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# Differential Equations MTH 235

Donny Graham
MSU
GPA 3.57

Gabriel Nagy

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COURSE
PROF.
Gabriel Nagy
TYPE
Class Notes
PAGES
102
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 102 page Class Notes was uploaded by Donny Graham on Saturday September 19, 2015. The Class Notes belongs to MTH 235 at Michigan State University taught by Gabriel Nagy in Fall. Since its upload, it has received 5 views. For similar materials see /class/207298/mth-235-michigan-state-university in Mathematics (M) at Michigan State University.

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Date Created: 09/19/15
Review for Exam 2 gt 6 Problems 55 Minutes in Recitation Rooms 100 Grading Attempts V V Problems Similar to Homeworks V Integration Table Provided in Handout V No Notes No Books No Calculators lVlLC lVlTl l 235 Exam 2 Review httpmathmsuedumlc Exam Covers Variation of Parameters 27 Undetermined Coefficients 26 Applications 25 Homogeneous Constant Coefficients 23 24 Special Second Order Reduction Order Method 22 Second Order Variable Coefficients 21 First order Nonlinear Equations 16 Applications 15 V V V VVVVVVV Review for Exam 2 Notation for webwork Consider the equation y l 311 l 32y 2 0 Let r r be the roots of the characteristic polynomial gt If r gt r real then gt First fundamental solution y1t e gt Second fundamental solution y2t e gt If ri 2 ad i6 complex then gt First fundamental solution y1t eat cos6t gt Second fundamental solution y2t eat sin6t gt If r r r real then gt First fundamental solution y1t e gt Second fundamental solution y2t te t Review for Exam 2 gt Exam Covers V Variation of Parameters 27 Undetermined Coefficients 26 Applications 25 Homogeneous Constant Coefficients 23 24 Special Second Order Reduction Order Method 22 Second Order Variable Coefficients 21 First order Nonlinear Equations 16 Applications 15 VVVVVVV Variation of Parameters 27 Example Find a particular solution of the equation X2 y 6xy 10y 2X107 2 are solutions to the knowing that y1 X5 and y2 X homogeneous equation Solution We first need to divide the equation by X2 6 10 yy2y2x87 X X Then the source function is fX 2 2X8 We now compute the Wronskian of y1 y2 X5 5X4 2X Y1 12 W 2X6 7 5X6 Y1 YQ Hence W 2 3X6 Variation of Parameters 27 Example Find a particular solution of the equation X2 y 6xy 10y 2X107 2 are solutions to the knowing that y1 X5 and y2 X homogeneous equation Solution y1 2 X5 y2 2 X2 1 2 2X8 W 2 3X6 Now we find the functions U1 and U2 2 8 ygf X2X 2 4 2 5 1 W 3x6 3X 1 15X 5 8 u 2amp2 X2X EX7 gt u22 3X8 2 W 3X6 3 2 2 2 1 1 VP U1Y1UQYQ X5X5 8 2 10 15 x X X E 2 8 5 1 that is yp X10Tgt hence yp EX lt1 Variation of Parameters 27 Example Use the variation of parameters to find the general solution of y 4y 4y 2 X 2 6 Solution We find the solutions of the homogeneous equation r24r40 gt HE 4j 16 16 gt ri 2 Fundamental solutions of the homogeneous equations are y1 e y2 xe We now compute their Wronskian 2x eizx 1 7 2X e4quot 2X 6 4quot 64quot X e 725 1 7 2x Y1 Y2 yf y Hence W 6quot W Variation of Parameters 27 Example Use the variation of parameters to find the general solution of y X72 72X Solution y1 6 y2 xe zx g x 2 6 W 6quot Now we find the functions U1 and U2 ygg xe 2X X 2 e 2X 1 4 gt u1 lnx W e X X ylg e 2X x 2 e 2X 2 1 22 2 4X gt u2 W e X X 72x 1 72x 72X ypu1y1u2y2 lnxe Xe 1lnxe Since fp ln X e 2X is solution y c1 CQX ln 6 lt1 Review for Exam 2 gt Exam Covers gt Variation of Parameters 27 gt Undetermined Coefficients 26 gt Applications 25 r Homogeneous Constant Coefficients 23 24 gt Special Second Order Reduction Order Method 22 gt Second Order Variable Coefficients 21 gt First order Nonlinear Equations 16 gt Applications 15 Undetermined Coefficients 26 Guessing Solution Table f Source K m a b given yp Guess k not given Keat keat KmtmKo kmtmko K1 cosbt K2 sinbt k1 cosbt k2 sinbt Kmtm K0 eat kmtmk0 eat K1 cosbt K2 sinbt eat k1 cosbt k2 sinbt eat Kmt R1 cosbt R2 sinbt kmtm k0 fq cosbt IQ sinbt Undetermined Coefficients 26 Example Find all the solutions to the inhomogeneous equation y 3y 4y 2 25int Solution We know that the general solution to homogeneous equation is yt 2 C16 cge t Following the table Since 1 2sint then we guess yp 2 k1 sint k2 cost This guess satisfies Lyp 7A 0 Compute y 2 k1 cost k2 sint y 2 lt1 sint k2 cost Lyp klsint k2 cost 3k1 cost k2 sint 4k1 sint k2 cost 2sint7 Undetermined Coefficients 26 Example Find all the solutions to the inhomogeneous equation y 3y 4y 2 2sint Solution Recall Lyp klsint k2 cost 3k1 cost k2 sint 4k1 sint k2 cost 2sint7 5k1 3k sint 3k1 5kg cost 2 sin A t This equation holds for all t E R In particular at t t 0 5 5lt1 3k2 27 k1 7 gt 3k1 5k207 k 3 2 17 Undetermined Coefficients 26 Example Find all the solutions to the inhomogeneous equation y 3y 4y 2 2sint 3 1739 So the particular solution to the inhomogeneous equation is 5 Solution Recall k1 E and k2 1 ypt 5 sint 3 cost The general solution is 1 ya cle cze t E 5 sint 3 cost Undetermined Coefficients 26 Example Use the undetermined coefficients to find the general solution of y 4y 2 3sin2X 63X Solution Find the solutions of the homogeneous problem r240 gt rii2i y1 cos2X7 y2 sin2X Start with the first source f1X 3 sin2X The function fpl 2 k1 sin2X k2 cos2X is the wrong guess since it is solution of the homogeneous equation We guess yp xlk1 sin2X k2 COS2X y 2 k1 sin2X k2 cos2x 2Xlt1 cos2X k2 sin2X y illlt1 cos2X k2 sin2X 4X k1 sin2X k2 COS2X Undetermined Coefficients 26 Example Use the undetermined coefficients to find the general solution of y 4y 2 3sin2X 63X Solution Recall y1 sin2X and y2 cos2X illlt1 cos2X k2 sin2X 4X k1 sin2X k2 cos2X 4Xk1 sin2X k2 cos2x 3 sin2X7 Therefore illlt1 cos2X k2 sin2X 3sin2X Evaluating at X 0 and X 7r4 we get 41207 gt 1207 Therefore yp1 XCOS2X Undetermined Coefficients 26 Example Use the undetermined coefficients to find the general solution of y 4y 3sin2X 63X 3 Solution Recall ypl Zxcos2x We now compute yp2 for f2X 63X We guess yp2 ke3X Then ygg 9 63X 1 1 94lte3Xe3X gt kz gt ymz e3x Therefore the general solution is yX c1 sin2X c2 2X cos2X 63X Undetermined Coefficients 26 Example gt For y 3y 4y 2 3621 sint guess ypt 2 k1 sint k2 cost 6 gt For y 3y 4y 2 212 637 guess ypt 2 k0 l k1t k2t2 631 gt For y 3y 4y 2 3t sint guess ypt 1 klt k2 sint k3 cost Undetermined Coefficients 26 Example Find a particular solution to y 2y 2y ei4it Using this solution find particular solutions to the equations y 2y 2y 2 cos 4t7 y 2y 2y 2 sin 4t Solution Since the source is and exponential ft 2 674 we guess as particular solution the exponential ypt k 6 4 We now check whether yp is solution at the homogeneous eq 1 r22r 20 gt ri 2l48 gt Realroots Hence yp is not solution of the homogeneous equation Undetermined coefficients 26 Example Find a particular solution to y l 2y 2y 2 674 Using this solution find particular solutions to the equations y l 2y 2y 2 cos 4t7 y 2y 2y 2 sin 4t Solution Recall ypt k 6 4i2 2 4i 2 ke 4it e 4 t gt 16 8i 2k 1 1 1 1 9 4i19 4i 188i 294i94 29242 Hence ypt 1 9 4067413 292 42 Undetermined Coefficients 26 Example Find a particular solution to y 2y 2y ei4it Using this solution find particular solutions to the equations y 2y 2y 2 cos 4t7 y 2y 2y 2 sin 4t 1 292 42 For the second part of the problem we need to compute the real and imaginary parts of or solution Solution Recall ypt g 4i ei4it ypnr 292 1429 4icos4t isin4t ypr W 9 cos4t 4sin4t ypl W 4 cos4t 95in4t Review for Exam 2 gt Exam Covers V Variation of Parameters 27 Undetermined Coefficients 26 Applications 25 Homogeneous Constant Coefficients 23 24 Special Second Order Reduction Order Method 22 Second Order Variable Coefficients 21 First order Nonlinear Equations 16 Applications 15 VVVVVVV Applications 25 Example Find the movement of a 5Kg mass attached to a spring with constant k 5KgSecs2 moving in a medium with damping constant d 5KgSecs with initial conditions y0 2 and yO 0 Solution The equation is my dy ky 0 with m 5 k 5 d 5 The characteristic roots are 2 2 d 1 k riz wdztiwd wm wdz zi7 too 31 1 1 1 3 ri 2 E I Z 1 E l ig Under damped oscillations Wt A cos t 15 eat2 Applications 25 Example Find the movement of a 5Kg