POWER SERIES METHODS CHAPTER 8 SECTION 8.1 INTRODUCTION AND REVIEW OF POWER SERIES

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1 CHAPTER 8 POWER SERIES METHODS SECTION 8. INTRODUCTION AND REVIEW OF POWER SERIES The power series method osists of substitutig a series y = ito a give differetial equatio i order to determie what the oeffiiets { } must be i order that the power series will satisfy the equatio. It might be poited out that, if we fid a reurree relatio i the form + = (), the we a determie the radius of overgee of the series solutio diretly from the reurree relatio lim lim. ( ) I Problems we give first a reurree relatio that a be used to fid the radius of overgee ad to alulate the sueedig oeffiiets,,, i terms of the arbitrary ostat. The we give the series itself.. ; it follows that ad lim ( ).! 4 4 y( ) e 6 4!!! 4!. 4 4 ; it follows that ad lim! 4. y ( ) e!!! 4!. ; it follows that y ( ) 8 6 8! ad lim. 47 Copyright 5 Pearso Eduatio, I.

2 474 INTRODUCTION AND REVIEW OF POWER SERIES e!!! 4! / 4. Whe we substitute y = ito the equatio y' + y =, we fid that ( ). Hee = whih we see by equatig ostat terms o the two sides of this equatio ad. It follows that Hee ad. k ( ) 5 odd ad k. k! y ( ) e!!! 5. Whe we substitute y = ito the equatio y y, we fid that ( ). Hee = = whih we see by equatig ostat terms ad -terms o the two sides of this equatio ad. It follows that k+ = k+ = ad k. k 6 () k k! Hee y ( ) !!! e ad. ( /) 6. ; it follows that ad lim. 4 y ( ) Copyright 5 Pearso Eduatio, I.

3 Setio ; it follows that ad lim. 4 y ( ) () ; it follows that lim. 4 5 y ( ) Separatio of variables gives y ( ). 9. ( ) ; it follows that ( ) ad y ( ) Separatio of variables gives y ( ). ( ) lim.. ( ) ; it follows that lim. 4 y ( ) / Separatio of variables gives y ( ) ( ). I Problems 4 the differetial equatios are seod-order, ad we fid that the two iitial oeffiiets ad are both arbitrary. I eah ase we fid the eve-degree oeffiiets i terms of ad the odd-degree oeffiiets i terms of. The solutio series i these problems are all reogizable power series that have ifiite radii of overgee.. ; ( )( ) it follows that k ad k ( k)! (k )! y ( ) oshsih! 4! 6!! 5! 7! Copyright 5 Pearso Eduatio, I.

4 476 INTRODUCTION AND REVIEW OF POWER SERIES 4. ; it follows that ( )( ). k ( k)! k ad k y ( ) k. (k )! ( ) ( ) ( ) ( ) ( ) ( ) ( )! 4! 6!! 5! 7! osh sih 9 ; it follows that ( )( ) k ( ) ( k)! k k ad k y ( ) k k ( ). (k )! ( ) ( ) ( ) ( ) ( ) ( ) ( )! 4! 6!! 5! 7! os si 4. Whe we substitute y = ito y'' + y = ad split off the terms of degrees ad, we get ( + ) + (6 + ) + [( )( ) ] =. Hee,, ad for. 6 ( )( ) It follows that y ( )! 4! 6!! 5! 7! ! 4! 6!! 5! 7! os ( )si. 5. Assumig a power series solutio of the form y =, we substitute it ito the differetial equatio y y ad fid that ( + ) = for all. This implies that = for all, whih meas that the oly power series solutio of our differetial equatio is the trivial solutio y ( ). Therefore the equatio has o o-trivial power series solutio. Copyright 5 Pearso Eduatio, I.

5 Setio Assumig a power series solutio of the form y =, we substitute it ito the differetial equatio y y ad fid that for all. This implies that,, 4,, ad hee that = for all, whih meas that the oly power series solutio of our differetial equatio is the trivial solutio y ( ). Therefore the equatio has o o-trivial power series solutio. 7. Assumig a power series solutio of the form y =, we substitute it ito the differetial equatio y y. We fid that = = ad that + = for, so it follows that = for all. Just as i Problems 5 ad 6, this meas that the equatio has o o-trivial power series solutio. 8. Whe we substitute ad assumed power series solutio y = ito y' = y, we fid that = = = ad that + = / for. Hee = for all, just as i Problems 5 7. I Problems 9 we first give the reurree relatio that results upo substitutio of a assumed power series solutio y = ito the give seod-order differetial equatio. The we give the resultig geeral solutio, ad fially apply the iitial oditios y() ad y() to determie the desired partiular solutio. 9. k k k k ( ) ( ) for, so k ad k. ( )( ) ( k)! (k )! y ( )! 4! 6!! 5! 7! y() ad y(), so y ( ) ! 5! 7! 5 7 ( ) ( ) ( ) ( ) si.! 5! 7!. k k for, so k ad k. ( )( ) ( k)! (k )! y ( )! 4! 6!! 5! 7! y() ad y(), so 4 6 ( ) ( ) ( ) y( ) osh.! 4! 6! Copyright 5 Pearso Eduatio, I.

