g 98ms2 b Air resistance drag Yv where T is the drag coefficient and v velocity ms 4 NowF ma m mgYv where drag inhibits downward motion of object and is thus subtracted ii Example 2 Falling object but now with numbers 1 m 10kg T 2 kgs vO 0 mswhat is vvelocity in this situation 2 Consider the general solution a vt491e39 392tis this a solution a dt acceleration i 98equotquot2t is the first derivative of vt which is also Differential Equations Guilliams 824159 21 2 3 F ma mdt ngv a LHS left hand side m i LHS 1098e 2t b RHS Right hand side mgYv i RHS 1o9s 2491e39 392t 98e39 392t c LHSRHS thus vt491e39 392t can be considered a solution II First Order Differential Equations dy a a fyt where y is a function of t b Baby Model I dt ft ii yf ftdt C III Section 21Linear Differential Equations with variable coefficients a Method 1 Integration factors d I Ex d Z ay b where ab are constants d d 1 d Z ay b remInds one of the md mgYv ii General Form 1 Pt Qty Gt where PQG are given Differential Equations Martell 82615 21 Continued Linear Differential Equations with variable coef cients a Method 1 Integration factors General forms dy dy 1 pt E QtyGtremember E Y39 d 2 dt dy Pty Gtthis form isolates dt and is easier than the above form to use Integration Factors ut 1 Simple form ut equot IF where Pt comes from general dy form 2 where E is isolatedYOU MUST isolate dy E for this to work 2 Long form dy dy a Step 0 isolate E to get 5 Pty Gt b Step 1 Mt c Step 2 multiply step 0 equation by ut dy i ut E utPty ult Gt d Step 3 product rule and compare to step 2 i d I d quot0 dt 3 ii utPty du iii a ptPt iv Therefore 1 du 1 Mt E Ptnowintegrate du 2 IL lnpt IF dt dt 3 lnut IF dt this equation is changed to exponential form Differential Equations Martell 82615 21 Continued 4 mt equot IF hmm this looks identical to simple form e Step 4 substitute ut with equot IF and solve as shown below iii examples using integration factors 1 1 1 y39 5 y 5 et 3step 0 is already done for us because y39 is isolatednow to step 2 1 1 2 uty39 ut E y ut E e 3 here we multiply by ut d dy du 3 E uty E y and we can compare with equation i from step 3 1 4 uty39 ut E y and now onto the product rule in step 3 a From 3iii and 3iv du i E utPt which for our equation becomes du i ii dt Mt 2 1 du 1 iii 0 E 5 integration time du J 1 t 5 5 becomes lnut 5 becomes ut dt et2 6 Now multiply in the integration factor and solve for y 1 1 a et2yi et2 E y E et2 et3 1 1 b et2yl et2gtlt E y e5t6 c E et 2y 5 e5 6 now integrate each side Differential Equations Martell 82615 21 Continued 1 6 i et 2y 5 g equot 6 C and solve for y by dividing by et 2 ii y g e 3 Ce39t 2 iv more examples using Integration factors 18 pg40 1 ty39 2y sint where y 1 and t gt0time to isolate y39 2 sint 2 y39 y t now time to nd the integration factors the hard way because Dr Martell likes it D so we multiply by mu 2 ut y ut t y ut t dy d i 3 pt dt dt y dt uty d 2 3 dt l1t t 1 du 2 b 0 E 7 nowintegratetoget c lnut 2nt d l1tt2 now multiply by your integration factor 2 sint 4 t2y39 7 yt2 T t2 5 t2 y39 2yt sint t d a E t2 y sintt and we integrate each sideremember right side needs integration by parts which I am