mass attached to a spring with constant k 5KgSecs2 moving in a medium with damping constant d 5KgSecs with initial conditions y0 2 and yO 0 3 Solution Recall yt Acos t lt15 642 Hence 1 it2 42 yt 2 Asm 2 t q5e 2Acos 2 t e The initial conditions 2 y0 Acosqb7 0 yO g Asinq5 Acos b 1 7T tan E gt b 67 gt A We conclude yt 2cos t e t2 lt1 Review for Exam 2 gt Exam Covers Variation of Parameters 27 Undetermined Coefficients 26 Applications 25 Homogeneous Constant Coefficients 23 24 Special Second Order Reduction Order Method 22 Second Order Variable Coefficients 21 First order Nonlinear Equations 16 Applications 15 V VVVVVVV Special Second Order Reduction Order Method 22 Example Find the solution y of the IVP yy 4v2 07 y0 17 yO 7 Solution This is an equation of the form y fyy 1 missing Introduce the function vt yt that implies vt y t so 2 v yv4v20 gt v 4 7 v07 y The difficulty is that y still appears in the equation We now look only for invertible solutions functions I gt yt that is we have the inverse function y gt ty For this type of solutions introduce the function Wy vty Special Second Order Reduction Order Method 22 Example Find the solution y of the NP yy 4v2 07 y0 17 yO T 2 Solution Recall v 4V with v0 7 and Wy vty The initial conditions for W are obtained as follows yt01 ltgt ty107 Chain rule on W always implies the equation t 2 1 MWZM W4W4u Wy y W y Special Second Order Reduction Order Method 22 Example Find the solution y of the NP yy 4v2 07 y0 17 yO 7 Solution Recall W 2 4K with W1 7 and Wy vty y W 4 l 2 4 l 4 W y 5 nW nyC ny C We obtain the solution Wy Ey 4 The initial condition implies 7 W1 E hence Wy 7y 4 We now consider y as function oft and we recall that Special Second Order Reduction Order Method 22 Example Find the solution y of the NP yy 4v2 07 y0 17 yO T 7 Solution Recall t with 0 1 y W y ys fly7 gt 7tc The initial condition fixes the integration constant 1 y5t 1 7 5 C7 gt 5 5 We then obtain the solution of the NP as yt 535 t 1 Special Second Order Reduction Order Method 22 Example Find a fundamental set of solutions to tzy 213 2y 2 07 knowing that y1t t is a solution Solution Express y2t vt y1t The equation for v comes from tzyz 213 2y2 0 We need to compute Y2Vt7 ygztvv7 ytv 2v So the equation for v is given by t2tv 2v 2ttvv 2tv 0 t3 v 2t2 2t2v2t 2tv 0 4 t3v 4t2v 0 gt VlV 0 Special Second Order Reduction Order Method 22 Example Find a fundamental set of solutions to t2y 213 2y 2 07 knowing that y1t t is a solution 39 N 4 Solution Recall v Ev O 4 This is a first order equation for W v given by w EW 2 0 so w 4 gt lnw 4lnt c0 gt wt C11647 c1 6 R w Integrating w we obtain v that is v c2153 c3 with c27 c3 6 R Recalling that y2 2 tv we then conclude that y2 c2152 c316 Choosing c2 1 and c3 0 we obtain the fundamental solutions 1 y1t t and lt1 Review for Exam 2 gt Exam Covers Variation of Parameters 27 Undetermined Coefficients 26 Applications 25 Homogeneous Constant Coefficients 23 24 Special Second Order Reduction Order Method 22 Second Order Variable Coefficients 21 First order Nonlinear Equations 16 Applications 15 V VVVVVVV Second Order Variable Coefficients 21 Example Find the Wronskian of two solutions of the equation t2y tt2y l t l 2y0 tgt0 Solution Write the equation as in Abel s Theorem Abel s Theorem says that the Wronskian satisfies the equation 2 12t 1 Wy1y2t 0 This is a first order linear equation for Wylyg The integrating factor method implies At tt 1ds 2lnt t to Second Order Variable Coefficients 21 Example Find the Wronskian of two solutions of the equation t2y tt l 2y l t l 2y0 tgt0 t 1 Solution At 2lnt t to lnt 2 t to 0 t2 The integrating factor is u t OZ 6707150 Therefore MtWy1y2tl0 MtWy1y2t MtoWy1y2to0 2 t 7 so the solution IS Wyly2t Wyn2a t2 et to 0 Denoting c Wy1y2t0t02 640 then Wy1y2t ctzet lt1 Review for Exam 2 gt Exam Covers V Variation of Parameters 27 Undetermined Coefficients 26 Applications 25 Homogeneous Constant Coefficients 23 24 Special Second Order Reduction Order Method 22 Second Order Variable Coefficients 21 First order Nonlinear Equations 16 Applications 15 VVVVVVV First order Nonlinear Equations 16 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y2y3 y01 Solution First notice that the equation is linear So it is simple to find the solution following Section 11 3 e 2ty 2y 2 3 672 gt e zty 2 6721 c7 3 t 2t y ce 2 The initial condition implies 5 1yoc yrgt e In the next slide we use Picard Lindelo39f s idea First order Nonlinear Equations 16 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y2v3 y01 Solution We first transform the differential equation into an integral equation t t ysws 2ys 3 ds 0 0 t 0 Using the initial condition y0 1 yt 1 t2ys 3 ds 0 ya ylto 2ys 3 ds This is the integral equation First order Nonlinear Equations 16 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y 2y3 y01 1 Solution Integral equation yt 2 1 2 ys 3 ds We now define the sequence of approximate solutions 1 yo y017 yn1t1 2yns 3 ds7 n 2 0 0 We now compute the first elements in the sequence t t n07 y1t12y0s3ds15ds15t 0 0 Soyozl and y115t First order Nonlinear Equations 16 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y 2y3 y01 t Solution Integral equation yt 1 2 ys 3 ds 0 And yo 1 and y1 1 516 Let s compute y2 y21At2y1s3ds1At2153ds t y21 510ds15t5t2 0 So we39ve got y2t15t5t2 First order Nonlinear Equations 16 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y 2y3 y01 1 Solution Integral equation yt 1 2 ys 3 ds 0 And yo 1 and y1 1 51 and y2 15t5t2 Now y3 t t y312y2s3ds1 215523ds 0 0 t 10 y31 510102ds15t5t2t3 0 2 10 3 Sowevegoty3t15t5t t Rewrite y3t 1 g 215 First order Nonlinear Equations 16 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y2y37 y01 Solution y3t 2 1 gums 233 By computing few more terms one finds n k Ynt1g Hence the limit n gt 00 is given by 5 00 2tk 5 2t Vltlznlmwyquott1 k 21 e 1 00 k X X 5 2t 3 Since 6 370 We conclude yt Ee lt1 Review for Exam 2 gt Exam Covers V Variation of Parameters 27 Undetermined Coefficients 26 Applications 25 Homogeneous Constant Coefficients 23 24 Special Second Order Reduction Order Method 22 Second Order Variable Coefficients 21 First order Nonlinear Equations 16 Applications 15 VVVVVVV Review for Exam 3 gt V V V V 6 Problems 55 Minutes in Recitation Rooms 100 Grading Attempts Problems Similar to Homeworks Integration Laplace Transform Tables in Handout No Notes No Books No Calculators gt lVlLC lVlTl l 235 Exam Review httpmathmsuedumlc V Exam Covers V Power Series Solutions 31 Euler Differential Equation 32 Definition of Laplace Transform 41 Solving lVP Using LT 42 Discontinuous Sources 43 Generalized Sources 44 Convolutions and LT 45 VVVVVV Review for Exam 3 V V V V V V V Power Series Solutions 31 Euler Differential Equation 32 Definition of Laplace Transform 41 Solving lVP Using LT 42 Discontinuous Sources 43 Generalized Sources 44 Convolutions and LT 45 Power Series Solutions 31 Example Using a power series centered at X0 0 find the three first terms of the general solution of 4 X2y 2y 2 0 0 Solution We look for solutions y Z 3 X Therefore n0 y Z nn 1anX 2 n0 The differential equation is then given by 4 X2 Znn 1anx 2 2 Zanx 07 70 70 i4nn 1anx 2 i0 nn 1aX 23 X 0 quot0 n0 n0 Power Series Solutions 31 Example Using a power series centered at X0 0 find the three first terms of the general solution of 4 X2y 2y 2 0 Soglution 00 00 Z4nn 1anX 2 Z nn 1aX 223 X 0 n2 n0 n0 Re label the first sum m n 2 and then switch back to n 0 Z4n 2n1an2X Z nn 1aX Z2anx 0 n0 n0 n0 4n 2n 1a2 nn 1a7 23 X 0 M8 3 l l o 4n 2n 1a2 n2 n 2a O Power Series Solutions 31 Example Using a power series centered at X0 0 find the three first terms of the general solution of 4 X2y 2y 2 0 Solution 4n 2n 1a2 n2 n 2a 0 Notice n2 n 2 n 2n l 1 hence n 2 a 4n2n1a2 n 2n l 1a7 0 gt an2 For n even the power series terminates at n 2 since 2 322 307 3407 3607 8 a3 31 F dd 1 or n 0 a3 12 7 35 20 12x20 1 1 1 1 2 3 5 lt1 y aol 4X 31 X 12X 1220X Review for Exam 3 V Power Series Solutions 31 Euler Differential Equation 32 Definition of Laplace Transform 41 Solving lVP Using LT 42 Discontinuous Sources 43 V V V V V Generalized Sources 44 Convolutions and LT 45 V Euler Differential Equation 32 Summary 39 X X02y l X X0P0y l Joy 2 0 gt Find ri