Math Week 5 concepts and homework, due Friday February 10

Transcription

1 Mt Week 5 concepts nd omework, due Fridy Februry 0 Recll tt ll problems re good for seeing if you cn work wit te underlying concepts; tt te underlined problems re to be nded in; nd tt te Fridy quiz will be drwn from ll of tese concepts nd from tese or relted problems 3 Solving initil vlue problems for liner omogeneous second order differentil equtions, given bsis for te solution spce Finding generl solutions for constnt coefficient omogeneous DE's by sercing for exponentil or oter functions Superposition for liner differentil equtions, nd its filure for nonliner DE's,, (in use initil vlues y 0 = 0, y 0 = 5 rter tn te ones in te text), 0,, 2, 4 (In 4 use te initil vlues y =, y = 7 rter tn te ones in te text), 7, 8, 27, 33, 39 w53) In 3 bove, te text tells you tt y x = e 2 x x = e 3 x re two independent solutions to te second order omogeneous differentil eqution y y y = 0 Verify tt you could ve found tese two exponentil solutions vi te following guessing lgoritm: Try y x = e r x were te constnt r is to be determined Substitute tis possible solution into te omogeneous differentil eqution nd find te only two vlues of r for wic y x will stisfy te DE (See Teorem 5 in te text) w53b) In 30 bove, te text tells you tt y x = e 5 x x = x e 5 x re two independent solutions to te second order omogeneous differentil eqution y 0 y 25 y = 0 Follow te procedure in prt of trying for solutions of te form y x = e r x, nd ten use te repeted roots Teorem in te text, to recover tese two solutions 32 Testing collections of functions for dependence nd independence Solving IVP's for omogeneous nd non-omogeneous differentil equtions Superposition, 2, 5, 8,,3,, 2, 25, 2 Here re two problems tt explicitly connect ides from sections 3-32 wit liner lgebr concepts from Mt 2270 w54) Consider te 3 rd order omogeneous liner differentil eqution for y x y x = 0 nd let W be te solution spce w54) Use successive ntidifferentition to solve tis differentil eqution Interpret your results using vector spce concepts to sow tt te functions y 0 x =, y x = x x = x 2 re bsis for W Tus te dimension of W is 3 w54b) Sow tt te functions z 0 x =, z x = x 2, z 2 x = 2 x 2 2 re lso bsis for W Hint: If you verify tt tey solve te differentil eqution nd tt tey're linerly independent, tey will utomticlly spn te 3-dimensionl solution spce nd terefore be bsis w54c) Use liner combintion of te solution bsis from prt b, in order to solve te initil vlue

2 problem below Notice ow tis bsis is dpted to initil vlue problems t x 0 = 2, weres for n IVP t x 0 = 0 te bsis in would ve been esier to use y x = 0 y 2 = 7 y 2 = 3 y 2 = 5 w55) Consider te tree functions y x = cos 2 x x = sin 2 x, y 3 x = sin 2 x w55) Sow tt ll tree functions solve te differentil eqution y 4 y = 0 w5b) Te differentil eqution bove is second order liner omogeneous DE, so te solution spce is 2 dimensionl Tus te tree functions y, y 3 bove must be linerly dependent Find liner dependency (Hint: use trigonometry ddition ngle formul) w55c) Explicitly verify tt every initil vlue problem y 4 y = 0 y 0 = b y 0 = b 2 s solution of te form y x = c cos 2 x c 2 sin 2 x, nd tt c, c 2 re uniquely determined by b, b 2 (Tus cos 2 x, sin 2 x re bsis for te solution spce of y 4 y = 0: every solution y x s initil vlues tt cn be mtced wit liner combintion of y, but once te initil vlues mtc te solutions must gree by te uniqueness teorem, so y spn te solution spce; y re linerly independent becuse if c cos 2 x c 2 sin 2 x = y x 0 ten y 0 = y 0 = 0 so lso c = c 2 = 0) w55d) Find by inspection, prticulr solutions y x to te two non-omogeneous differentil equtions y 4 y = 28, y 4 y = x Hint: one of tem could be constnt, te oter could be multiple of x w55e) Use superposition (linerity) nd your work from c,d to find te generl solution to te nonomogeneous differentil eqution y 4 y = 28 x w55f) Solve te initil vlue problem, using your work bove: y 4 y = 28 x y 0 = 0 y 0 = 0