6 478 INTRODUCTION AND REVIEW OF POWER SERIES. for ; with y() ad y(), we obtai ( ),, 4, 5, 6. Evidetly, 6! 4 4! 5! ( )! so y e!! 4!!! 4! ( ).. for ; with y() ad y(), we obtai ( ) ,, 4, 5. Apparetly, so! 4! 5 5!! 4 5 ( ). y e!! 4! 5!. = = ad the reursio relatio ( + ) + ( ) = for imply that = for. Thus ay assumed power series solutio y = must redue to the trivial solutio y ( ). 4. (a) The fat that y( ) = ( + ) satisfies the differetial equatio ( ) y y follows immediately from the fat that y ( ) ( ). (b) Whe we substitute y = ito the differetial equatio ( ) y y we get the reurree formula ( ). + = ( ) /( + ). Sie = beause of the iitial oditio y() =, the biomial series (Equatio () i the tet) follows. () The futio ( + ) ad the biomial series must agree o (, ) beause of the uiqueess of solutios of liear iitial value problems. 5. Substitutio of ito the differetial equatio y y y via shifts of summatio to ehibit leads routiely -terms throughout to the reurree formula ( )( ) ( ), Copyright 5 Pearso Eduatio, I.

7 Setio ad the give iitial oditios yield F ad F. But istead of proeedig immediately to alulate epliit values of further oeffiiets, let us first multiply the reurree relatio by!. This trik provides the relatio ( )! ( )!!, that is, the Fiboai-defiig relatio F F F where F!, so we see that F /! as desired. 6. This problem is pretty fully outlied i the tetbook. The oly hard part is squarig the power series: (b) The roots of the harateristi equatio r = are r =, r = = ( + i )/, ad r = = ( i )/. The the geeral solutio is y( ) Ae Be Ce. (*) Imposig the iitial oditios, we get the equatios A + B + C = A + B + C = A + B + C =. The solutio of this system is A = /, B = ( i )/, C = ( + i )/. Substitutio of these oeffiiets i (*) ad use of Euler's relatio e i = os + i si fially yields the desired result. SECTION 8. SERIES SOLUTIONS NEAR ORDINARY POINTS Istead of derivig i detail the reurree relatios ad solutio series for Problems through 5, we idiate where some of these problems ad aswers origially ame from. Eah of the differetial equatios i Problems is of the form Copyright 5 Pearso Eduatio, I.

8 48 SERIES SOLUTIONS NEAR ORDINARY POINTS (A + B)y'' + Cy' + Dy = with seleted values of the ostats A, B, C, D. Whe we substitute y =, shift idies where appropriate, ad ollet oeffiiets, we get Thus the reurree relatio is A( ) B( )( ) C D. A ( C A) D B ( )( ) for. It yields a solutio of the form y = y eve + y odd where y eve ad y odd deote series with terms of eve ad odd degrees, respetively. The evedegree series overges (by the ratio test) provided 4 that 4 A lim. B Hee its radius of overgee is at least B / A 4 5, as is that of the odd-degree series. (See Problem 6 for a eample i whih the radius of overgee is, surprisigly, greater tha B / A.) I Problems 5 we give first the reurree relatio ad the radius of overgee, the the resultig power series solutio. ; ;. ; 4 5 y( ) ( ) ( ). ; ; ; y( ) ( ) ( ). ; ; ( ) Copyright 5 Pearso Eduatio, I.

9 Setio ( ) ( ) ; ( )() 4! ( ) ( ) ()() 5 ()!! y( )! ( )!! ( ) ( ) 4 4. ; 6 4 ( ) ( ) ( ) ( ) 5 y( ) ( ) ( ) ( ) ( ) ; ; ( ) () () (5) () () y( ) ( ) ( )( 4) 6. ( )( ) The fator ( ) i the umerator yields 5 7 9, ad the fator ( 4) yields 6 8. Hee y eve ad y odd are both polyomials with radius of overgee. y( ) ( 6 ) ( ) 4 7. ( 4) ; ( )( ) The fator ( 4) yields 6 8, so y eve is a 4th-degree polyomial. We fid first that / ad 5 /, ad the for that Copyright 5 Pearso Eduatio, I.

10 48 SERIES SOLUTIONS NEAR ORDINARY POINTS (5) (7) 5 ( )() ()() (6)(7) (5)!! (5)!! ( ) 9 ( ) ()() 7 6 ()! [( 5)!!] ( ) y( ) 9 7 ( )! ( 4)( 4) 8. ; ( )( ) We fid first that 5 / 4 ad 5 7 /, ad the for that 5 (5)() (7)() 9 ( )( ) ( )( ) (6)(7) (5)!!()() 9 7 5! 7 (5)!!()!! 4 ()( ) ()! (5)!!()!! ()! 5 7 (5)!!()!! y( ) ( )! 9. ( )( 4) ; ( )( ) ()() ()( ) 4 ( )() ()( ) ()() ()() ( )() 45 ( )( ) ( )() ()() y( ) ( )() ( )() ( 4). ; ( )( ) The fator ( 4) yields 6 8, so y eve is a 4th-degree polyomial. We fid first that / 6 ad 5 / 6, ad the for that Copyright 5 Pearso Eduatio, I.