solutions of rr 1 por q0 0 gt If r 7E r and both are real then fundamental solutions are y IX Xol y IX Xo lt gt If ri 2 ad i6 then real valued fundamental solutions are y X X0C cos ln X X07 y X X0C sin ln lX Xol gt If r r and both are real then fundamental solutions are y Ix xOrt y IX Xol In IX Xol Euler Differential Equation 32 Example Find real valued fundamental solutions of X 22y5X 2y8y0 Solution This is an Euler equation Find r solution of rr 15r8 0 that is r24r8 0 1 ri E 4ix16 32 gt ri 2i2i Real valued fundamental solutions are yX X 2 2 cos2 ln X 27 yX X 2 2 sin2lnX 2 lt1 Review for Exam 3 V Power Series Solutions 31 Euler Differential Equation 32 Definition of Laplace Transform 41 Solving lVP Using LT 42 Discontinuous Sources 43 V V V V V Generalized Sources 44 Convolutions and LT 45 V Laplace Transforms Review Summary gt Main Properties fltquotgtt s ft sltquot1gtfo WW0 18 e CS ft uct ft C 13 ft H tie no 14 gt Convolutions f gtl ftl gtl gt Partial fraction decompositions completing the squares Definition of Laplace Transform 41 Example Use the definition of the LT to find the LT of ft cosht Solution Recall that cosht 2 et e t2 and that 75 6t 671 cosht 20 e fdt N cosht Nim O 6571teis1tdt gtltgto I 1 eisilt eis1t N LlCOShltllZAliT wEh 5 1 51l 039 1 1 1 ls1s 1 lCOShltll ls 1 51l 2 s2 1 S We conclude cosht lt1 Definition of Laplace Transform 41 Example F39dth39 L T f fF i In e Inverse 3p ace rans orm O S 5 32 Solution We start rewriting function F is s iss33l Flsl e 2 m 6 2 m Fs2672s 5 3 7251 e 5 Fs 6 25 cos5ts 3 g 6 25 sin5ts 3 Recall 14 fts c eCt ft Fs 6 25 e3t cos5t g 6 25 e3t sin5t Definition of Laplace Transform 41 Example 6455 Find the inverse Laplace Transform of Fs Solution 3 Recall Fs e725 e3t cos5t E e725 e3t sin5t Recall 13 e cs ft ut c ft Fs ut 2 e304 cos5t 2 3 ut 2 e304 sin5t 2 ft ut 2 e304 cos5t 2 2 sin5t 2 lt1 Review for Exam 3 V Power Series Solutions 31 Euler Differential Equation 32 Definition of Laplace Transform 41 Solving lVP Using LT 42 Discontinuous Sources 43 V V V V V Generalized Sources 44 Convolutions and LT 45 V Discontinuous Sources 43 Example 0 tge 1 Find the LT of ft 5 3 t 2 6 Solution We need to rewrite the function f in terms of functions that apear in the LT table We need a box function for the first part and a step function for the second part NIH ft ut ut 6 3 ut 6 ms 5 W g 3 ut 6 gum t6 t 6 Discontinuous Sources 43 Example ogtge Find the LT of ft WMIH 1326 Solution Recall ft t ut t 6 ut mm ammo wt 6 um o1 ft gm an two We conclude that ft 0 e765 Discontinuous Sources 42 43 Example Use LT to find y solution of 07 t lt 27 3 1 02020 1 y y g y y g 602 122 Solution Express g using step functions gt 2 L120 4 uct ft c e CS ft Therefore gte 254et gt Discontinuous Sources 42 43 Example Use LT to find y solution of y 3y gm y0 No 0 gm 07 t lt 27 6tT27 t 2 2 6725 Solution Recall gt 725 llll 3 lYl lgtl s23 y1 1 a bsc Hss 152 3 5 1 s2 3 5 1 t WHZW39 1 352 3bscs 1 Discontinuous Sources 42 43 Example Use LT to find y solution of 0 t lt 27 y 3ygt7 y0y007 gte t27 92 Solution Recall 1 352 3 bs cs 1 1asz3ab52cs bs c 1ab52c bs3a c ab07 c b07 3a c1 3b b1 gt b a1 7 47 Hltsgtil Zl Discontinuous Sources 42 43 Example Use LT to find y solution of 0 t lt 27 7 l 3 t7 0 007 Iquot y y g y y 539 413722 1 1 51 725 4ls 1523l39 lyl e Hlsl Solution Recall Hs 1 1 s 1 Hlsl1l sz3 ml7 Hs et cos t sinSt Hs Eltet cosSt sin Discontinuous Sources 42 43 Example Use LT to find y solution of d tlta 3 t7 020207 1 y y g y y g 602 92 Solution Recall Hs 2 LE er cos3t sin3t i sin 1 Hs ht x3 y e725 Hs e725 ht u2t ht ht 2 261quot cos3 t We conclude yt U2t ht 2 Equivalently u t t 1 yt 2T a 2 cos t 2 E Sln t 2L Review for Exam 3 V Power Series Solutions 31 Euler Differential Equation 32 Definition of Laplace Transform 41 Solving lVP Using LT 42 Discontinuous Sources 43 V V V V V Generalized Sources 44 Convolutions and LT 45 V Generalized Sources 44 Example Use Laplace Transform to find y solution of y 21 2y 6t 2 y017 yO 3 Solution Compute the LT of the equation llquotl MM 2 y 6t 2 6 25 W 2 y sy0 yOL bl s ly y0 52 2S 2 ly sy0 yO 2y0 6 25 52 2s2 y s 1e 25 51 1 725 lyl 52 2s2 52 2s2e Generalized Sources 44 Example Use Laplace Transform to find y solution of y 21 2y 6t 2 y017 yO 3 51 1 6725 s2 2s2 s2 2s2 Solution Recall y 1 52 2s20 gt si 2l4 87 complexroots 52 2s252 2s1 12s 121 51 1 2 2 S y s 121s 121e 1 1 1 1 5M s e725 s 121 s 121 Generalized Sources 44 Example Use Laplace Transform to find y solution of y 2y 2y 6t 2 y017 yO 3 s 1 2 1 725 5 121 6 Solution Recall y s 12 1 5 1 1 72s 1 2 s l21e s l217 lyls 12l 1 cosat sinat s a 2a27 2a27 y costlsil 2 sintlsil 6 25 sintlsil Generalized Sources 44 Example Use Laplace Transform to find y solution of y 2y 2y 2 6t 27 y017 yO 3 Solution Recall y costsil 2 sintsil 6 25 sintsil and ftsic ed ft Therefore y et cost 2 et sint 6 25 et sint Also recall e CS ft uct ft Therefore y et cost 2 et sint U2t 6tT2 sint Yt cost 2sint et L12 sin 2 6072 lt1 Generalized Sources 44 Example Sect 65 Probl7 Find the solution to the initial value problem y y 61 7TCosl 7 y0 07 yO 0 Solution Compute the Laplace Transform of the equation llquotl ly 6t 7T COSt To compute the right hand side above we need the definition of the LT Given any smooth function f and a constant c holds tC 6t cft 000 e ft6t c dt e stft ce We have used that 6t c gt dt gc C76 We obtain the formula 6t cft fc e cs Generalized Sources 44 Example Sect 65 Probl7 Find the solution to the initial value problem My 6a weasel yo 0 yo 0 Solution Recall 6t cft fce cs Hence 2 y y 6t 7r cost cos7re 7rs e 7rs 1 Z iws Z y e 21 e 7rs L sint Recall the property 13 e cs ft ut c ft c y ut 7r sint 7r gt yt ut 7r sint 7r ltl Review for Exam 3 V Power Series Solutions 31 Euler Differential Equation 32 Definition of Laplace Transform 41 Solving lVP Using LT 42 Discontinuous Sources 43 V V V V V Generalized Sources 44 Convolutions and LT 45 V Convolutions and LT 45 Example 725 s 1s2 339 Solution One way to solve this is with the splitting Use convolutions to find 1 satisfying ft 5 1 1 S1 1 Llfltllze 2 523s 1e 2 523s 17 ft e45 sin3 15 et 1 mm quza sin3t 2 et W 0t U2739 sin3739 2 60 dT Review for Final Exam V V V V V V V 15 problems 120 minutes in Computer Labs 200 Grading Attempts Problems Similar to Homeworks Integration and LT Tables Provided No Notes No Books No Calculators MLC MTH 235 Review Sunday httpmathmsuedumlc Exam Covers First Order differential Equations Chptr 1 Second Order Linear Equations Chptr 2 Power Series Solutions Chptr 3 Laplace Transforms Chptr 4 Systems of Linear Equations Chptr 5 Eigenvalue Eigenfunction BVP Chptr 7 Heat Eq and Fourier Series Chptr V VVVVVV Review for Final Exam V V V V V V V 15 problems 120 minutes in Computer Labs 200 Grading Attempts Problems Similar to Homeworks Integration and LT Tables Provided No Notes No Books No Calculators MLC MTH 235 Review Sunday httpmathmsuedumlc Exam Covers First Order differential Equations Chptr 1 Second Order Linear Equations Chptr 2 Power Series Solutions Chptr 3 Laplace Transforms Chptr 4 Systems of Linear Equations Chptr 5 Eigenvalue Eigenfunction BVP Chptr 7 Heat Eq and Fourier Series Chptr V VVVVVV First Order Differential Equations Summary gt Linear first order equations y pty qt Use the integrating factor method Mt ef pm gt Separable non linear equations hy y 2 gt Integrate with the substitution u yt du yt dt that is hu dugtdtc The solution can be found in implicit of explicit form gt Homogeneous equations can be converted into separable equations First Order Differential Equations Summary gt Bernoulli equations y pty qt y with n E R A Bernoulli equation for y can be converted into a linear equation for v ynil gt Exact equations and integrating factors Nx7 y y MX7 y 0 The equation is exact iff 8XN 8yM If the equation is exact then there is a potential function w such that N 8y and M 8gb The solution of the differential equation is X7yX C First Order Differential Equations Advice In order to find out what type of equation is the one you have to solve check from simple types to the more difficult types 1 Linear equations Just by looking at it y aty bt Bernoulli equations Just by looking at it y aty bt y Separable equations Few manipulations hy y 2 gt l JO 4 