3 33: using te lgoritm for finding te generl solution to constnt coefficient liner omogeneous differentil equtions: rel roots, Euler's formul nd complex roots, repeted roots; solving ssocited initil vlue problems 33: 3, 9,, 23, 27 NOTE: w5 nd w57 re postponed until next week (Fridy) We will tlk bout section 33 tis Fridy Feb 0 w5) Do te following problems for omogeneous liner differentil equtions by nd Tey re testing your bility to use te lgoritm for finding bses for te solution spces, bsed on te crcteristic polynomil Ceck your work wit Mple (or oter softwre) In Mple you will wnt to use te "dsolve" commnd to ceck differentil eqution solutions, nd my wnt to use te "fctor" commnd to ceck your fctoriztions of te crcteristic polynomils Hnd in printout of your computer verifictions for te differentil eqution solutions, long wit your written work ) Find te generl solution to te differentil eqution for y x y 3 5 y 3 y 9 y = 0 Hint: Find root r of te cubic crcteristic polynomil, ten divide it by r r to get quotient qudrtic polynomil b) Find te generl solution to te differentil eqution for x t x 4 x 3 x = 0 Hint: completing te squre works well ere - probbly better tn te qudrtic formul c) Solve te initil vlue problem for te differentil eqution in b, wit x 0 = 0, x 0 = 9 d) Find te generl solution to te differentil eqution for y x y 4 8 y = 0 e) Find te generl solution to te differentil eqution for y x y 5 y 3 9 y = 0 w57) Euler's formul e i = cos i sin is extremely useful in iger mtemtics/science/engineering In clss we discuss ow tis definition is motivted by Tylor series Amzingly, te rule of exponents is true for suc expressions In oter words e i i = e i e i Ceck tis identity by rewriting te left side s e i nd ten using Euler's formul to expnd bot sides You'll notice tt te identity is true becuse of te ddition ngle formuls for cos nd sin (And so tis gives good wy to recover te trig identities if you ppen to forget tem) 24-2 Numericl metods problems re t te end of tis ssignment w5) Runge-Kutt is bsed on Simpson's rule for numericl integrtion Simpson's rule is bsed on te fct tt for subintervl d, d of lengt, te prbol y = p x wic psses troug te points

4 d, y 0, d 2, y, d s integrl d d p x dx = y 0 4 y y 2 w5) Te integrl pproximtion bove follows from one on te intervl, by n ffine cnge of vribles So first consider te intervl, We wis to find te prbolic function q x = x 2 b x c wit unknown prmeters, b, c We wnt q = y 0, q 0 = y, q = y 2 Tis gives 3 equtions in 3 unknowns, to find, b, c in terms of y 0, y Write down tese liner equtions nd find, b, c w5b) Compute q x dx for tes vlues of, b, c you find in prt, nd verify te identity q x dx = 3 y 0 4y y 2 Note, te formul for generl intervl follows from cnge of vribles, s indicted below: Remrk: If you've forgotten, or if you never tlked bout Simpson's rule in your Clculus clss, ere's ow it goes: In order to pproximte te definite integrl of f x on te intervl, b, you subdivide, b into n subintervls of widt x = b n = Ten dd te midpoints of ec subintervl Lbel tese x vlues (incluidng midpoints) s x 0 =, x = 2, x 2 =, x 3 = x 3 0 2, x 4 = x 0 2, x 2 n = x 0 2n = b, wit corresponding y vlues y i = f x i, i = 0,2 n On ec successive pir of intervls x 2 k, x 2 k use te prbolic estimte x 2 k x 2 k f u du f x 2 k 4 f x 2 k f x 2 k 2 = y 2 k 4 y 2 k y 2 k 2 bove, estimting te integrl of f by te integrl of te interpolting prbol on te subintervl Tis

5 yields te very ccurte (for lrge enoug n) Simpson's rule formul b f x dx y 4 y y y 4 y y y 4 y y, n 2 2 n 2 n ie b f x dx b n y 0 4 y 2 y 2 4 y 3 2 y 4 2 y 2 n 2 4 y 2 n y 2 n ttp://enwikipediorg/wiki/simpson's_rule w52) (Fmous numbers revisited, section 2, pge 35, of text) Te mty numbers e, ln 2, cn be well-pproximted using pproximte solutions to differentil equtions We illustrte tis on Wednesdy Feb 4 for e, wic is y for te solution to te IVP y x = y y 0 = Apply Runge-Kutt wit n = 0, 20, 40 subintervls, successively doubling te number of subintervls until you obtin te trget number below - rounded to 9 deciml digits - twice in succession We will do tis in clss for e, nd you cn modify tt code if you wis w52) ln 2 is y 2, were y x solves te IVP (since y x = ln x ) y x = x y = 0 w52b) is y, were y x solves te IVP 4 y x = x 2 y 0 = 0 (since y x = 4 rctn x ) Note tt in,b you re ctully "just" using Simpson's rule from Clculus, since te rigt sides of tese DE's only depend on te vrible x nd not on te vlue of te function y x For reference: > Digits : #ow mny digits to use in floting point numbers nd clculultions evlf e ; #evlute te floting point of e evlf ; ()