11 Setio (5) () ()( ) ()() (7)(6) 5 (5)!!( ) ()( ) (7)(6) 6 6 ()( ) (7)(6)5! ()! 5! ( 5)!!( ) ( 5)!!( ) 4 5 ( 5)!!( ) y( ) ( )! ( 5). ; 5( )( ) The fator ( 5) yields 7 9, so y odd is a 5th-degree polyomial. We fid first that, 4 / ad 6 / 75, ad the for 4 that (7) (5) () 5( )() 5()() 5(8)(7) 6 5 ( )() (8)(7) 75 ( 7)!! 75 5 ( )( ) (8)(7) 6! 5 ( )! 5 6! ( 7)!! ( 7)!! ( 7)!! y ( ) ( )! 5. ; Whe we substitute y = ito the give differetial equatio, we fid first that, so the reurree relatio yields 5 8 also. y( ) 5 ( )!. ; Whe we substitute y = ito the give differetial equatio, we fid first that, so the reurree relatio yields 5 8 also. ( ) ( ) y( )! 4 () Copyright 5 Pearso Eduatio, I.

12 484 SERIES SOLUTIONS NEAR ORDINARY POINTS 4. ; ( )( ) Whe we substitute y = ito the give differetial equatio, we fid first that, so the reurree relatio yields 5 8 also. The ( ), ( )() ()(4)! ()(4) 5 ( ). ()( ) ()() 4! ()() 4 ( ) ( ) y( )! 5 ( )! 4 () 5. 4 ; ( )( 4) Whe we substitute y = ito the give differetial equatio, we fid first that, so the reurree relatio yields 6 ad 7 also. The ( ) 4, (4 )(4) (44)(45) 4 4! (4)(45) 5 ( ). (4)(4 ) (4)(44) 54 4! (4)(4) ( ) ( ) 4! 7 (4 ) 4! 59 (4) y( ) 6. The reurree relatio is for. The fator ( ) i the umerator yields 5 7. Whe we substitute y = ito the give differetial equatio, we fid first that, ad the the reurree relatio gives Hee 5 ( ) y ( ) ta. With = y() = ad = y'() = we obtai the partiular solutio y() =. Copyright 5 Pearso Eduatio, I.

13 Setio The reurree relatio ( ) ( )( ) yields = = y() = ad 4 = 6 = =. Beause = y'() =, it follows also that = = 5 = =. Thus the desired partiular solutio is y() = The substitutio t = yields y'' + ty' + y =, where primes ow deote differetiatio with respet to t. Whe we substitute y = t we get the reurree relatio. for, so the solutio series has radius of overgee. The iitial oditios give = ad =, so odd = ad it follows that 4 6 t t t y, ( ) ( ) ( ) ( ) ( ) y ( ). 4 46! 9. The substitutio t = yields ( t )y'' 6ty' 4y =, where primes ow deote differetiatio with respet to t. Whe we substitute y = t we get the reurree relatio 4. for, so the solutio series has radius of overgee, ad therefore overges if < t <. The iitial oditios give = ad =, so eve = ad Thus y () t ()( ), ad the -series overges if < <.. The substitutio t = yields (t + )y'' 4ty' + 6y =, where primes ow deote differetiatio with respet to t. Whe we substitute y = t we get the reurree relatio ( )( ) ( )( ) Copyright 5 Pearso Eduatio, I.

14 486 SERIES SOLUTIONS NEAR ORDINARY POINTS for. The iitial oditios give = ad =. It follows that odd =, = 6 ad 4 = 6 = =, so the solutio redues to y = 6t = 6( ).. The substitutio t = yields (4t + )y'' = 8y, where primes ow deote differetiatio with respet to t. Whe we substitute y = t we get the reurree relatio 4( ) ( ) for. The iitial oditios give = ad =. It follows that odd =, = ad 4 = 6 = =, so the solutio redues to y = t = 4( ).. The substitutio t = + yields (t 9)y'' + ty' y =, with primes ow deotig differetiatio with respet to t. Whe we substitute y = t we get the reurree relatio ( )( ) 9( )( ) for. The iitial oditios give = ad =. It follows that eve = ad = 5 = =, so y = t = + 6. I Problems 6 we first derive the reurree relatio, ad the alulate the solutio series y ( ) with ad as well as the solutio series y ( ) with ad.. Substitutio of y = yields so ( )( ), ( )( ), for y( ) ; y( ) Substitutio of y = yields ( ) ( )( ), Copyright 5 Pearso Eduatio, I.

15 Setio so ( ) ( )( ), for y( ) ; y( ) Substitutio of y = yields so 6 ( ) ( )( ), ( ) ( )( ), for y( ) ; y( ) Substitutio of y = yields so 6 ( ) ( )( ) ( )( ), 4 ( )( ) ( )( ) 4 4 5, for y( ) ; y( ) Substitutio of y = yields so ( 6 ) ( ) ( )( ), ( ),, for. ( )( ) With y() ad y(), we obtai y ( ) Copyright 5 Pearso Eduatio, I.

