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# Ordinary Differential Equation MATH 365

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This 99 page Class Notes was uploaded by Vernie McCullough on Thursday October 15, 2015. The Class Notes belongs to MATH 365 at Millersville University of Pennsylvania taught by J. Buchanan in Fall. Since its upload, it has received 10 views. For similar materials see /class/223525/math-365-millersville-university-of-pennsylvania in Mathematics (M) at Millersville University of Pennsylvania.

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Date Created: 10/15/15

Impulse Functions MATH 365 Ordinary Differential Equations J Robert Buchanan Department of Mathematics Summer 2008 Introduction A special case of discontinuous forcing functions is that of large magnitude but short duration forces eg impact from a strike or a collision Suppose gt is large for to 7 T lt tlt to T and zero otherwise The impulse of gt is g 739 co Tt l gtdti gtdt 77 Example 1 of 2 Determine the impulse for a force of the form l if76lttlt6 d6 26 O if t ZE where e gt O Example 2 of 2 Note that 5 1 t 1 672d12 independent of 5 Thus 151 has the following two properties Example 2 of 2 Note that 5 1 t 1 672d12 independent of 5 Thus 151 has the following two properties 0 lira 5 1 Example 2 of 2 Note that 5 1 t 1 672d12 independent of 5 Thus 151 has the following two properties 0 in 5 1 9 Hr 151 O for all 17 0 irc Delta Function i We can use 151 to define a generalized unit impulse function Definition The Dirac delta function is the function 6 having the properties that 61 O forty Oand 6tdt 1 Properties of the Dirac Delta Fncti For any to 61710 O forty toand 6titodt 1 Laplace Transform of the DiraCDeltnction 7 5TTO rg weU ToH 00 t Iim e sd tit dt 6H00 5 0 95 1 Iim e Stidt Ho 05 25 71 Iim ie St Ho 25 i 1 675m 7 65t075gt m Ho 253 I eistg ese 7 6755 IIm 7 Ho es 2 eisto m slnhes eisto Ho es ther Useful Result 1 of 2 H is continuous then f 617 to 1 dt 10 iAppmerUseMIResuH20fm EEHIIIIIIIIIIIIIIIIIIIIIII W6U7m 0w glr uimaow 1 00 95 n i ftd earng Zeto t is lira fc25 for some to 7 e g c g to e by IMVT Aw 10 Solving an IVP Solve the following IVP y y 614 Solution 710 Solve the following IVP y 5y 4y 614 Homework 0 Read Section 65 0 Pages 344346 1 11 odd Existence and Uniqueness of Solutions MATH 365 Ordinary Differential Equations J Robert Buchanan Department of Mathematics Spring 2009 Backg ound We have studied methods for solving first order 0 linear o separable and o exact ODEs and IVPs The question remains whether the general first order IVP dy 7 E 7 My We yo has a solution and it so is that solution unique an Result Consider the l VP dy 7 E i any y0 0 fl and 8f8y are continuous in a rectangle H t y M g a M g b then there is some interval M g h g a in which there exists a unique solution y gtt of the l VP Integral Equation 1 of 2 If gtt solves the NP then MT ff7 gtf t t wwYmmmw o o t m Arw w Integral Equation 1 of 2 It gtt solves the NP then MT ff7 gtf t t gt sds rs7 gtsds o o t W A rs7 sds Now suppose there exists a function gtt for which gtt 0t fs gts ds then 0 0 0rs sdso and t W A rs7 gtsdsra r it fis continuous Integral Equation 2 of 2 We have just shown that solving the initial value problem My y0 0 is equivalent to solving the integral equation W firms ds To prove the existence and uniqueness theorem it is more convenient to find a unique solution to the integral equation Integral Equation 2 of 2 We have just shown that solving the initial value problem My y0 0 is equivalent to solving the integral equation W firms ds To prove the existence and uniqueness theorem it is more convenient to find a unique