KEY. Math 334 Midterm II Fall 2007 section 004 Instructor: Scott Glasgow

Transcription

1 KEY Math 334 Midtem II Fall 7 sectio 4 Istucto: Scott Glasgow Please do NOT wite o this exam. No cedit will be give fo such wok. Rathe wite i a blue book, o o you ow pape, pefeably egieeig pape. Wite you ame, couse, ad sectio umbe o the blue book, o o you ow pile of papes. Agai, do ot wite this o ay othe type of ifomatio o this exam.

2 . Detemie a lowe boud fo the adius of covegece of the powe seies epesetatio of the geeal solutio of the followig diffeetial equatio about the poit x = : ( ) x x + 4 y + xy + x y =. (.) 5 poits Solutio Equatio (.) ca be educed to the equatio ( ) x + 4 y + y + xy = (.) which has sigulaities at the zeoes of the leadig coefficiet, which ae x = ± i. I the complex plai the distace of the sigulaities to the expasio poit x = = + i is ( m ) + = 3. Thus, eve alog the eal axis, we caot guaatee a adius of covegece beyod 3 without moe ifomatio.. Fid a (paticula) solutio of the followig diffeetial equatio by the method of udetemied coefficiets: 7 poits Solutio y y + y = 3t+ 5. (.3)

3 The usual explaatio of the asatz fo developig a paticula solutio to a liea costat coefficiet diffeetial equatio (with a RHS that is i the ull space of a liea costat coefficiet diffeetial opeato) is to fist fid a basis fo the spa of the RHS togethe with all its deivatives. The, baig the pheomea of esoace (which is that oe o moe elemets of such a basis ae i the ull space of the specific diffeetial opeato i questio), oe the foms a geeal elemet of the space spaed by the basis, which geeal elemet costitutes the method of udetemied coefficiets asatz fo a solutio of the equatio i questio. Fo the poblem at had, ad sice the RHS of (.3) is spaed by the set of fuctios { t,}, whose fist deivatives ae both spaed by the set of fuctios { } { t,}, so that all subsequet/highe deivatives ae spaed by{ t,}, the elevat basis fo a paticula solutio of (.3) is, baig esoace, { t,}. Oe soo fids that eithe of these is a solutio of the homogeeous vesio of(.3), so that thee is o esoace, ad the asatz fo a solutio of (.3) is, togethe with elevat deivatives, y = At + B y = t+ A y = t+. (.4) Weighted appopiate fo the equatio (.3), the equatios (.4) ae y = At + B y = t A + y = t+ (.5) which sum to y y + y = A+ At+ B= At+ B A = 3t+ 5. (.6) The latte equatio holds uifomly i t if ad oly if A = 3 ad B A= 5 B= 5+ A= 5+ 3=. Thus the solutio sought is y = At+ B= 3t+. (.7) 3. A 5kilogam mass stetches a spig /5 mete. If the mass is set i motio fom the equilibium positio at 5metes pe secod upwad, ad thee is o dampig, detemie the displacemet ut () of the mass above the equilibium positio at ay subsequet time t. Use that the acceleatio of gavity is 49/5 metes pe secod pe secod. 9 poits

4 3 Solutio The elevat vesio of Newto s secod law is = mu + ku = 5kgu + ku. (.8) Hee we may detemie the spig costat k fom k F s ma s mete mete = / = / = 5kg49 / 5 / sec /(/ 5 ) = 5 7 kg / sec, (.9) so that (.8) is 5kg 5 7 kg / sec 7 / sec = u + u = u + u. (.) Redeig (.) uit-less, by measuig time i secods, this is the geeal solutio to which beig The iitial data specifies that 7 = u + u, (.) u = Acos(7 t) + Bsi(7 t). (.) u() = = A, u () = 5 = 7B A=, B= 5 / 7, (.3) so that the equied solutio to the iitial value poblem is u = Acos(7 t) + Bsi(7 t) = 5/ 7 si(7 t). (.4) 4. Fid the geeal solutio of the followig Eule equatio, oe that is valid fo x > : =. (.5) xy xy y poits Solutio