16 488 SERIES SOLUTIONS NEAR ORDINARY POINTS Fially, =.5 gives y(.5) y(.5) Whe we substitute y = ad e ( ) /! ad the ollet oeffiiets of the terms ivolvig,,, ad, we fid that,, 4, 5. 6 With the hoies, ad, we obtai the two series solutios y( ) ad y( ) Whe we substitute y = ad os ( ) /( )! ad the ollet 6 oeffiiets of the terms ivolvig,,,,, we obtai the equatios, 6,,, , 95 46, Give ad, we a solve easily for,,, 8 i tur. With the hoies, ad, we obtai the two series solutios y( ) ad y( ) Whe we substitute y = ad si = () + /( + )!, ad the ollet 5 oeffiiets of the terms ivolvig,,,,, we obtai the equatios, 6, 4, , Give ad, we a solve easily for,,, 6 i tur. With the hoies, ad, we obtai the two series solutios Copyright 5 Pearso Eduatio, I.

17 Setio y( ) ad y( ) Substitutio of y = i Hermite's equatio leads i the usual way to the reurree formula ( ). ( )( ) Startig with, this formula yields ( ) ( )( 4), 4, 6,.! 4! 6! Startig with, it yields ( ) ( )( ) ( )( )( 5), 5, 7,.! 5! 7! This gives the desired eve-term ad odd-term series y ad y. If is a iteger, the obviously oe series or the other has oly fiitely may o-zero terms. For istae, with 4 we get y( ) , 4 ad with 5 we get y( ) The figure below shows the iterlaed zeros of the 4th ad 5th Hermite polyomials. y - H 4 - H 5 Copyright 5 Pearso Eduatio, I.

18 49 SERIES SOLUTIONS NEAR ORDINARY POINTS 4. Substitutio of y = i the Airy equatio leads upo shift of ide ad olletio of terms to ( )( ). The idetity priiple the gives ad the reurree formula. ( )( ) Beause of the -step i idies, it follows that 5 8. Startig with, we alulate , 6, 9,.!! 56 6! 6! 89 9! Startig with, we alulate , 7,,. 4 4! 4! 67 7! 7! 9! Evidetly we are buildig up the oeffiiets 4 (k ) 5 (k ) ad ( k)! (k )! k k that appear i the desired series for y( ) ad y( ). Fially, the Mathematia ommads A[] = /6; A[k_] := A[k - ]/( k*( k - )); B[] = /; B[k_] := B[k - ]/( k*( k + )); = 4; y = + Sum[A[k]*^( k), {k,, }]; y = + Sum[B[k]*^( k + ), {k,, }]; ya = y/(^(/)*gamma[/]) - y/(^(/)*gamma[/]); yb = y/(^(/6)*gamma[/]) + y/(^(-/6)*gamma[/]); Plot[{yA, yb}, {, -.5, }, PlotRage -> {-.75,.5}] Copyright 5 Pearso Eduatio, I.

19 Setio Ai Bi 5.5 produe the figure above. But with 5 (istead of 4 ) terms we get a figure that is visually idistiguishable from Figure 8.. i the tetbook. 5. (a) If ( )!! y a z! ( )!! where a, the the radius of overgee of the series i! z is a ()!!/! ( ) a ( )!!/ ( )! lim lim lim 4. Thus the series i z overges if 4 z 4, so the series y ( ) overges if, ad thus has radius of overgee equal to. (b) If where b! y b z ( )!!!, the the radius of overgee of the series i z is ()!! b!/ ()!! () lim lim lim 4. b ( )!/ ()!! Hee it follows as i part (a) that the series y ( ) has radius of overgee equal to. Copyright 5 Pearso Eduatio, I.

20 49 REGULAR SINGULAR POINTS SECTION 8. REGULAR SINGULAR POINTS. Upo divisio of the give differetial equatio by we see that P() = ad Q() = (si )/. Beause both are aalyti at = i partiular, (si ) / as beause 4 6 si ( ) ( ) ()! ()!! 5! 7! it follows that = is a ordiary poit.. Divisio of the differetial equatio by yields Beause the futio e y y y. e!!!! 4! is aalyti at the origi, we see that = is a ordiary poit.. Whe we rewrite the give equatio i the stadard form of Equatio () i this setio, we see that p() = (os )/ ad q() =. Beause (os ) / as it follows that p() is ot aalyti at =, so = is a irregular sigular poit. 4. Whe we rewrite the give equatio i the stadard form of Equatio (), we have p() = / ad q() = ( )/. Sie q() is ot aalyti at the origi, = is a irregular sigular poit. 5. I the stadard form of Equatio () we have p() = /( + ) ad q() = /( + ). Both are aalyti =, so = is a regular sigular poit. The idiial equatio is r(r ) + r = r + r = r(r + ) =, so the epoets are r = ad r =. 6. I the stadard form of Equatio () we have p() = /( ) ad q() = /( ), so = is a regular sigular poit with p = ad q =. The idiial equatio is r + r =, so the epoets are r =,. Copyright 5 Pearso Eduatio, I.