not showing b t2 y tcost sint C now solve for y Differential Equations Martell 82615 21 Continued c y t t2 t2 now we solve for C with initial value of y 3 1 i CgtIlt4 d 1 0 H2 H2 and solving for C becomes H2 e C I 1 H2 cost Lnt 1 6 SOLUTION y t 2 4 Differential Equations Martell 82615 21 Continued Linear Differential Equations with variable coefficients a Method 1 Integration factors i General forms 1 2 NW Qlty Gtremember y ii Integration Factors pt I 2 Pty Gtthis form isolates and is easier than the above form to use Simple form pt equotf Pt where Pt comes from general form 2 where is isolatedYOU MUST isolate for this to work Long form a Step 0 isolate to get Pty Gt b Step 1 pt c Step 2 multiply step 0 equation by pt d I ut d Z utPtv ut Gt d Step 3 product rule and compare to step 2 i I dtut v ii du dt pt Pt iv Therefore 1 39 I 1 Mt dt Ptnow Integrate 1 du 2 IE E lnlllltll fPtdt 3 lnlt f Ptdt this equation is changed to exponential form 4 pt equotf Pt hmm this looks identical to simple form e Step 4 substitute pt with equotf Pt and solve as shown below iii examples using integration factors 1 2 3 y y et3step 0 is already done for us because y is isolatednow to step 2 pty pt y pt et3 here we multiply by pt pty pt and we can compare with equation i from step 3 pt and now onto the product rule in step 3 a From 3iii and 3iv i ptPt which for our equation becomes du 1 dt pt 2 i du 1 I III Mt dt 2 Integration time i Q l E t2 0 dt f2 becomes lnpt 2 becomes pt e Now multiply in the integration factor and solve for y Differential Equations Martell 82615 21 Continued 1 1 a et2y et2 E et2 et3 v b et2yl et2 y l e5t6 d Eet2y 2 est 5 now Integrate each Side NIHNI i et2y g e5t6 C and solve for y by dividing by et2 3 t3 t2 H y 5 e Ce iv more examples using Integration factors 18 pg40 1 ty 2y sint where y 1 and t gt0time to isolate y 2 39 t 2 y y now tIme to find the Integration factors the hard way because Dr Martell likes it so we multiply by pt 2 39 t 3 ut v ut v ut 2 4 t 2 Ex 1 t If we simplyjust did tenf EulteZIntt2 39 391 dt dt v dt quot y du 2 It would have been far easner a a pt E b now integrate to get c lnlt 2lnt d ptt2 now multiply by your integration factor 2 Ea 2 2 5 t y t y t t t 6 t2 y 2yt sint t a t2 y sintt and we integrate each sideremember right side needs integration by parts which I am not showing b t2 y tcost sint C now solve for y cost sint C n c y t tAZ tA z now we solve for C With InItIaI value of y2 1 4 04 d 1 0 and solvmg for C becomes HA2 1tquot2 HA2 4 1 HA2 1 cost sint T 7 SOLUTION y t m m C i OMS 32 Manual 1 LamWow fad P1902414 d a 39 A 14H quot deg lo 50M d 5 739 ESO CL ILG a4quot 392 5 l B bk 5quot 739 1L FAQJT OA C wr 38 QM it 53 Solve Ll my t fly 1 l I H i I quot K 0 jc M i i 39 I jlwm mio aw M i 394 MHMKH r J T w J J 2 AH I WWW 5 a as 4 v 213 a x i 1 V Mm M4 waw a q 39 quot 1 g PHXH AMA 1 awJ 3 SPHM J2 max cum5g 1L0 s SPHM I ExpowJCak i J 393 AH PM w jL i39quot 74 A A 39C AKi a a f Iquot j H SPGW I 954 59600 4 i 39 I 39 Haw 06 Md I Lng 5 A P A P I 1 3 9 5 1 1 Odd in MUG MMM R 1 WWW 5PM n quotJ J 5 31 8 a 3 H A t W F 394 394 quot SPHW A A U V 6 39 L J 13 2 m LJ JIIquot f M671 L ilM CV 1 