5 4 The diffeetial equatio (.5) defies a liea diffeetial opeato L x, i tems of which (.5) ca be witte Lx[ y ] =. O a fuctio y = x oe fids that = + = + = +, (.6) L [ y ] ( ) 3 3 x 4 3 x 3 x x so that complex solutios of (.5) ae clealy the + 3i 3ilx y+ 3i = x = x e = x ( cos(3l x) + isi(3l x) ) ad 3i 3ilx y 3i = x = x e = x ( cos(3l x) isi(3l x) ). Idepedet complex liea combiatios of these liealy idepedet complex valued solutios gives the followig eal-epesetatio of the geeal solutio: ( cos(3l ) si(3l )) = +. (.7) y x A x B x 5. Solve the followig iitial value poblem: poits Solutio y 4y + 9y = ; y() =, y () = 3. (.8) This liea homogeeous diffeetial equatio is associated with the followig chaacteistic (polyomial) ad chaacteistic expoets : = 4+ 9 = = 5i = ± 5. i (.9) Accodig to the usual theoy, a eal-epesetatio of the geeal solutio, ad its coespodig fist deivative, ae ( cos5 si 5 ) = + t y e C t C t ad (( ) ( ) ) = + 5 cos si5. t y e C C t C C t (.) Isetig t = ito (.), ad usig the iitial data give i (.8), oe obtais

6 5 y() = C = ad y () = C + 5C = 3, (.) the solutio to which beig C = ad C = / 5. Thus the solutio to the iitial value poblem is the y = e cos5t+ si 5t 5. (.) t 6. Give that y = t is a solutio of = (.3) t y ty y 4 poits Solutio fo t >, fid a secod, liealy idepedet solutio y of (.3) fo t > by makig the D Alembet asatz y = vy = vt. Usig D Alembet s asatz i (.3) gives ( 4 3 ) ( 6 4 ) = t y 3ty + 4y = t vt 3t vt + 4vt = t vt + 4vt + v 3t vt + tv + 4vt tv t t v t t t v = = + = + = tv tv tu tu t tu u, (.4) ad whee we defied u = v. By ay oe of a umbe of stadad techiques, oe fids that the fist ode homogeeous equatio (.4) has a otivial solutio u = t = v v = l t. Thus a secod, liealy idepedet solutio is y = vt = t t (.5) l. 7. Fid the geeal solutio of the followig Eule equatio, oe that is valid fo x > :

7 =. (.6) xy xy y 6 poits Solutio The diffeetial equatio (.6) defies a liea diffeetial opeato L x, i tems of which (.6) ca be witte Lx[ y ] =. O a fuctio y = x oe fids that L [ y ] = ( ) 5+ 9 x = 6+ 9 x = 3 x, (.7) x 3 so that a solutio of (.6) is clealy the y3 = x. To fid the geeal solutio to this secod ode diffeetial equatio we eed to fid a secod, liealy idepedet 3 solutio. Sice the asatz y = x oly poduces solutios depedet upo y3 = x, we must use aothe asatz. Fotuately the stuctue of (.7), togethe with the fact that the diffeetial opeatos d d ad Lx commute, suggest such a alteative asatz: applyig d to both sides of (.7), ad usig the idicated commutivity, oe obtais d d [ ] 3 l ( 3) Lx y = x x+ x, (.8) d d y 3 x l x x l x 3 = 3 so that = = is clealy a secod, liealy idepedet solutio of d = (.6). Thus the geeal solutio to this liea homogeeous equatio is ( l ) 3 y = A+ B x x. (.9) 8. Fid the fist two ozeo tems (if thee ae that may) i the seies epesetatio of oe of the liealy idepedet solutios of the equatio 8 poits about the poit x =. 3 x + x y x+ 6x y + + 6x y =. (.3)

8 7 Solutio The poit x = is a sigula poit, so that the equied seies solutio is ot quite a Taylo seies: iset y = ax + (with the assumptio that a = fo <, ad that the sum is ove the iteges, ad that a )i (.3) to obtai 3 = x + x + ( + ) a x x+ 6x + a x + + 6x a x ( ) ax 6( ) ax ( + ) a x + + ( + ) ax = ax + 6ax + + ( + ) ( + ) ax + ( + )( + ) a x + + = ( + ) ax 6( + ) a x ax + 6a x = { ( + ) ( + ) ( + ) + a + ( + )( + ) 6( + ) + 6 a } x = { ( ) 6 6 } = a a x = [ ( ) + ] ax + = Evidetly we equie ( + ) ( + ) ( + ) + a x + + ( + ) a (.3) 3 = + = + =. (.3) Sice these oots diffe by a itege, accodig to the geeal theoy we should oly use the lage of the two oots. Howeve, hee a miacle occus ad it tus out that both will wok (so that you do ot eed to be awae of the geeal theoy): eve fo the wog choice = we get that (.3) becomes