Homogeneous equations Several manipulations y Fyt 5 Exact equations Check one equation Ny M 0 and 8tN 8yM 6 Exact equation with integrating factor Very complicated to check First Order Differential Equations Example x2 xy y2 xy 39 Solution The sum of the powers in X and y on every term is the same number two in this example The equation is homogeneous Find all solutions of y x2xyy2 1x2 1 2 Y 2 gt y y 39 Xy 1X y 1 2 VXZ yZ x v 1vv2 yxv7 yxvv xvv v 1vv2 1 l vv2 v2 1v xv v xv V V V First Order Differential Equations Example 2 2 Find all solutions of y Xy 1v Solution Recall v This is a separable equation v vX 1 vX dX d 1lVXVX X 1VXVX X XC Use the substitution u 2 1 v hence du VX dX L1duc 1 duc u lnulnX l c gt 1v ln1vlnXc VZZ gt 1m lnl1 llnlxlc lt1 X X X First Order Differential Equations Example Find the solution y to the initial value problem 1 y y 62Xy3 07 V0 Solution This is a Bernoulli equation y y 62Xy n 3 3 y 1 2x DiVide by y That is 3 2 e y y 1 y 1 2X Let v 2 Since v 2 3 we obtain v v e y y 2 We obtain the linear equation vl 2v 2 262X Use the integrating factor method uX e ZX 7 7 7 e 2Xvl 2e 2Xv2 gt 6 2Xv 2 First Order Differential Equations Example Find the solution y to the initial value problem 1 yy 62Xy3 07 V0 1 Solution Recall V 2 and e 2X V 2 y 1 e zxv22xc gt VX2XC62X gt 22xcezx y 1 67X 2 gt X E y e2x 2X C yi x2x c The initial condition y0 13 gt 0 implies Choose y 1 5 Y0 7X 6 gt 2 le N2x9 lt1 gt c9 i E First Order Differential Equations Example Find all solutions of 2xy2 2y 2X2yy 2xy 0 Solution Re write the equation is a more organized way 2x2y 2xy 2xy2 2y 2 0 N 2X2y 2X gt 8XN 4xy 2 2 gt 8XN ByM M 2xy 2y gt 8yM 4xy2 The equation is exact There exists a potential function 1b with 8y N7 8gb M aw 2X2y 2X gt We y X2y2 2Xy ggtlt 2xy2 2ygx 8gb M 2xy2 2y gt gX 0 1W7 y X2y2 2Xy c7 x2 y2X 2Xyx c 0 lt1 First Order Differential Equations Example Assume that r r0 r and q are constants If r 2 litersmin q 0 V0 2 200 liters 20 V0 1gramsliter find t1 such that qt1 Qt1Vt1 is 1 the initial value Solution This problem is a particular case q 0 of the previous Example Since Qt Q0 quo ei39TV0 quo we get Q Q0 GertVo Since Vt r r0 t V0 and r r0 we obtain Vt 2 V0 So qt Qt Vt is given by qt e rtVo Therefore 0 1 1 Q0 Q0 rtlVO ltlVO 1 70 e gt e 10070 First Order Differential Equations Example Assume that r r0 r and q are constants If r 2 litersmin q 0 V0 2 200 liters 20 V0 1gramsliter find t1 such that qt1 Qt1Vt1 is 1 the initial value 1 39 irtlVo Solution Recall e 100 Then 1 r In100 gt t1 In100 V0 V We conclude that t1 To In100 In this case t1 100 In100 lt1 Review for Final Exam V First Order Differential Equations Chptr 1 Second Order Linear Equations Chptr 2 V V Power Series Solutions Chptr 3 V Laplace Transforms Chptr 4 V Systems of Linear Equations Chptr 5 Eigenvalue Eigenfunction BVP Chptr 7 Heat Eq and Fourier Series Chptr V V Second Order Linear Equations Summary Solve y 31 y aoy 2 gt First find fundamental solutions yt e to the case g 0 where r is a root of pr r2 alr 20 a If r1 7E r2 real then the general solution is yt C1 erlt C2 6 b If r1 7E r2 complex then denoting ri a l i complex valued fundamental solutions are mt Ari301 lt3 mt cosww i isinwmi and real valued fundamental solutions are y1t 2 eat cos t7 y2t 2 eat sin t c If r1 r2 r real then the general solution is yt C1 C2t e Second Order Linear Equations Remark Case c is solved using the reduction of order method Summary Non homogeneous equations g 7A 0 i Undetermined coefficients Guess the particular solution yp using the guessing table g gt yp ii Variation of parameters If yl and y2 are fundamental solutions to the homogeneous equation and W is their Wronskian then yp ulyl uzyz where E W u1 u Second Order Linear Equations Guessing Solution Table f Source K m a b given yp Guess k not given Keat keat KmtmKo kmtmko K1 cosbt K2 sinbt k1 cosbt k2 sinbt KmtmK0 eat kmt vvk0 eat K1 cosbt K2 sinbt eat k1L cosbt k2 sinbt eat Kmt R1 cosbt R2 sinbt kmtm k0 lg cosbt IQ sinbt Second Order Linear Equations Example Knowing that y1X X2 solves X2 y 4xy 6y 2 0 with X gt 0 find a second solution y2 not proportional to yl Solution Use the reduction of order method We verify that y1 X2 solves the equation x2 2 4x 2x 6X2 0 Look for a solution y2X vX y1X and find an equation for v y2 X2v7 y X2V 2XV7 y2 X2v 4XV 2v X2X2V 4XV 2v 4X XZV 2XV 6X2V 0 X4v 4X3 4X3 v 2X2 8X2 6X2 v 0 v 0 gt v C1lC2X gt y2 c1y1czxy1 Choose c1 0 C2 1 Hence y2X 2 X3 and y1X 2 X2 lt1 Second Order Linear Equations Example Find the solution y to the initial value problem y 2y 3y 3 e47 Y017 yO Solution 1 Solve the homogeneous equation yt e 7 pr r2 2r 3 0 3 ri2l 412 22ix16lzli2 gt fl 71 r 31 it and y2t e 2 Guess yp Since gt 3 6 gt ypt ke t Fundamental solutions y1t 6 But this yp k 6 is solution of the homogeneous equation Then propose ypt kt e t Second Order Linear Equations Example Find the solution y to the initial value problem y0 17 t 7 y 21 3y 3 6 yO Solution Recall ypt kt 671 This is correct since te t is not solution of the homogeneous equation 3 Find the undetermined coefficient k y ke t kt e 7 y 2k 6 kt e t 2k 6 kt e t 2k e t kt e t 3kt 677 3 6 2t 22t 3tket3et gt 4k3 gt k We obtain ypt te t Second Order Linear Equations Example Find the solution y to the initial value problem y 21 3y 3 6 y017 yO 3 4 Solution Recall ypt te 4 t 3 1quot te 4 Find the general solution yt c1 63 cz e 5 Impose the initial conditions The derivative function is 3 yt 3C1 e3t C2 6 Ze t te t 3 3C1 C2 Hi4 1 yO 1y0c1C27 Z gt 1 1 3 1 l 1 1 C1lC217 31 C21 C1 C2 Second Order Linear Equations Example Find the solution y to the initial value problem 1 1 y 21 3y 36 t7 Y017 yO Z 3 Solution Recall yt 2 c1 63 C2 6 Zte t and 1 1 c11gt c1i 1 1 112 3 1 cz1 C2 4 3 1 1 4239 1 1 Since c1 E and C2 we obtain 1 3 yt e3te t Zte t lt1 lIl Second Order Linear Equations Example Find the movement of a 5Kg mass attached to a spring with constant k 5KgSecs2 moving in a medium with damping constant d 5KgSecs with initial conditions y0 2 and yO 0 Solution The equation is my dy ky 0 with m 5 k 5 d 5 The characteristic roots are T 2 d 1 k riZ wdil dd 407 wd 7 do ri llll 1 l l i Under damped oscillations 2 4 2 2 Wt A cos t 15 Git2 Second Order Linear Equations Example Find the movement of a 5Kg mass attached to a spring with constant k 5KgSecs2 moving in a medium with damping constant d 5KgSecs with initial conditions y0 xS and yO 0 Solution Recall yt Acoslt t lt15 e tZ Hence 1 t2 t2 yt 2 Asm 2 t q5gte 2Acos 2 1 Q5 6 The initial conditions 1 xS y0 Acosqb7 0 y 0 7 Asmq5 E Acosq5 tanq5 gt 152 gt A22 7 42 We conclude yt 2cos7 t 6 lt1 Review for Final Exam V First Order Differential Equations Chptr 1 Second Order Linear Equations Chptr 2 Power Series Solutions Chptr 3 V V V Laplace Transforms Chptr 4 V Systems of Linear Equations Chptr 5 Eigenvalue Eigenfunction BVP Chptr 7 Heat Eq and Fourier Series Chptr V V Power Series Solutions Example Using a power series centered at X0 0 find the three first terms of the general solution of 4 X2y 2y 2 0 0 Solution We look for solutions y Z 3 X Therefore n0 y Z nn 1anX 2 n0 The differential equation is then given by 4 X2 Znn 1anx 2 2 Zanx 07 70 70 i4nn 1anx 2 i0 nn 1aX 23 X 0 quot0 n0 n0 Power Series Solutions Example Using a power series centered at X0 0 find the three first terms of the general solution of 4 X2y 2y 2 0 Soglution 00 00 Z4nn 1anX 2 Z nn 1aX 223 X 0 n2 n0 n0 Re label the first sum m n 2 and then switch back to n 0 Z4n 2n1an2X Z nn 1aX Z2anx 0 n0 n0 n0 4n 2n 1a2 nn 1a7 23 X 0 M8 3 l l o 4n 2n 1a2 n2 n 2a O Power Series Solutions Example Using a power series centered at X0 0 find the three first terms of the general solution of 4 X2y 2y 2 0 Solution 4n 2n 1a2 n2 n 2a7 0 Notice n2 n 2 n 2n l 1 hence n 2 a 4n2n1a2 n 2n l 1a7 0 gt an2 For n even the power series terminates at n 2 since 2 322 307 3407 3607 8 a3 31 F dd 1 or n 0 a3 12 7 35 20 12x20 1 1 1 1 2 3 5 lt1 y aol 4X 31 X 12X 1220X Power Series Solutions Summary 39 X X02y l X X0P0y Joy 2 0 gt Find ri solutions of rr 1 por q0 0 gt If r1 7E r and both are real then fundamental solutions are y IX Xol y IX Xo lt gt If ri al i then real valued fundamental solutions are y X X0C cos ln X X07 y X X0C sin ln lX Xol gt If r r and both are real then