solution to the integral equation The procedure for solving the integral equation is called the method of successive approximations or sometimes Picard iteration Method of Successive Approximtion Define a sequence of functions gtnt 20 by setting gto 0 t and mm fs gtnsds for n e N 0 Note gtn0 O ie each member of the sequence satisfies the initial condition of the NP Questions to be Answered 7 7 0 Are all elements of the sequence gtnt 200 defined Questions to be Answered 7 7 0 Are all elements of the sequence gtnt 200 defined 9 Does the sequence have a limit Questions to be Answered 7 7 0 Are all elements of the sequence gtnt 200 defined 9 Does the sequence have a limit 0 It the limit exists does it solve the NP Questions to be Answered 7 7 0 Are all elements of the sequence gtnt 200 defined 9 Does the sequence have a limit 0 It the limit exists does it solve the NP 9 Is the solution to the NP unique Example 1 of 3 Consider the NP Find the first four terms of the sequence of successive approximations Example 2 of 3 gt01 O t 4W 08702ds12 t M0 0s7s222dsr272lor5 gt3t railr Lgs L111 160 7 4400 Example 3 of 3 Is the sequence q5nt defnd 0 We have assumed that fty is continuous on H a closed and bounded set Is the sequence qbnt fo defnd 0 We have assumed that fty is continuous on H a closed and bounded set 0 By the Extreme Value Theorem there exists a positive number Msuch that iftyi g Mtor a y 6 H Is the sequence qbnt fo defind 7 0 We have assumed that fty is continuous on H a closed and bounded set 0 By the Extreme Value Theorem there exists a positive number Msuch that iftyi g Mtor a y 6 H 0 Thus 7M g y g Mtorall y 6 H Restricting t Define h mina bM then if t gtt e Hfor iti g hthen gtn1t is defined WW Does the sequence qbnt cnv A t fs 0 ds 0 t fs0ds o t Mds 0 Mt 8f Slnce 87 IS contlnuous on H there exlsts another posltlve constant K such that 8f 5 gK fortyeH 8f Slnce 7 IS contlnuous on H there exlsts another posltlve constant K such that 8f 5 gK tortyeH By the Mean Value Theorem fty1fty2 if 2 ywyz 8y for some 2 between y1 and y2 Apply the MVT 7 gt2t7 gt1t Af smsndsiAt s Ws A t 7 0 fs gt1s7fs0 ds A t 0 Kws 70 ds A t KMs ds 0 Induction Step The Principle of Mathematical Induction can be used to show that MKquot n1 n1 i gtn10 i gtn0i S forneN Telescoping Summation 1 of 2 Consider the infinite series defined as 00 gt1T k1t M0 k Its nth partial sum is E E N E vx H 2 x 33 N SE w SE 2 Am 06 S coszEsm 960898 Uniform Convergence Hence the sequence gtt 200 converges uniformly Define at quotEwart and note that quotng0 mt Iqu Otfs7 gtn1sds t lim s n1sds 0 NHOO t fs lim n1sds 0 nHOO t at 0 rs7 sds which means gtt solves the integral equation and hence also solves the NP Is the Solution Unique 7 7 0 Suppose gtt and 11 both solve the NP t o Define ut l gts7wslds 0 0 Note that u0 O and ut 2 O for all t 10 l gt1 0l Uniqueness e KtuU e KtuU 0 g ut lA lA lA l A l A lA lA WT iwmi on H by the MVT Which implies ut O for all twhich is equivalent to gtt 11 for all t The continuity of ST is essential for uniqueness Consider the NP Show that y1t O and y2t IS2 are both solutions to the NP Homework 0 Read Section 28 0 Pages 117119 3 7 Differential Equations with Discontinuous Forcing Functions MATH 365 Ordinary Differential Equations J Robert Buchanan Department of Mathematics Summer 2008 Background 1 of 2 Recall from our discussion of step functions and the Laplace transform the following two theorems ilfFs E TO exists fors gt a 2 O and ifc is a positive constant then E uctft 7 0 e CSE ft e CSFs fors gt 3 Conversely ifft E 1 Fs then uctft 7 c E 1 e CSFs Background 2 of 2 IfFs ft fors gt a 2 O and ifc is a constant then E e ft Fs 7 c fors