fo QW39J39th MaC39DLS 59399611 Cbl 39 Ezgau b mg 3 39 L A z I 4 AX l UV g WCCL 1106 9 11721 a U 012 m I l lJJ AX v 3 6k MCKHUA F M r K j WHX 4 M5 so i Lo 6 464 3 are gorable examupk S M 0 L90 Siegwad x3 1L 39 0 9 D0 m ASE wwtxde lKQAQ 39339 589 169 I 39 oi X XOIT W k 939 ts to SIMS K39 39 wadj q MK 93 2 Wquot W b kx 06A 5 01123 3 M J39 m f IM KJK VJ L U 5 3 3 41me aim7i Sa lw I Ex I f 42 W a 9 E My 3935 w 39 3 a Mm quotH z quot d 5a 5amquot QM 10 EOSS WFM E 3 x A L 393 l 3 C quot I P 3 P 2 39 P 3 A quotF C f 3 J 734 A g r 5L 3 C 2 0i h ml Cw Exawuw 61 r ma g 50 Cluck UMMQ which quot39 z ME 52 x l 14926quot I 430G gtlt quot 3 Q J M E ii a 3 sh smock x Md 1 9 I aa 9 6 U 3 1 A 56gt mme on M g h 1 X a S t nL szxoC e 0quot WAV C 2 gt 39 U b 10 K 860 w r M if a A I a i 59W 3 quot 2 C W s0516 P 50quot lt9 393 3 3 3 if 9 quotg C7 3 9130 MW 0 cfovst ym i Jemwlok a 93 Jh 77 ALX m WW a 28 e a w 93 3C it 2 OS OYX w C05 3 93 3 w k was 9109352 7 o 0quot f 05 2Xwoamp 9 3gt2 0 39 R I 3 Te MWquot bQf WWW Whamg oes OSQCE SB 10 k a v Q l 39 h 7 W Wk A A g 1 1 Sig 1 w A k grQNJ J 114 Tr ukm e e 3quot ELI1509 0Q5W5C39Wamp170 ivrvasz w isqwg f pg m f L v a Qquot aquot A m e as 20 cog h A 5 15Lw2 3 i ib fCl L w LSMWCM MK 1 5 61g5 74 a v gwx 60 e 393 quotEMW 55 35 dkHC WSQZK Mgi 5a MLij b5 x quot1M Clix 392 392quot 605933 J L 663533 5059 53 leaksWV jwcanw 3 5605 W in i460 99n Wt 0quot 2 3 w 30306 r we S M C1Kgtj f a C 7 L m Izr r X39 V 39 l39 1 SOler39txOm 7 1 K 01C MMJJJJJJJmewJIImwwmleI 232 13 0 E emmrxle 3quot 9 Mi quotatx quot 3ka r l I v uq39f atl quotO 1 valxe J a FCLJ nk a39 MIMI Sdbikb 3 dg CD 18quotan39 501m H O g X 0 3 0A9 295 valw mp d0a vm JF Y S J i J 24 2xquot Vquot l 1 v v MU b r 69quotquot erQ is A0 mm M 3 WWW i 1 MOL VPS 0 Ni no u a all w L P 3x N 36 3 erapg 701 68 6003 gt J J quotx f 395 ZVgtltLI35 quotWJSH5X 1 533 i 333 ng l x 3 quot3 A 22quot S S W f5quot 3 2 7lt 4 X S WLC 500 mm 3911 9 3 me 6 member 103 2 u W 3 m3202 41 4C u J 01 c r39 quot Q 35 W 5 WqK QEWWWW SbM ftx rx 7 L fewk CVVLbQP IfN RAULQPH S V r AxMil 36 A4 T39hd 5le K 01 w if 23 In quot 5M 339 th lt3 25 V BJK 3 0 4 0 0392 f MOW Bf mc L414 and M I a 27 v U Mot I SOIVKWLUOA and W domm vx CUT1 Cit S luh n 6K 169 5 3303 34 I Xquot 2lt3 quot lt9 We 3 0 30239 gtlt2lt39 amp L 1197 9 3a quot391 MK 3 A M 393 O Xgtlt 9 0 f r rcarPk of 5136 9 X S W I A gtlt2lt 3 quot 9 K 6 r1gt 0 H i 393 IT A 39 C 1 30 x Wkva M20 x K W 3 5 Q 1 xrxiqg camka K1Man A 55 39 399 X L quot X f x 9 7756 I 3 HULKC f4 lt29 2 my Ni 1 gtlt 2 U YQ 0 1 O UJMJMJLJN bJ pHNNb lleHJUJAWMJLHHHHHMWHULWH Q C 3 gj 7 ampfw4 JUN K WW3 ISSOCFCK LG AP we 20 114 198M104 151 f Tao M t eflCJ I312 w I k W Sc llVhbn 5 119 1 l u K 3 93 3990 130639 A U quotI R 00 140 Oamp L KELLOO F C A f 1 4 394 H 7 g 7 l h i A C H 7H H WW AA gt 7 i D r A ii A I

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Jim McGreen Ohio University

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Anthony Lee UC Santa Barbara

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

Steve Martinelli UC Los Angeles

Forbes

#### "Their 'Elite Notetakers' are making over \$1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com