fundamental solutions are y X X0r7 y X X0r ln X X0 Power Series Solutions Example Find real valued fundamental solutions of X 22y5X 2y8y0 Solution This is an Euler equation Find r solution of rr 15r8 0 that is r24r8 0 1 ri E 4ix16 32 gt ri 2i2i Real valued fundamental solutions are yX X 2 2 cos2 ln X 27 yX X 2 2 sin2 ln X Review for Final Exam V First Order Differential Equations Chptr 1 Second Order Linear Equations Chptr 2 V V Power Series Solutions Chptr 3 V Laplace Transforms Chptr 4 V Systems of Linear Equations Chptr 5 Eigenvalue Eigenfunction BVP Chptr 7 Heat Eq and Fourier Series Chptr V V Laplace Transforms Summary gt Main Properties f t s ft AH rm f 10 18 e CS ft uct ft c 13 ft Sic eCt ft 14 gt Convolutions f gtl lftl gtl gt Partial fraction decompositions completing the squares Laplace Transforms Example Use LT to find the solution to the NP y 9y U5t7 y0 37 yO 2 755 Solution Compute y 9 y U5t 2 es and recall Lv 2 Lv sy0 yO cw 52 y 3s 2 6755 529 y 3s 2 35 2 6755 1 ly s2 9 552 9 s 2 3 5 1 WWW 52965m Laplace Transforms Example Use LT to find the solution to the IVP y 9y U5t7 y0 37 yO 2 s 2 3 55 1 529 529e ss2939 Solution Recall y 3 y 3 cos3t Sin3tl l 6755 Partial fractions on Hsi bSlC 3529bsc5 29 s 529 5529 7 1a5293b52cs3b52csga 1 1 32 7 C207 DZ 3 Laplace Transforms Example Use LT to find the solution to the IVP y 9y U5t7 y0 37 yO 2 Solution So y 3 cos3t sin3t 6 55 Hs and 45 2 aqua cos3tl 6 55 Hs 3e SS ut 6 55 cos3t e755 Hs 2 amino U5t cos3t 5 y 3 cos3t sin3t U5ti U5t cos3t75gt Laplace Transforms Example Use LT to find the solution to the IVP y 9y U5t7 y0 37 yO 2 Solution y 3 cos3t sin3t 3 U5t i u t cos3t75 Therefore we conclude that yt 3 cos3t g sin3t us t 1 cos3t Review for Final Exam V First Order Differential Equations Chptr 1 Second Order Linear Equations Chptr 2 V V Power Series Solutions Chptr 3 V Laplace Transforms Chptr 4 V Systems of Linear Equations Chptr 5 Eigenvalue Eigenfunction BVP Chptr 7 Heat Eq and Fourier Series Chptr V V Systems of Linear Equations Summary Find solutions of x Ax with A a 2 x 2 matrix First find the eigenvalues A and the eigenvectors v of A a If A1 7E A2 real then v1v2 are linearly independent and the general solution is xX C1 V1e1t C2 v2 e Vt b If A1 7E A2 complex then denoting Ai a l i and Vi a l bi the complex valued fundamental solutions xi a l bi ewj m xi 2 eat a l bi cos t isin t xi 2 eat a cos t bsin t tieat a sin tbcos t Real valued fundamental solutions are x1 2 eat a cos t bsin t7 x2 2 eat a sin t bcos t Systems of Linear Equations Summary Find solutions of x Ax with A a 2 x 2 matrix First find the eigenvalues A and the eigenvectors v of A c If A1 A2 A real and their eigenvectors v1v2 are linearly independent then the general solution is xX C1 V1 ext C2 v2 ext d If A1 A2 A real and there is only one eigendirection v then find w solution of A Alw v Then fundamental solutions to the differential equation are given by x1 v e w7 x2 v t l w ext Then the general solution is xC1ve tC2vtwe t Systems of Linear Equations Example 3 1 4 Find the solution to x Ax7 x0 2 7 A 2 1 Solution p 1 14Al 11 82 1 87 p2 90 gt A1513 Case 3 2 4 1 2 2 A 3I 2 4 gt0 0 gtv1 2V2gtv 1 Case 3 4 4 1 1 7 1 A3l 2 2 gt 0 0 gt v1 V2 gt v 1 Systems of Linear Equations Example 3 1 4 Find the solution to x Ax7 x0 2 7 A 2 1 Solution Recall i l3 V vli 1 The general solution is xt c1 631 C2 1 6737 The initial condition implies l3lxlt0gtch lilwl fl i ll Zill3l ll22l1l11lll liil lil We conclude xt 2 2 6315 1 673139s lt1 Review for Final Exam V First Order Differential Equations Chptr 1 Second Order Linear Equations Chptr 2 V V Power Series Solutions Chptr 3 V Laplace Transforms Chptr 4 V Systems of Linear Equations Chptr 5 EigenvalueEigenfunction BVP Chptr 7 Heat Eq and Fourier Series Chptr V V Eigenvalue Eigenfunction BVP Example Find the positive eigenvalues and their eigenfunctions of yo 07 y8 0 Solution Since gt 0 introduce 2 with u gt 0 y Ay07 yX erX implies that r is solution of prr2u20 gt riziui The general solution is yX C1 cosuX C2 sinuX The boundary conditions imply 0 y0 C1 0 y8 C2 sinu87 gt yX C2 sinuX C2 7A 0 gt sinu8 0 2 uzn g7 E7 yXsin 7 n1727 8 8 lt1 Eigenvalue Eigenfunction BVP Example Find the positive eigenvalues and their eigenfunctions of y M 07 y0 07 yB 0 Solution The general solution is yX C1 cosuX C2 sinuX The boundary conditions imply 0 Y0 C1 gt yx C2 sinix 0 yB C2M cosu87 C2 7A 0 gt cosu8 0 7r 2n17r Then for n 12 holds 2n17r 2 A l 16 l 7 yX sinW Eigenvalue Eigenfunction BVP Example Find the non negative eigenvalues and their eigenfunctions of y M 07 yO 07 yB 0 Solution Case gt 0 Then yX C1 cosuX C2 sinux Then yX C1u sinux C2u cosux The BC imply 0 yO C2 gt yX C1 cosuX7 yX C1u sinux 0 y8 2 CW sin187 C1 7A 0 gt WT 81 2 mr gt u 8 Then choosing C1 1 for n 12 holds sinu8 0 8 quota ynltxgtcosltmgt Eigenvalue Eigenfunction BVP Example Find the non negative eigenvalues and their eigenfunctions of y M 07 MW 07 yB 0 Solution The case 0 The general solution is yX C1 C2X The BC imply 0 yO c2 gt yX C17 yX 0 Then choosing C1 1 holds 07 YOX 1 lt1 Eigenvalue Eigenfunction BVP Example Find the solution of the BVP y y 07 y017 Y7T3 0 Solution yX e implies that r is solution of prr2u20 gt rizii The general solution is yX C1 cosX C2 sinX Then yX C1 sinX l C2 cosX The BC imply 1 yO C2 gt yX C1 cosX sinX sin7r3 0 y7r3 C1 cos7r3 sm7r3 gt C1 W C1 2 gt yX 2 cosX sinX lt1 Review for Final Exam V First Order Differential Equations Chptr 1 Second Order Linear Equations Chptr 2 V V Power Series Solutions Chptr 3 V Laplace Transforms Chptr 4 V Systems of Linear Equations Chptr 5 Eigenvalue Eigenfunction BVP Chptr 7 Heat Eq and Fourier Series Chptr V V Fourier Series Example Find the Fourier series of the odd periodic extension of the function fx 1 for X 6 10 Solution The Fourier series is fX 2 quot213 cosnLLX bn sinnLLX Since 1 is odd and periodic then the Fourier Series is a Sine Series that is an 2 0 bn fxsinnLLX dx OL fxsinnLLX dx 1 b 20 1sinn7rxdx 2 bn 3 chosn7r 1 gt bn 1 1 77139 Fourier Series Example Find the Fourier series of the odd periodic extension of the function fX 1 for X E 170 2 Slt39 R b 1 1 oqun eca n mr 2 If 2k h 12k 1 0 n tenb2lt 2k7r fn2k 1 2 4 2k71 b2k 12k 17r 1 1 2k 17r39 w df 4OO 1 392k1 econcu e X sm uX lt1 k1 Fourier Series Example Find the Fourier series of the odd periodic extension of the function fX 2 X for X E 07 2 Solution The Fourier series is fX quot213 cosnLLX bn sinnLLX Since f is odd and periodic then the Fourier Series is a Sine Series that is 3 0 1 L mrX 2 L mrX bnzzil fxsmT dX 20 fX539nT dxi L27 bn 2 022 X sin dXa Fourier Series Example Find the Fourier series of the odd periodic extension of the function fX 2 X for X E 07 2 Solution bn 202 sin dX 02X sin dX sin dX 2 73 cos7 The other integral is done by parts u 2 X7 v sin 2 mrX I XsmT dX7 2 mm u 17 v cos mr 2 cosltquot Xgt lt gt a4 Fourier Series Example Find the Fourier series of the odd periodic extension of the function fX 2 X for X E 07 2 Solution I 2X COSlt gt 2 COSlt gt dX mr 2 n 2 2X mrX 2 2 mr I cos sm So we get mr 2 mt 2 gtlt a Fourier Series Example Find the Fourier series of the even periodic extension of the function fX 2 X for X E 07 2 Solution The Fourier series is fX 2 quot213 cosnLLX bn sinnLLX Since 1 is even and periodic then the Fourier Series is a Cosine Series that is bn O 2 2 aozl fXdX 2 XdXW gt 202 2 2 0 2 1 L mrX 2 L mrX an ZXL fX cosT dX ZO 1 XcosTdX7 L 27 2 2 022 X cos dX Fourier Series Example Find the Fourier series of the even periodic extension of the function fX 2 X for X E 07 2 2 2 Solution 3 2 cos dX X cos dX 0 2 0 2 cosmTX dX 2 sinmTX 2 mt 2 7 The other integral is done by parts nva u X7 v cos2 l nva d xcosT X7 2 I mm u 17 v sm mt 2 Fourier Series Example Find the Fourier series of the even periodic extension of the function fX 2 X for X E 07 2 Solution Recall I 2 X sin 3 sin dX MT 2 mr 2 2X mrX 2 2 mrX I sm cos So we get n 2 mr 2 Fourier Series Example Find the Fourier series of the even periodic extension of the function fX 2 X for X E 07 2 4 Solution Recall bn 0 30 2 an 2 W1 1 4 2k lfn2k then 32kW1 1 0 If n 2k 1 then we obtain 4 2k71 8 32k 1 2k 127T2 l11 l 2k 127T2 8 00 1 2k 17rx We conclude fX 2 1 kg W cos lt1 The Heat Equation Example Find the