gt a 0 Conversely ifft E 1 Fs then e ft r1 Fs 7 0 Solve the following initial value problem y 9y sintu7rtsint77r V0 0 V 0 0 W 1 1 1 1 i sm3t gsmt u7rt i sm 313977139 gsmt Y0 015 010 005 1 t 2 4 6 Solve the following initial value problem Wty U1tU2t OOOO Letht 71 geosht geost then VU 1tht 1 2tht 2 Y0 Homework 0 Read Section 64 0 Pages 337339 1 13 odd Series Solutions Near an Ordinary Point MATH 365 Ordinary Differential Equations J Robert Buchanan Department of Mathematics Summer 2009 Ordinary Points 1 of 2 Consider the second order linear homogeneous ODE Pty Qty Rty o where P Q and R are polynomials Definition A point to such that Pt0 7 O is called an ordinary point If Pt0 0 then to is called a singular point Ordinary Points 2 of 2 W i Pty Qty Rty o If to is an ordinary point then by continuity there exists an interval ab containing to on which Pt 7 O for all t e a b 7 Qt 7 Rt Thus the functions p t 7 Pm and qt 7 Pm are defIned and continuous on a b and the ODE can be written as y 190 qty 0 If the initial conditions are yt0 yo and y t0 y then there exists a unique solution to the ODE satisfying the initial conditions Power Series Solutions 1 of i 7 7 Consider the ODE y i 4y 0 and find a power series solution with positive radius of convergence centered at an ordinary point Power Series Solutions 1 of Consider the ODE y i 4y 0 and find a power series solution with positive radius of convergence centered at an ordinary point Let to 0 be the ordinary point for simplicity and yt 2 ant n0 y t zjnantn 1 n1 8 Substitute into the ODE Power Series Solutions 2 of 0 y 7 4y Z nn 71at 2 7 4 2 ant quot2 n0 2m 2n 1a2t 7 4 Zant n0 quot0 Z a2n 2n 1 7 4a t n0 O an2n2n 1 7 4617 forn 012 223 an2 n n2n 1 This last equation is called a recurrence relation Power Series Solutions 3 of 7 i if 7 Let ao be arbitrary then a i 22 a 22a 2 2x1 0 2 0 a i 22 a 24a 4 4X3 2 1 7 22 26 as Wa4 aquot 22n a2n 7 a0 Power Series Solutions 4 of i if 7 7 Let 31 be arbitrary then 33 as 37 a2n1 22 a 2728 32 1 3 1 22 a 246 54 3 5 1 22 26 76quotquot5 is 22n Power Series Solutions 5 of Thus the general solution to y 7 4y O can be written as Y Za2nt2n Za2n1t2n1 n70 n0 7 i 22 2n i 22 2n1 80 t 31 it quot0 2n quot0 2n1 0 2t2n a1 00 2t2n1 a i 7 0 2n 2 n02n1 a0 cosh2t 3 sinh2t 2 We can confirm this series converges for all t e R Airy s Equation 1 of 6 7 Find a power series solution about the ordinary point to O to Airy s equation y 7ty 0 Airy s Equation 1 of 6 7 Find a power series solution about the ordinary point to O to Airy s equation y 7 ty O yt 2 ant n0 y t zjneintn 1 n1 8 Substitute into the ODE Airy s Equation 2 of 6 7 i if 7 7 nn 71at 2 7 t ant n0 n 2n 1a2t 7 Z ant 1 n0 232 Zn 2n 1a2t 7 Zam n1 n1 H M8 3 H N H M8 3 H o 232 i n 2n 1a2 7 371 t n1 Airy s Equation 3 of 6 7 Thus 0 282 O n2n1an27an1 forn12 From the last equation we obtain the recurrence relation an71 an2 n2n139 Airy s Equation 3 of 6 7 Thus 0 282 O n2n1an2 76174 forn 12 From the last equation we obtain the recurrence relation a an 4 quott2 n2n139 Sincea20thena5a3a11O Airy s Equation 4 of 6 7 Let ao be arbitrary then i 33 23 a 7 i3 30 5 56 6532 a is 60 9 897986532 6390 83quot 2356 3397 7 4X3 7 3X3 713n39 Airy s Equation 5 of 6 7 Let 31 be arbitrary then a i 781 4 34 a 7 347 a1 7 67 7643 a 7 a7 7 31 1 91071097643 31 33 3467 3n 7 33n 7 23n3n 139 Airy s Equation 6 of 6 7 7 7 Thus the solution to Airy s equation y7ty0 co t3n ym lt1 moxmwyoniownimoninwmgt t3n1 zmmwmim4mmww Homework 0 Read Section 52 0 Pages 259260 1 2 3 5 6 21 Review We have determined a method for solving an ODE ofthe form Pty Qty Rty o where P Q and R are polynomials The solution is a power series ofthe form yt Zena 7 to n0 where to