21 Setio I the stadard form of Equatio () we have p() = (6 si )/ ad q() = 6, so = is a regular sigular poit with p = q = 6. The idiial equatio is r + 5r + 6 =, so the epoets are r = ad r =. 8. I the stadard form of Equatio () we have p() = /(6 + ) ad q() = 9( )/(6 + ), so = is a regular sigular poit with p = 7/ ad q = /. The idiial equatio simplifies to r + 5r =, so the epoets are r =, /. 9. The oly sigular poit of the differetial equatio y y y is =. Upo substitutig t =, = t + we get the trasformed equatio t ( t) y y y t t, where primes ow deote differetiatio with respet to t. I the stadard form of Equatio () we have p( t) ( t) ad qt () t( t). Both these futios are aalyti, so it follows that = is a regular sigular poit of the origial equatio.. The oly sigular poit of the differetial equatio y y ( ) y is =. Upo substitutig t =, = t + we get the trasformed equatio y y y, where primes ow deote differetiatio with respet to t. I the t t stadard form of Equatio () we have pt ( ) ad qt ( ). Both these futios are aalyti, so it follows that = is a regular sigular poit of the origial equatio.. The oly sigular poits of the differetial equatio y y y = + ad =. are = +: Upo substitutig t =, = t + we get the trasformed equatio ( t ) y y y, where primes ow deote differetiatio with respet to tt ( ) tt ( ) ( t ) t t. I the stadard form of Equatio () we have pt () ad qt (). t t Both these futios are aalyti at t =, so it follows that = + is a regular sigular poit of the origial equatio. = : Upo substitutig t =, = t we get the trasformed equatio ( t ) y y y, where primes ow deote differetiatio with respet to tt ( ) tt ( ) ( t ) t t. I the stadard form of Equatio () we have pt () ad qt (). t t Copyright 5 Pearso Eduatio, I.

22 494 REGULAR SINGULAR POINTS Both these futios are aalyti at t =, so it follows that = is a regular sigular poit of the origial equatio.. The oly sigular poit of the differetial equatio y y y is ( ) =. Upo substitutig t =, = t + we get the trasformed equatio ( t ) y y y, where primes ow deote differetiatio with respet to t. I t t ( t ) the stadard form of Equatio () we have pt ( ) ad qt (). Beause q t is ot aalyti at t =, it follows that = is a irregular sigular poit of the origial equatio.. The oly sigular poits of the differetial equatio = + ad =. y y y are = +: Upo substitutig t =, = t + we get the trasformed equatio y, t 4 y t y where primes ow deote differetiatio with respet to t. I the t stadard form of Equatio () we have pt () ad qt ( ) t. Both these t 4 futios are aalyti at t =, so it follows that = + is a regular sigular poit of the origial equatio. = : Upo substitutig t =, = t we get the trasformed equatio y y, t t 4 y where primes ow deote differetiatio with respet to t. I the t stadard form of Equatio () we have pt () ad qt (). Both these t 4 futios are aalyti at t =, so it follows that = is a regular sigular poit of the origial equatio. 4. The oly sigular poits of the differetial equatio are = + ad =. 9 4 y y y ( 9) ( 9) = +: Upo substitutig t =, = t + we get the trasformed equatio t 6t t 6t 8 y y y, where primes ow deote differetiatio with t ( t 6) t ( t 6) Copyright 5 Pearso Eduatio, I.

23 Setio t 6t respet to t. Beause pt () is ot aalyti at t =, it follows that = tt ( 6) is a irregular sigular poit of the origial equatio. = : Upo substitutig t =, = t we get the trasformed equatio t 6t t 6t 8 y y y, where primes ow deote differetiatio with t ( t 6) t ( t 6) t 6t respet to t. Beause pt () is ot aalyti at t =, it follows that = tt ( 6) is a irregular sigular poit of the origial equatio The oly sigular poit of the differetial equatio y y y is ( ) ( ) =. Upo substitutig t =, = t + we get the trasformed equatio t4 t4 y y y, where primes ow deote differetiatio with respet to t. I t t the stadard form of Equatio () we have pt ( ) ( t 4) ad qt ( ) t 4. Both these futios are aalyti, so it follows that = is a regular sigular poit of the origial equatio. 6. The oly sigular poits of the differetial equatio y y y ( ) ( ) are = ad =. = : I the stadard form of Equatio () we have p ( ) ad ( ) q ( ). Sie p is ot aalyti at =, it follows that = is a irregular sigular poit. = : Upo substitutig t =, = t + we get the trasformed equatio t 5 t y y y, where primes ow deote differetiatio with respet ( t) ( t) t(t 5) t to t. Both pt () ad qt () are aalyti at t =, so it follows ( t ) ( t ) that = is a regular sigular poit of the origial equatio. Eah of the differetial equatios i Problems 7 is of the form Ay'' + By' + Cy = Copyright 5 Pearso Eduatio, I.

24 496 REGULAR SINGULAR POINTS with idiial equatio Ar + (B A)r =. Substitutio of y = +r ito the differetial equatio yields the reurree relatio C Ar B A r ( ) ( )( ) for. I these problems the epoets r = ad r = (A B)/A do ot differ by a iteger, so this reurree relatio yields two liearly idepedet Frobeius series solutios whe we apply it separately with r = r ad with r = r. 7. With epoet r : 4 ( ) ( ) os 4 7 ( )! y With epoet r : 4 ( ) / y( ) si 6 54 ( )! 8. With epoet r : 4 y( ) 6 68!( )!! With epoet r : 4 / y( ) 6 9 5!()!! 9. With epoet r : 4 y( ) 8 6!( )!! With epoet r : 4 / / y( ) !( )!! Copyright 5 Pearso Eduatio, I.