solution to the IBVP 8tu 2 283m 1 gt 07 X 6 01 2 x e 012 u0X 07 X 6 1271 8Xut0 07 ut1 0 Solution Let 1t X vt WX Then V 2W V W X The equations for v and W are vt 2A vt 07 WX A WX 0 We solve for v using the integrating factor method eZAt Vt 2A eZAt Va 2 0 62M Vt 0 The Heat Equation Example Find the solution to the IBVP 8tu 282u7 t gt 07 X 6 01 2 x e 012 u0X 07 X 6 1271 8Xut0 07 ut1 0 Solution Recall 62M vt 0 Therefore Vt v0 6 2 Next the BVP W X A WX 0 with wO 07 Wl O Nonzero solutions for A gt 0 so introduce A 2 Then pr r2 u2 0 gt ri Eui The general solution WX c1 cosuX C2 sinuX The boundary conditions imply 0 wO ucz gt WX c1 cosuX The Heat Equation Example Find the solution to the IBVP 8tu 282u t gt 0 X 6 01 2 x e 012 u0X 07 X 6 1271 8Xut0 0 ut 1 0 Solution Recall Vt V0 e ZM and WX c1 cosuX 0 W1 c1 cosu c1 7A 0 gt cosu 0 Then u 2n Choosing c1 1 we conclude An 2n1w2 WnX cos 2n 217TX n 1 2 00 2n711r2t 2n 17TX utX Z vn0 e 2 cos 2 n1 The Heat Equation Example Find the solution to the IBVP 8tu 2 283m 1 gt 0 X 6 01 2 x e 012 u0X 07 X 6 1271 8Xut0 0 ut 1 0 Solution 00 1 M 2n 1 X Recall utX quot271 Vn0e 2 Qt COS 2 7T 00 2n 1 7TX The Initial condition is 1 00 2 E Vn0 COS 2 Where n1 X Z 2 x e 012 f 0 XE 121 The Heat Equation Exampm Find the solution to the IBVP 8tu 2 283m 27 X E 07127 U07X 07 X 6 1271 8Xut70 07 ut71 0 tgtQ XEWJL Recall fX 0 v0 cosm 2 n1 Multiply by cos and integrate on 071 1 m 7TX 0fxcosgt dX ivnm 01COSM CO 2m 17TX 2 S dX n1 The Heat Equation Example Find the solution to the IBVP 8tu 28 u 27 x e 07127 U07X 07 X 6 1271 8Xut70 07 ut 1 0 tgtQ XEWJL The orthogonality of the cosine functions implies L n 7rx m 7TX 0cos2 21 cos2 1 dX 2 MI 0 Since L 1 we get 1 2m 17TX 0 fX cos 22H f 21 it 20 equot 6963 S 2 z gto grad MM aer 2 40 arm M law 30 Zlto Igt K5 to sgt lrh l B HEB 3 law o 2 gto arer v H lam Prune 12m1 atm 2 T fmwud e 25 I 2 2 yy way 2 wy wy eo y39Li I MPH1 I ywb 3er 02 e gt 12ml r equot L Review for Exam 1 V On Webwork Exam Server in Recitation Rooms and Times V 100 grading attempts V 6 problems 55 minutes V Problems similar to webwork homework problems V Integration table provided in the handout V No notes no books no calculators no phones MLC Exam Review for MTH 235 today 730pm B 117 WH MTH 235 Exam 1 covers Linear equations 11 12 Bernoulli equation 12 Separable equations 13 Euler homogeneous equations 13 Exact equations 14 Exact equations with integrating factors 14 Applications 15 Not Covered Picard Lindelof iteration 16 Not Covered V V V VVVVVVV Exam Overview Remark gt Exam problems will be Solve this equation We don t tell you if the equation is linear separable etc You must find that out gt If you know what type of equation is then the equation is simple to solve gt The difficult part in Exam 1 is to know what type of equation is the one you have to solve Exam Overview Advice In order to find out what type of equation is the one you have to solve check from simple types to the more difficult types 1 Linear equations Just by looking at it y aty bt Bernoulli equations Just by looking at it y aty bt y Separable equations Few manipulations hy y 2 gt 3 JG 4gt Euler homogeneous equations Several manipulations y Fyt 5 Exact equations Check one equation Ny M 0 and 8tN 8yM 6 Exact equation with integrating factor Could be very complicated to check Review Exam 1 Example Find every solution y to the equation I2 y2t yy 2 0 Solution Rewrite the equation in a more standard way t2 y2t2 2 2 2 2 t t t20 ltgt yyy y y 12 y Not linear Not Bernoulli Not Separable Not Euler homogeneous So the equation must be exact or exact with integrating factor Nt2yy3 gt 8tN2ty Mt3ty22 8yM2ty The equation is exact 8tN 8yM Review Exam 1 Example Find every solution y to the equation t2 y2t yy 2 0 Solution atN 8yM t2 y2y y t2 y2t 2 0 There exits a potential function 1p such that ayw N7 8M 2 M y4 2 ay t2yy3 gt t2y3fgt W2gt8t Mt3t322 4 t gt t32 gt gt I2L 122 Y4 174 t7y ty t7ytC ltl Review Exam 1 Example Find the explicit solution y to the NP 1 t2 et V 7 Y0 Solution Not linear Bernoulli with n 3 Numerator depends only on t denominator depends only on y Separable 4y3y2t3tet gt 4y3ydtt3tetdtc The usual substitution u yt implies du yt dt 4 4u3dut3tetdtc gt u4tetdtc Review Exam 1 Example Find the explicit solution y to the IVP t t2et t4 Solution Recall U4 I tet dt c Integration by parts 1 t et f 17 g 17 gt tetdt tet etdt t 1et 7 g e 7 14 We obtain y4t I t 1et c The initial condition 400 1c 4 1c c5 14 We conclude y4t I t 1et 5 Implicit form Review Exam 1 Example Find the explicit solution y to the IVP m2 er 2 0 2 y 4y3 7 y 14 Solution Recall y4t I t 1et 5 Implicit form The explicit form of the solution is one of 14 4 yt it 1et5 The initial condition implies y0 2 lt 0 We conclude that the unique solution to the IVP is t4 14 yt I l t 1et5 Review Exam 1 Example 3y2 t2 Find every solution y of the equation y 3 1 Solution Not linear Bernoulli n 1 y y Not separable Every term on the right hand side is of the form tnym with n m 2 Euler homogeneous 1 y 2 y 2 H E 1 y 2 3 1 i 2e 39 t2 t We introduce the change of unknown v2 gt ytv gt yvtv Review Exam 1 Example 3y2 t2 Find every solution y of the equation y 2 3Z 1 y Solutionyt y v yzvtv 2H t VtV3v2 1 3v2 1 3v2 1 2v2 2 2v 2v tv V21 gt 2V v 2v v2 1 t 2 1 This is a separable equation for v V dt dt C v v2 1 Review Exam 1 Example 3y2 t2 2ty 39 Find every solution y of the equation y 2 1 Solution 2 V v dt dtc v 1 t The substitution u v2 1 implies du 2v v dt So d 1 7u dtc gt nu lntc gt u c1 where c1 6 Substitute back v2 1 2 c1 Finally v yt 12 2 2 3 lt 2 1lc1t y tc1t lt1 Review Exam 1 Not Covered in Exam 1 Example A water tank initially has V0 2 100 liters of water with Q0 grams of salt At to 0 fresh water is poured into the tank The salt in the tank is always well mixed Find the rates r and r0 such that a The tank water volume is constant b The time to reduce the salt in the tank to one percent of the initial value is t1 25 min Solution Part a Water volume constant implies r r0 r Then Vt 0 so Vt 2 V0 d Part b First find the salt in the tank Qt d C rq roqot Incoming fresh water q 0 Mixing qot Qt Vt L rtVo dt V0Qt gt Qt Qoe Review Exam 1 Not Covered in Exam 1 Example A water tank initially has V0 2 100 liters of water with Q0 grams of salt At to 0 fresh water is poured into the tank The salt in the tank is always well mixed Find the rates r and r0 such that a The tank water volume is constant b The time to reduce the salt in the tank to one percent of the initial value is t1 25 min Solution Recall Qt Q0 e rtVO Condition for r 1 Qt1 amp gt Q0 Trt1V0 1 gt rtl ln 100 0 70 W 11 V0 ln100 gt r ln100 gt r 4 ln100 lt1 V0 11 Review Exam 1 Example Find the solution y to the NP yy ty27 y27r27r7 tgt0 Solution Not linear Bernoulli for n 2 Divide by y2 21 39 t 1 V Zm V s Veil y ty t y y V2Vsint V2V sint t t t t We solve the linear equation with the integrating factor method At dt 2lnt lnt2 gt pt t2 Review Exam 1 Example Find the solution y to the IVP 2 sint y2 yz y 2 2 t 0 t t y7T 7T7 gt 7 Solution Recall Mt t2 Then t2v g v t2 gt 1 2 v tsint Integrating t2 v tsintdt C The right hand side can be computed integrating by parts 5 till Review Exam 1 Example Find the solution y to the IVP 2 I t yy fy27 y27r27r7 tgt0 Solution tsint dt tcost cost dt Then 1 t2 v tcost sint C gt t2 tcost sint C y 1 The initial condition 4722 2 27r cos27r 0 C so C 47T 7t 12 y sint tcost47r ltl Review Exam 1 Example Find the integrating factor that converts the equation below into an exact equation where x3ey i y 2X26y 1 0 y Solution We first verify if the equation is not exact 1 N x3ey l 3 gt 8XN 3x2ey y y M2x2eY1o 8yM2xzey So the equation is not exact We now compute 1 1 2y 2y 2y ayMaXV2xe 3xey xe y 1 N X3ey l i xxzeyl X y y Review Exam 1 Example Find the integrating factor that converts the equation below into an exact equation where x3ey i y 2X26y 1 0 y Solution Recall w l Therefore N x MX 1 l 1 W X moi nx nX W X 1 1 So the equation xzey l y 2xey 0 is exact Indeed y x N Xzey gt 8X1 2xey7 N N y 8XN ayM 1 M 2xey gt 8yM 2xey7 Review Exam 1 Example Find the integrating factor that converts the equation below into an exact equation where x3ey i y 2X26y 1 0 y Solution We first verify if the equation is not exact N x3ey l i gt 