is an ordinary point Review We have determined a method for solving an ODE ofthe form Pty Qty Rty o where P Q and R are polynomials The solution is a power series ofthe form yt Zena 7 to n0 where to is an ordinary point Today we extend this work to a broader range of functions than polynomial P Q and R Differentiation of Power Serie Suppose gtt Zena 7 to has a positive radius of n0 convergence then we may differentiate the series termbyterm n i gtO t i gt to o 2at7 to ac n0 1 nat 7 to 1 31 noo1 2 nn 7 1at 7 to 2 2612 a 2 3 M8 nn7 1 n7 m 1at 7 to quot mlam nm Solution of an ODE 7 7 7 Suppose gtt solve the ODE y pty qty 0 then 0 gt tPt gt tqt gtt t WT P gtT 7T gtT gt to Pto gtto 7To t0 2a2 7pt0a1iqtoao Thus we may find 32 in terms of ac and a1 Continued Differentiation 1 02 i gt t PUWt 70 5 PtWt gtWT0 P To 3la3 PToa1 2PTo32 qToao QToa1 Substituting the previously determined value of 32 we may find 33 in terms of ac and a1 Continued Differentiation 2 02 i if 7 We can proceed by repeated differentiation to find 34 a5 provided Continued Differentiation 2 02 i 7 7 We can proceed by repeated differentiation to find 34 a5 provided 0 pt and qt have derivatives of all orders and Continued Differentiation 2 02 We can proceed by repeated differentiation to find 34 a5 provided 0 pt and qt have derivatives of all orders and 0 we can show the resulting power series converges Analytic Functions 7 7 7 7 If pt and qt are analytic functions at to in other words have Taylor series expansions about to which converge to pt and qt respectively then p and q will have derivatives of all orders at to W 219707 to n0 W an 7 to n0 Ordinary and Singular Points evisited i Suppose Pty Qty Rty O H 00 R0 y W Wy 7 0 y pty qty 0 and pt and qt are analytic at to then we say that to is an ordinary point ofthe ODE Otherwise to is a singular point Main Result r If to is an ordinary point of the ODE Pty Qty Rty 0 then the general solution of the ODE is Y Zena toquot aoY1t a1V2t n0 Where a0 and a1 are arbitrary and y1 and y2 are linearly independent series solutions that are analytic at to Further the radius of convergence for each of y1 and y2 is at least as large as the minimum of the radii of convergence for the series pm camPm andqt RmPm Radius of Convergence W If pt QtPt and qt RtPt and pt and qt are analytic at to then from the theory of complex variables we have the result that the radius of convergence of pt and similarly qt is at least as large as the minimum distance from to to any root of Pt in the complex plane The value to 1 is an ordinary point ofthe ODE t2y 1 ty 3Inty 0 Find the radius of convergence ofpt and 3Int q T 12 The value to O is an ordinary point ofthe ODE 1 t4y 4ty y 0 Find the radius of convergence ofpt 1724 and 1 t 7 q 1 t4 The value to 1 is an ordinary point ofthe ODE 1 t4y 4ty y 0 Find the radius of convergence ofpt 1724 and 1 t 7 q 1 t4 Assuming thaty gtt is a solution to the NP y tzy sinty O V0 1 VO 1 find the first four nonzero terms in the power series representation of gtt Homework 0 Read Section 53 0 Pages 265267 1 7 odd 10 11 22 29 First Order Linear Equations and Integrating Factors MATH 365 Ordinary Differential Equations J Robert Buchanan Department of Mathematics Spring 2009 Introduction A first order linear ordinary differential equation has the general form dy iplttgt y glttgt d C lfpt and gt are constanm as in d l t a y b d t we say the first order linear equation is a constant coefficient ordinary differential equation Integrating Factor Given the first order linear ordinary differential equation gummy am an integrating factor is a function of the form M t ejpsdls If we multiply both sides of the ODE by p t