25 Setio With epoet r : 4 ( ) y( ) 5 6! 5 () With epoet r : 4 / / ( ) y( ) 4 546! 4 ( ) The differetial equatios i Problems 4 are all of the form with idial equatio Substitutio of y = +r A y'' + By' +(C + D )y = () (r) = Ar + (B A)r + C =. () ito the differetial equatio yields r r r () () r ( r) ( r ) D. I eah of Problems 4 the epoets r ad r do ot differ by a iteger. Hee whe we substitute either r = r or r = r ito Equatio (*) above, we fid that is arbitrary beause () r is the zero, that = beause its oeffiiet ( r ) is the ozero ad that D D (4) ( r) A( r) ( B A)( r) C for. Thus this reurree formula yields two liearly idepedet Frobeius series solutios whe we apply it separately with r = r ad with r = r.. With epoet r :, ( ) ! 7 (4) y ( ) With epoet r :, ( ) 4 6 / y( ) 7! 5 (4) Copyright 5 Pearso Eduatio, I.

26 498 REGULAR SINGULAR POINTS. With epoet r :, ( 5) y ( ) 4 6 / / ! 9 (45) With epoet r :, ( 5) ( ) y( ) ! 7 (45). With epoet r :, (6 7) 4 6 / y ( ) ! 9 (7) With epoet r :, (6 7) 4 6 / y / ( ) 68 8! 57 (7) 4. With epoet r :, ( ) 4 6 / ( ) y( ) ! 7 (6) With epoet r :, () 4 6 ( ) y( ) ! 5 (6) 5. With epoet r : 4 / ( ) y( ) ! With epoet r : 4 ( ) y ( ) 5 5 ( )!! e / Copyright 5 Pearso Eduatio, I.

27 Setio With epoet r :, / / y( ) e ! With epoet r :, y( ) (4 ) The differetial equatios i Problems 7 9 (after multipliatio by ) ad the oe i Problem are of the same form () above as those i Problems 4. However, ow the epoets r ad r = r do differ by a iteger. Hee whe we substitute the smaller epoet r = r ito Equatio (), we fid that ad are both arbitrary, ad that is give (for ) by the reurree relatio i (4). Thus the smaller epoet r yields the geeral solutio y( ) y ( ) y ( ) i terms of the two liearly idepedet Frobeius series solutios y ( ) ad y ( ). 7. Epoets r ad 9 r ; with r : ( ) y ( ) y ( ) os si os The figure below shows the graphs of the idepedet solutios y( ) ad si y( ). y y y Π 4 Π Copyright 5 Pearso Eduatio, I.

28 5 REGULAR SINGULAR POINTS 8. Epoets r ad 4 r ; with r : ( ) y ( ) y ( ) osh sih The figure below shows the graphs of the two idepedet solutios osh sih y( ) ad y( ). y 8 6 y 4 y 9. Epoets r ad r ; with r : 4 ( ) y ( ) y ( ) os si The figure at the top of the et page shows the graphs of the idepedet solutios os / si / y( ) ad y( ). Copyright 5 Pearso Eduatio, I.

29 Setio 8. 5 y.5 y y Π 4 Π.5. The give differetial equatio r r r r y y y 4 has idiial equatio ( ), so its epoets are r ad r. Takig r =, substitutio of the power series y gives (4 8 ) (4 5 ) (4 4 ) We see that ad (4 5 ) (4 48 ) (4 6 ) for 4. ( ) Hee the odd subsripts all vaish, ad we obtai y ( ) y( ) os si. The figure below shows the graphs of the idepedet solutios ad y( ) si. y y ( ) os y Π Π y Copyright 5 Pearso Eduatio, I.

30 5 REGULAR SINGULAR POINTS. The give differetial equatio has idiial equatio 4y 4 y ( 4 ) y 4r 8r (r)(r), so its epoets are r / ad r /. With r = /, the reurree relatio / ( ) yields the geeral solutio / / y ( ) y( ) osh sih. The figure below shows the graphs of the idepedet solutios y ( ) osh ad y ( ) sih. y y y. The two idiial epoets are r = ad r = /. With r = : Substitutio of y i the differetial equatio yields (5 ) 4 ( 7 ) ( 44 ) ( 65 ) Hee we see that /5 ad 4 5. Thus the series termiates ad we obtai the polyomial solutio ( ) y. 5 5 With r = /: We substitute y ad obtai the Frobeius solutio / y( ) Remark: With appropriate iterpretatio of its result, the Mathematia DSolve futio yields the two losed form solutios ( ) y ad Copyright 5 Pearso Eduatio, I.