8XN 3xzeyl y y M2xzey1O 8yM2x2ey So the equation is not exact We now compute 1 1 2y 2y 2y ayMaXV2xe 3xey xe y 1 N X3ey l i xxzeyl X y y Review Exam 1 Example Find the integrating factor that converts the equation below into an exact equation where x3ey i y 2X26y 1 0 y Solution Recall w l Therefore N X u x 1 1 1 W X moi lnx nX ux X 1 1 So the equation xzey y 2xey 0 is exact Indeed y x NX2eyl gt 9X122xey7 N N V 8XN8yM 1 M 2xey gt 8yM 2xey7 x Review Exam 1 Example Find every solution y of the equation xzey y 2x ey 0 y X Solution The equation is exact We need to find the potential f t39 unCIonw aywzm axwzm From the first equation we get 2 1 2 8y x ey l y gt wzx eylnygx Introduce the expression for w in the equation 8gb M that is 1 1 2xeygx8X M2xey gt gx Review Exam 1 Example Find every solution y of the equation xzey y 2x ey 0 Solution Recall gx Therefore gx lnx The potential function is w X26y lny lnx The solution y satisfies xzey lnyx l lnx c Verification Compute the implicit derivative in the equation above and you should get the original differential equation 1 1 2xeyxzeyyl yl O y x Review Exam 1 Example Find every solution of the initial value problem y 4Xy xTL y0 4 Solution The equation is Not linear It is a Bernoulli equation y 4xy 4xy with n 12 y It IS separable y W7 The equation is not homogeneous It is not exact Although the equation is both separable and Bernoulli it is not simple to integrate using the separable equation method Indeed y dt 4xdxc gt dy 2x2c y7 y7 39 The integral on the left hand side requires an integration table Review Exam 1 Example Find every solution of the initial value problem y 4Xy xTL y0 4 Solution We find solutions using the Bernoulli method y 4xy llxyl2 gt 4xy12 4x y Change the unknowns v 1yn 1 with n 212 That is 1 12 13 Vzm Vy 7 V y12 2v 4xv 4x gt v 2xv 2x The coefficient function is ax 2x so Ax x2 and the integrating factor is ux e XQ Review Exam 1 Example Find every solution of the initial value problem y 4Xy xTL y0 4 Solution Recall v 2XV 2X and uX e XQ 72 72 2Verif 72 72 er 2xer2xeX gty er2xeX 2 72 72 72 7 er22xexdxc gt erz eX l c We conclude that v ceX2 1 The initial condition for y implies the initial condition for v that is vX wyX implies v0 2 2v0c 1 gt c3 gt vX3eX2 1 We finally find y 2 v2 that is yX 3eX2 12 lt1 Review Exam 1 Example Find the domain of the function y solution of the NP 2t y Solution We first need to find the solution y The equation is separable 2 yy 2t gt yydt 2tdtc gt t2c 2 gy 1 lc gt c3 gt ytl23 t2 The domain of the solution y is D 7 The points l do not belong to the domain of y since y and the differential equation are not defined there lt1 Review Exam 1 Example Find the domain of the function y solution of the IVP 21 y 7 yto yo gt 0 y y2 Solution The solution y is given as above 3 t2 c The initial condition implies 2 2 2 2 2 y y to y y y 30 T t c gt c 61 gt 3 t2t 3 The solution to the IVP is yt M2063 t2 y3 The domain of the solution depends on the initial condition to yo 2 2 42 yo 2 yo 0 t037 t03 lt1 Review Exam 1 Example 2x 3y T3x4y39 Find every solution y to the equation y Solution The equation is not linear not Bernoulli not separable It is homogeneous Multiply numerator and denominator on the right hand side by Is it exact 3x 4y y 2x 3y 2 0 implies 8XN 3 8yM So the equation is exact We choose here the exact equation method Finding the potential function is sometimes simpler that solving homogeneous Eqs We need to find the potential function w ayzp N w 3xy 2y2 gX 8X M gt 3ygx 2x3y gt gx 2x2 We conclude wx7 y 3xy 2y2 x2 and xyx c lt1 Review Exam 1 Example 2x 3y 3x4y39 Find every solution y to the equation y Solution If we solve the problem using that the equation is homogeneous it is more complicated than the previous calculation We just start the calculation to see the difficulty lt2x3ygt i MG 3x4y 1 34z X X The change v yx implies y xv and y vxv Hence 23v 23v 23v 3v4v2 gt xv VXV 34V 34v 34v 34v 1 We conclude that v satisfies 2 4v2v Tx39 Review Exam 1 Example 2 3 Find every solution y to the equation y X y 3x4y 3 4v 1 S l t R II oqun eca 24V2v X This equation is complicated to integrate 3v 4vv 1 mdx24V2 dxdxcnxc The usual substitution u vx implies du v dx so 3du 4udu nXC 2 4u2 2 4u2 39 The first integral on the left hand side requires integration tables This is why the exact method is simpler to use in this case lt1 Review Exam 1 Not Covered in Exam 1 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y2y3 y01 Solution First notice that the equation is linear So it is simple to find the solution following Section 11 3 e 2ty 2y 2 3 672 gt eizty 2 6T2t c7 In the next slide we use Picard Lindelo39f s idea Review Exam 1 Not Covered in Exam 1 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y2v3 y01 Solution We first transform the differential equation into an integral equation t t ysws 2ys 3 ds 0 0 t 0 Using the initial condition y0 1 yt 1 r2ys 3 ds 0 ya y0 2ys 3 ds This is the integral equation Review Exam 1 Not Covered in Exam 1 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y 2y3 y01 1 Solution Integral equation yt 1 2 ys 3 ds We now define the sequence of approximate solutions 1 Yo Y017 yn1t1 2Yn5 3 57 n 2 0 0 We now compute the first elements in the sequence t t n07 y1t12y0s3ds15ds15t 0 0 Soy01 and y115t Review Exam 1 Not Covered in Exam 1 Example Use the proof of Picard Lindelo39f s Theorem to find the solution to y2y3 y01 1 Solution Integral equation yt 1 2 ys 3 ds 0 And yo 1 and y1 15t Let s compute y2 t t y21 2y1s3ds1 215s3ds 0 0 t y2 1 510ds15t5t2 0 So we ve got y2t 2 1 5t 512 5 Pl l 41 Lt 1 cm ye ye 2 40 1 139P 1 z yy 9 1 ta b 7 t R1 R8 249439 I 51 e e M M WV armH a 44pr Lo MN 8 g rj b 419 e A 4 aCf 3 fl K hd ImZL HPfwj Em 2 sgm b antPf e 1 1 II 39P 5 2 2m 2 9 red 2259 e a e39 tgto 5 4 gt 3er in e eggs Q 139 0579 3 latf 0 lt396 o M M f bIau 3643 Li lama uo 9 lt0 BEBE i k 5 9 SZLmLE 7 9 do Review for Exam 4 gt V V V V 6 problems 55 minutes in Recitation Rooms 100 Grading Attempts Problems Similar to Homeworks Integration Table Provided No Notes No Books No Calculators gt lVlLC lVlTl l 235 Exam Review httpmathmsuedumlc V Exam covers gt Linear Differential Systems 51 gt Diagonalizable LDS 52 gt 2 x 2 LDS 53 gt 2 x 2 Phase Portraits 54 gt BVP Eigenfunction Problems 71 gt Fourier Series 72 Review for Exam 4 gt 6 problems 55 minutes in Recitation Rooms V V V V 100 Grading Attempts Problems Similar to Homeworks Integration Table Provided No Notes No Books No Calculators gt lVlLC lVlTl l 235 Exam Review httpmathmsuedumlc V Exam covers gt Linear Differential Systems 51 gt Diagonalizable LDS 52 gt 2 x 2 LDS 53 gt 2 x 2 Phase Portraits 54 gt BVP Eigenfunction Problems 71 gt Fourier Series 72 Linear Differential Systems 51 Theorem First Order Reduction A function y solves the second order linear equation y pty qty gt7 iff the functions X1 y and X2 y solve the 2 x 2 system Theorem Second Order Reduction X Any 2 x 2 constant coeffICIents system x Ax With x 2 X1 2 can be written as the second order equation for X1 and for X2 X1 tr A X detAXl 0 X tr A X detAX2 0 Second Order Reduction Example Express as a single second order equation Xl X1 3X27 the 2 x 2 system and solve it X 2 X1 X2 Solution The system is x Ax with A 11 31 Then tr A 2 and detA 2 Hence X1 2X 2X1 2 07 X2 2X 2X2 2 0 1 r22r 2 0 gt ri E ab4 8 gt ri 1i Therefore X1clertczer t7 ngclerft l Eger t lt1 Review for Exam 4 gt 6 problems 55 minutes in Recitation Rooms 100 Grading Attempts V Problems Similar to Homeworks V V Integration Ta ble Provided V No Notes No Books No Calculators gt lVlLC lVlTl l 235 Exam Review httpmathmsuedumlc Exam covers gt Linear Differential Systems 51 gt Diagonalizable LDS 52 gt 2 x 2 LDS 53 gt 2 x 2 Phase Portraits 54 gt BVP Eigenfunction Problems 71 gt Fourier Series 72 V Diagonalizable LDS 52 Definition A system x Ax is diagonaizabe iff matrix A is diagonalizable Remarks gt A matrix A is diagonalizable iff there exist an invertible matrix P and a diagonal matrix D such that A PDP l gt How can we know if a matrix A is diagonalizable Given A do matrices P D exist How can we compute them Theorem Diagonalizability and Eigenvectors An n x n matrix A is diagonaizabe iff matrix A has a linearly independent set of n eigenvectors