the leftihand side of the ODE becomes an exact derivative 2 Linear nb dy Mt 7pltcgty ulttgtglttgt dt t s s d t s s eLDPW gummy are gm t d t t em 5 as l p c Jump 5 as y em 5 as g t dt d t t J tup s dls LDP s dls g t Y dc Example Find the general solution of the first order linear ordinary differential equation dy 7 72ty dc ln1 soln TableD olvey t Exp 2tyt y0 c yt t c 3 2 l 0 l 2 3H ln2 Plotyt solnl t 039 5 AxesLabela quottquot quotytquot Y0 150 100 50 Oul2 t 1 4 5 750 7100 Exa m p le Find the general solution of the first order linear ordinary differential equation d l t 3 t y c d 1 mp solnTableDSolvey39t 3 tyt t YlO c Yt t c 3 2 ll 039 ll 2 3H Lineannb 3 In4 Plotyt solnl t 039 l AxesLabela quottquot quotytquot Y0 15 10 5 om4 i i t 0 8 1 0 75 710 Exa m p le Find the general solution of the first order linear ordinary differential equation dy y 7 7 7 cos t dt C In5 soln TableDSolvey39 t yt t Cost y7r2 c yt t c 3 2 ll 039 ll 2 3H In6 Plotyt solnl t 7r4 4n AxesLabel gt quottquot quotyt quot Y0 0mm Homework Read Section 21 Pages 3941 work exercises lcil 10 odd 1319 odd 31 33 Exact Equations and Integrating Factors MATH 365 Ordinary Differential Equations J Robert Buchanan Department of Mathematics Spring 2009 Ba c kg ro u n d We have already explored separable ODEs whose general form is Mt dtNy dy 0 Today we would like to generalize this equation to the general form Mt y dtNt y dy 0 In general these types of ODEs are nonlinear and nonseparable I Motivating Example Consider the following ODE dy 2 l y d5 2 cos y We can reiwrite this ODE in the form 21ydt12 cosydy 0 You can verify that this ODE is neither linear nor separable However if we define the following function of two variables Ma y 12y siny We can see that 2 Exact nb aw 0y t2 cosy This enables us to write the ODE as We can reiwrite this equation as a w a w dy a c a y dc Then if we assume that y is implicitly a function of I the equation above can be reiwntten according to the multivariable Chain Rule as d 11 l O 7 c d t Y Integrating both sides of this equation yields the implicit solution for y W I y C where C is a constant n1 CornourPlott2 y Siny t 10 10 Y 4 7r 4 7r FrameLabel gt quot tquot quotyquot Contourshadinga False Contours gt 15 PlotPoints gt 50 O 4 0mm Exactnb 3 General Result Given a differential equation of the form dy M0 y Nt y a 0 if there exists a function 1 t y such that g M I y and g Nt y and y is implicitly a function of I then the ODE is said to be exact A proof of this theorem can be found here I Example 1 of 2 Show that the following ODE is exact and find is general solution dy Eli271 dtilr t2y 4 Exact nb In2 CornourPlott2 y22 t y t 5 5 Y 5 5 FrameLabel gt quot tquot quotyquot Contourshadinga False PlotPoints gt 50 Contours gt 15 gt 0 om2 I Example 2 of 2 Determme 1f the followmg ODE IS exact If 1t ls nd the solutlon to the followmg IVP 1 dc y dy mm EMF 32 32 ylt1gt1 Exactnb 5 In3 CornourPlott2 y2 2 t qrt2 qrt2 y qrt2 qrt2 FrameLabela quottquot quotyquot 0mm o o 1 1H1111111iiiiiiiiiiiiiii 11111 iiiiiiiiiiiiiiiiiiw 00 05 10 Integ rating Factors Sometimes an ODE is not exact but becomes exact after multiplying it by an integrating factor Integrating factors may be of the form1 0 11 ler1U I Example 1 of 2 Consider the ODE below Show that it is not exact as written Find a integrating factor 1 k which makes the ODE exact and then solve the ODE d 2ty27y 510 dc I Example 2 of 2 Consider the ODE below Show that it is not exact as written Find a integrating factor which makes the ODE exact and then solve d1 2 c e y71 dc Homework Read Section 26 Pages 99101 work exercises 115 odd 18 19 21 25 27

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