31 Setio 8. 5 y e ( ) / / 4 erf. Iquirig mids aturally wat to kow! The Mathematia Series ommad reveals the aswer that y ( ) y ( ).. Epoets r / ad r. With eah epoet we fid that is arbitrary ad we a solve reursively for i terms of y( ) y( ) Epoets r ad r /. With eah epoet we fid that = ad we a solve reursively for i terms of y ( ) y( ) Substitutio of r y ito the differetial equatio yields a result of the form r r r r ( ) ( ), so we see immediately that implies that r =. The substitutio of the power series y yields ( ) (4 ) (9 ) (6 4 ) 4 4 Evidetly, so if = it follows that! for. But the series! has zero radius of overgee, ad hee overges oly if =. We therefore olude that the give differetial equatio has o otrivial Frobeius series solutio. 6. (a) Substitutio of yields a result of the form r y ito the differetial equatio Ar r r r ( ) ( ), y Ay By Copyright 5 Pearso Eduatio, I.

32 54 REGULAR SINGULAR POINTS so we see immediately that A ad imply that r =. (b) Substitutio of yields a result of the form r y ito the differetial equatio r r r ( Ar B) ( ) ( ), y Ay By so we see immediately that implies that r = B/A. () Substitutio of yields a result of the form r y ito the differetial equatio B r r r ( ) ( ), y Ay By whih is impossible beause both ad B. It follows that o Frobeius series a satisfy this equatio. 7. Substitutio of r y yields a result of the form ito the differetial equatio ( r) ( ) ( ), r r r y y y so it follows that r =. But the substitutio of equatio yields y , ito the differetial so it follows that 4. Hee y( ). 8. Epoets r / ad r /; with r /: ( ) y ( ) ! 4! 6!! 5! 7! y ( ) os si 9. Epoets r ad r ; with r :, ( ) y ( ) Copyright 5 Pearso Eduatio, I.

33 Setio !! 4!! 6!4! 8 4!5! If = /, the ( )!( ) y ( ) J( ). Now, osider the smaller epoet r =. A Frobeius series with r = is of the form y with. However, substitutio of this series ito Bessel's equatio of order gives ( ) ( 8 ) ( 5 ), so it follows that =, after all. Thus Bessel's equatio of order does ot have a Frobeius series solutio with leadig term. However, there is a little more here that meets the eye. We see further that is arbitrary ad that ad / ( ) for. It follows that our assumed Frobeius series atually redues to y y ( ) But this is the same as our series solutio obtaied above usig the larger epoet r = + (allig the arbitrary ostat rather tha ). SECTION 8.4 METHOD OF FROBENIUS THE EXCEPTIONAL CASES Eah of the differetial equatios i Problems 6 is of (or a be writte i) the form y'' + (A + B)y' + Cy =. The origi is a regular sigular poit with epoets r = ad r = A, so if A is a iteger the we have a eeptioal ase of the method of Frobeius. Whe we substitute y = +r i the differetial equatio we fid that the oeffiiet of +r is [( + r) + (A )( + r)] + [B( + r) + C B] =. (*) Copyright 5 Pearso Eduatio, I.

34 56 METHOD OF FROBENIUS THE EXCEPTIONAL CASES Case : I eah of Problems 4 we have A ad B = C, so the larger epoet r = ad the smaller epoet r = A = N differ by a positive iteger. Whe we substitute the smaller epoet r = N i Equatio (*) above, it simplifies to ( N) + B( N) =. () This equatio determies,,, N i terms of, thereby yieldig the solutio y () = N ( N ), () provided it is possible to hoose N. But whe = N, Equatio () redues to N + N =, so N may be hose arbitrarily. With CN we get the termiatig Frobeius series solutio i (). For > N, Equatio () yields the reurree formula = B /, whih if C gives a seod (o-termiatig) Frobeius series solutio of the form N y () = N + N+ + N+ +. () Case : If A the the larger epoet r = A = N ad the smaller epoet r = agai differ by a positive iteger. I Problems 5 ad 6 we have this ase with B =. Whe we substitute the smaller epoet r = i Equatio (*), it simplifies to ( N) ( C ) =. (4) This equatio determies,,, N i terms of. Whe = N it redues to N (N C ) N =. (5) If either N C = or N = (the latter happes i Problem 5) the N a be hose arbitrarily, ad fially N+, N+, are determied i terms of N. Thus we get two Frobeius series solutios y = N N, y = N N + N+ N+ +. (termiatig) (ot termiatig) O the other had, if (as i Problem 6) either N C = or N =, the N aot be hose so as to satisfy Equatio (5), ad hee there is o Frobeius series solutio orrespodig to the smaller epoet r =. We therefore fid the sigle Frobeius series solutio by substitutig the larger epoet r = N i Equatio (*) ad usig the resultig reurree relatio to determie,,, i terms of. Copyright 5 Pearso Eduatio, I.

35 Setio Problems 4 orrespod to ase above. We give first the idiial roots ad the ritial ide N, the the reurree relatio that defies i terms of, for both the N-term solutio y ( ) i () ad the o-termiatig series solutio y ( ) i (). ; y( ) 4 45 ( )!. r, r, N, ; y ( ).. 4 r, r 4, N 4, ; y( ) 6 y( ) ( 4)! r, r 4, N 4, ; y( ) ( ) y( ) ( 4)! 4. r, r 5, N 5, y( ) ( ) y( ) ( 5)!5 Problems 5 ad 6 orrespod to ase desribed above. 5. ( 4) r 5, r, N 5, ( 5) for 5 y ( ) With = 5 the reurree relatio is 5 4. Beause 4 we a hoose 5 arbitrarily ad proeed: ( ) y( ) ( 5)! Copyright 5 Pearso Eduatio, I.