Furthermore A PDP l P v1 vn Ddiag1 An Where 7 v for i 17 7 n are eigenvalue eigenvector pairs ofA Diagonalizable LDS 52 Example Show that A 1 3 is diagonalizable 3 1 Solution We known that the eigenvalue eigenvector pairs are 1 147 V1 1 and 2 27 V2 2 11 Introduce P and D as follows 1 1 1 1 1 1 4 0 P ll 1i i P TEl 11l7DTlO 2l Then a Theorem above says A PDP l lt1 Diagonalizable LDS 52 Theorem Diagonalizable System lfn x n matrix A is diagonalizable with a linearly independent eigenvectors set v17 7v and eigenvalues A1 7 then the general solution xgen to the system x A x is given by the expression below where c17 C E R xgent c1v1 e 1t cnvn e t Remark gt The differential system for the variable x is coupled gt We transform the system into a system for a variable y such that the system for y is decoupled that is yt Dyt where D is a diagonal matrix gt We solve for yt and we transform back to xt Diagonalizable LDS 52 Theorem Exponential Expression The initial value problem for an n x n homogeneous constant coefficients linear differential system x Ax7 x0 x0 has a unique solution x for every n vector x0 A xt e 1quotx0 Remarks y gt If A PDP l then eAt PeDtP l 1 3 3 1 A 1 1 4 0 1 1 1 1 1 0 2 2 1 1 7 A1 1e4t 0111 6 110e 2t2 11 For example in the case A We know that Therefore Review for Exam 4 gt 6 problems 55 minutes in Recitation Rooms y p y 100 Grading Attempts Problems Similar to Homeworks Integration Table Provided No Notes No Books No Calculators lVlLC lVlTl l 235 Exam Review httpmathmsuedumlc Exam covers gt Linear Differential Systems 51 gt Diagonalizable LDS 52 gt 2 x 2 LDS 53 gt 2 x 2 Phase Portraits 54 D V BVP Eigenfunction Problems 71 Fourier Series 72 2x2LDsws Summary Find solutions of x Ax with A a 2 x 2 matrix First find the eigenvalues A and the eigenvectors v of A a If A1 7E A2 real then v1v2 are linearly independent and the general solution is xX C1 V1e1t C2 v2 e Vt b If A1 7E A2 complex then denoting Ai a l i and Vi a l bi the complex valued fundamental solutions xi a l bi ewi i xi 2 eat a i bi cos6t i sin tl xi 2 eat a cos t bsin t tieat a sin tbcos t Real valued fundamental solutions are x1 2 eat a cos t bsin t7 x2 2 eat a sin t bcos t 2X2LDS 3 Summary Find solutions of x Ax with A a 2 x 2 matrix First find the eigenvalues A and the eigenvectors v of A c If A1 A2 A real and their eigenvectors v1v2 are linearly independent then the general solution is xX C1 V1 e t C2 v2 ext d If A1 A2 A real and there is only one eigendirection v then find w solution of A Alw v Then fundamental solutions to the differential equation are given by x1 v e w7 x2 v t w ext Then the general solution is xC1ve tC2vtwe t 2x2LDsw3 Example 3 Find the general solution of x Ax where A 2 Solution Eigenvalues of A 3 A x5 p 2 23 2 0 A254O Ai 5i25 16 5i3 Hence 1 2 4 Eigenvector for 2 2 2 Al gt2 gt 2V1 x2V2 Choosing V1 2 and V2 2 2 we get v S fl 2X2LDS 3 Example Find the general solution of x Ax where A Solution Recall 1 A 4 and v 2 EIgenvector for 2 mmpl flag gla V1 2 2 V2 Choosing V1 2 2 and V2 2 1 so v Fundamental solutions x 64 x 6quot General solution x C1 e tC2 6quot lt1 l 2x2LDsws Example Find x solution of the IVP xAx7 x07 Azlj Solution Eigenvalues of A 34 4 KM l 1 14 1340 2210 gt Ai 2im1 Hence 1 Eigenvector for Ai 2 4 1 2 1 2 MHPL1JL2k0l v1 2 V2 Choosing v1 2 and V2 2 1 we get V 2X2LDS 3 Example Find x solution of the IVP 3 4 x Ax7 x0 3J7 A 1 1 Solution Recall i 1 and V Find w solution of A Iw v 2 4 W1 2 gt 2 4 2 gt 1 2 1 1 2 W2 1 1 2 1 0 0 0 Hence W1 2 2W2 1 that is w W2 lJ l 1 Choose W2 2 0 so w 01 2x2LDsws Example Find x solution of the IVP 2X2LDS 3 Example Find x solution of the IVP lt0gtlil x 7 1 Solution Recall x C1 2 e t C2 2 t 1 1 1 Initial condition39 1 C 2 C 1 3 1 1 2 0 2 1 C1 1 C1 that Is 0 1 C2 also C2 The solution is x 3 6 5 2 x Ax7 Review for Exam 4 gt 6 problems 55 minutes in Recitation Rooms 100 Grading Attempts V Problems Similar to Homeworks V V Integration Ta ble Provided V No Notes No Books No Calculators gt lVlLC lVlTl l 235 Exam Review httpmathmsuedumlc Exam covers gt Linear Differential Systems 51 gt Diagonalizable LDS 52 gt 2 x 2 LDS 53 gt 2 x 2 Phase Portraits 54 gt BVP Eigenfunction Problems 71 gt Fourier Series 72 V BVP Eigenfunction Problems 71 Example Find the positive eigenvalues and their eigenfunctions of yo 0 y8 0 Solution Since gt 0 introduce 2 with u gt 0 y Ay07 yx erX implies that r is solution of prr2u20 gt riziui The general solution is yx C1 cosuX C2 sinux The boundary conditions imply 0 y0 C1 0 y8 C2 sinu87 gt yx C2 sinux C2 7A 0 gt sinu8 0 2 uzn g E7 yxsin 7 n12 8 8 lt1 BVP Eigenfunction Problems 71 Example Find the positive eigenvalues and their eigenfunctions of y M 07 y0 07 yB 0 Solution The general solution is yX C1 cosuX C2 sinuX The boundary conditions imply 0 Y0 C1 gt yx C2 sinix 0 yB C2M cosL87 C2 7A 0 gt cosp8 0 7r 2n17r Then for n 12 holds 2n17r 2 A l 16 l 7 yX sinW Eigenvalue Eigenfunction BVP Example Find the non negative eigenvalues and their eigenfunctions of y M 07 yO 07 yB 0 Solution Case gt 0 Then yX C1 cosuX C2 sinux Then yX C1u sinux C2u cosux The BC imply 0 yO C2 gt yX C1 cosuX7 yX C1u sinux 0 y8 2 CW sin187 C1 7A 0 gt WT 81 2 mr gt u 8 Then choosing C1 1 for n 12 holds sinu8 0 8 quota ynltxgtcosltmgt Eigenvalue Eigenfunction BVP Example Find the non negative eigenvalues and their eigenfunctions of y M 07 MW 07 yB 0 Solution The case 0 The general solution is yX C1 C2X The BC imply 0 yO c2 gt yX C17 yX 0 Then choosing C1 1 holds A207 YOX1 lt1 A Boundary Value Problem Example Find the solution of the BVP y y 07 y017 Y7T3 0 Solution yX e implies that r is solution of prr2u20 gt rizii The general solution is yX C1 cosX C2 sinX Then yX C1 sinX l C2 cosX The BC imply 1 yO C2 gt yX C1 cosX sinX 0 Y7T3 C1 cos7r3 sin7T3 gt C1 T T C1 2 T gt YX cosx sinX lt1 12 Review for Exam 4 gt 6 problems 55 minutes in Recitation Rooms 100 Grading Attempts V Problems Similar to Homeworks V V Integration Ta ble Provided V No Notes No Books No Calculators gt lVlLC lVlTl l 235 Exam Review httpmathmsuedumlc Exam covers gt Linear Differential Systems 51 gt Diagonalizable LDS 52 gt 2 x 2 LDS 53 gt 2 x 2 Phase Portraits 54 F V V BVP Eigenfunction Problems 71 Fourier Series 72 Fourier Series 72 Example Find the Fourier Series of the odd extension of fx 1 for x 6 10 Solution The Fourier series is fx 2 cosnLLX bn sinnLLX Since 1 is odd then the Fourier Series is a Sine Series an 2 0 bn fxsinnLLX dx OL fxsinnLLX dx 1 sin mrx x 1 DP01 d 2m 1 07 cos mrx bn 2 cosmr 1 gt bn l 1quot 1 77139 77139 Fourier Series 72 Example Find the Fourier Series of the odd extension of fX 1 for X E 170 2 Slt39 R b 1 1 oqun eca n mr 2 If 2k h 12k 1 0 n tenb2lt 2k7r fn2k 1 2 4 2k71 b2k 12k 17r 1 1 2k 17r39 w df 4OO 1 392k1 econcu e X sm uX lt1 k1 Fourier Series 72 Example Find the Fourier series expansion of the odd extension fX 2 X for X E 02 Solution The Fourier series is fX 2 013 cosnLLX bn sinnLLX n1 Since 1 is odd then the Fourier Series is a Sine Series 2 0 1 L mrX 2 L mrX bnzzil fXsmTgt dX ZO fX539nTgt dxi L27 bn 2 022 X sin dXa Fourier Series 72 Example Find the Fourier series expansion of the odd extension fx 2 x for X E 02 Solution bn 202 sin dx 02X sin dx sin dx 2 73 cos7 The other integral is done by parts mm ux7 v sm 2 mrx I Xsm 2 dX7 2 mm 17 v cos mr 2 cosltquot Xgt lt gt 04 Fourier Series 72 Example Find the Fourier series expansion of the odd extension fx 2 x for X E 02 Solution I 77 3 cos coslt dx 2x mrx 2 2 mrx I cos sm So we get 7r mr 2 2 bquot 2 Cosll l iii Coslmiixll fr 2 Sinll bn cosn7r 1 cosn7r 0 gt bn We conclude fx g sm lt1 Fourier Series 72 Example Find the Fourier series expansion of the even extension fX 2 X for X E 02 Solution The Fourier series is fX quotElan cosnLLX bn sinnLLX Since 1 is even then the Fourier Series is a Cosine Series bn O 1 2 2 b h39ht a0 fXdX 2 XdX aseX Elg gt a02 2 2 0 2 L L isxltxgtcosltquotirgtdxO mew 2 nva an 20 2 X cosltT dX Fourier Series 72 Example Find the Fourier series expansion of the even extension fX 2 X for X E 072 2 2 Solution an 2 2 cos dX X cos dX 0 2 0 2 cosmTX dX 2 sinmTX 2 mt 2 7 The other integral is done by parts nva u X7 v cos 2 l nva d xcosT X7 2 I mm u 17 v sm mt 2

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