9 8 { [ ] } = a + ( ) 3+ 3 a x = [ ( 3) ] = = = a + ( 3) a =, =,3, K = a + a x [ ] = ( ) a + ( 3) a x + [ ] = + ( ) a + ( 3) a x ( 3) a = a, =,3, K. Thus we have (.33) ( 3) a = a = a, (3 3) a3 = a3 =, 3 (4 3) a4 = a4 = =, 4 M a 3 =. (.34) Thus the ifiite seies temiates, ad we get , (.35) y = a x = a x + a x + a x = a x+ a x + a x = a x+ a x + x which gives the geeal solutio of (.3). 9. Fid the geeal solutio of the followig liea but o-homogeeous diffeetial equatio by the method of vaiatio of paametes. Do ot use the (memoized) fomula/theoem (ivolvig a Woskia), athe geeate the elevat vesio of the fomula afesh by usig the D Alembet-like asatz that leads to that fomula. y y y e 3t 3 + = (.36)

10 9 poits Solutio The chaacteistic equatio of the homogeeous vesio of the costat coefficiet diffeetial equatio (.36) is 3 ( )( ) = + = (.37) so that the geeal solutio of the coespodig homogeeous equatio is y Ae Be t t = +, (.38) whee A ad B ae idepedet of t. But allowig the paametes A ad B to vay with t i (.38), we have also a asatz thee fo the solutio of the o-homogeeous equatio (.36): with such a asatz oe immediately has = (.39) t t t t y Ae Be e A e B But this asatz is iitially cosistet with A ad B idepedet of t if we choose hee that so that the (.39) becomes t ea t + e B=, (.4) t t = +. (.4) y Ae Be Diffeetiatig (.4) gives = (.4) t t t t y Ae 4Be e A e B Combiig these deivatives with the appopiate weights (dictated by the diffeetial equatio) we get the ledge t y = Ae + Be t t 3y = 3Ae 6Be t + y = Ae + Be + e A + e B t t t t 4. (.43) ad fom which it is clea that the diffeetial equatio demads that

11 t t 3t ea + e B = e. (.44) Combiig this with the cosistecy asatz (.4) we get which implies that A, (.45) e e B e t t e e t t = 3t t t e e 3t t 5t t 3t 4t e e e t e e e t A = = = e, B = = = e. (.46) t t 3t t t 3t e e e e e e e e e e t t t t t e t Solutios to (.46) iclude the pai A=, B= e, so that a solutio to (.36) is, accodig to (.38), ad the geeal solutio to (.36) is whee c ad c ae (tuly) costats ow. t t t e t t t 3t y = Ae + Be = e + e e = e, (.47) t t 3t y = ce + ce + e, (.48). Solve the iitial value poblem obtaied fom combiig the diffeetial equatio of poblem 9 with the iitial data y() =, y () =. I ode that eos do t t t 3t cascade, I will tell you that y = Ae + Be + e ( A ad B tuly costat) is the geeal solutio of the diffeetial equatio of poblem 9. (So ow if you just wite dow this solutio to 9 without vey covicig wok, you will get poits o poblem 9.) Thus, I am oly testig if you udestad the coect piciples eeded to costuct the solutio to the iitial value poblem give the geeal solutio to the associated diffeetial equatio. poits

12 Solutio Fom the ifomatio give we have y() = = A+ B+ 3 y () = = A+ B+, (.49) o, equivaletly, the followig augmeted matix fo the colum vecto ( AB, ), which, togethe with ow eductio, is Fom (.5) oe has that ad the solutio sought is 4 3. (.5) A= /, B=. (.5) y Ae Be e e e e t t 3t t t 3t = + + = +. (.5)