36 58 METHOD OF FROBENIUS THE EXCEPTIONAL CASES 6. Here A =, B =, C = /, r = N = 4, ad r =, so Equatio (4) above is ( 4) ( ) =. Startig with =, this equatio gives = /6, = /48, = /96. With = 4 it redues to 7 4, 96 so 4 aot be hose. We therefore start over by substitutig r = 4 i Equatio (*) above ad get the reurree relatio 5 ( 4) for the oeffiiets i 4 y. This yields the sigle Frobeius series solutio y ( ) ( 5)!!. 5!( 4)! 7. The idiial epoets are r,. Substitutio of differetial equatio leads to the reurree relatio y i the ( ) ( ) that redues to whe = so havig foud ad a be hose arbitrarily. With = ad = we get the termiatig Frobeius series y () = ( ). Startig afresh with = /!=/, the reurree relatio / for yields the seod Frobeius series solutio 4 5 ( ) ( ).! 4! 5! ( )! y 8. The epoets are r =, 4. Whe we substitute y (orrespodig to r = ) i the differetial equatio we get the reurree relatio. ( 4) ( ) = Copyright 5 Pearso Eduatio, I.

37 Setio for. Startig with =, we ompute =, =, ad =. Beause of the latter, we a selet 4 = ad get the termiatig Frobeius series solutio y () = + +. But we also a hoose 4 =. The our reurree formula above yields 5 =, 6 =, 7 = 4,. Hee the seod Frobeius series solutio is y () = 4 ( ) = 4 /( ), with the losed form omig from the derivative of the geometri series /( ) =. I Problems 5, we give first the Frobeius series solutio y ( ) orrespodig to the larger idiial epoet r of the give differetial equatio. The, writig the equatio i the form y P( ) y Q( ) y, we apply the redutio of order formula y( ) ep Pd ( ) y ( ) d to derive a seod idepedet solutio y ( ). 9. r r y P ( ) / ; ep Pd ( ) / y y d y d y d y yl Copyright 5 Pearso Eduatio, I.

38 5 METHOD OF FROBENIUS THE EXCEPTIONAL CASES. r r y P ( ) / ; ep Pd ( ) y y d y d y d y yl r r y P ( ) / ; ep Pd ( ) e y y e d y e 47 d y 49 d y d 6 8 y yl 4 8. r, r 4 y 68 Copyright 5 Pearso Eduatio, I.

39 P ( ) ; ep Pd ( ) e 4 4 y y e d y e d y d y d 7 5 Setio y 47 y [o logarithmi term]. r, r y 4 4 P ( ) / ; ep Pd ( ) e y y e d 4 y e 48 d y 48 d 4 4 y d y 4 yl 4. r, r 4 5 y P ( ) / ; ep Pd ( ) e Copyright 5 Pearso Eduatio, I.

40 5 METHOD OF FROBENIUS THE EXCEPTIONAL CASES y y e d y e d y d y d y y l Thus the seod solutio y () otais o logarithmi term. 5. r r ( ) J P ( ) / ; ep Pd ( ) / y( ) J( ) d J( ) d J( ) d J( ) l J ( )l y( ) J( )l Copyright 5 Pearso Eduatio, I.

41 Setio The idiial epoets are r. We start with the larger epoet r ad / substitute y a ito Bessel's equatio of order. We fid that, ad the the reurree relatio a a, ( ) implies that all odd subsripts vaish. Startig with a, this reurree relatio yields i the usual maer the first solutio / ( ).! 57 () 4 6 ( ) / y Now we start afresh with the smaller epoet r ad substitute / y b ito Bessel's equatio of order. This time, we fid that ( ) b b for. We a satisfy the ritial ase bb by hoosig b =, whih the implies that all odd oeffiiets vaish. The the reurree relatio b b ( ) yields routiely the seod solutio 4 6 / y( ) ( ) 4 ( ) 4 6 ( ) / ( ).! ( ) () 7. The give first solutio 4 5 ( ) y e 6 4 a be derived by startig with the sigle epoet r =, substitutig y ito the differetial equatio, ad alulatig suessive oeffiiets reursively as usual. We a verify the alleged seod solutio by applyig the method of redutio of order as i Problems 9 4: Copyright 5 Pearso Eduatio, I.

42 54 BESSEL S EQUATION P ( ) ; ep Pd ( ) e y y e e d y e d 4 5 y d y l l y y l y ( ) e l H! 8. Whe we substitute y() = Cy l + b i the differetial equatio y'' = we fid that b = b = C ad that () C ( ) b b!( )! for. To solve this reurree relatio we take C = ad substitute b = / ( )!!. The result is. ( ) Startig with = b =, it follows readily by idutio o that = (H + H + ). SECTION 8.5 BESSEL'S EQUATION Of ourse Bessel's equatio is the most importat speial ordiary differetial equatio i mathematis, ad every studet should be eposed at least to Bessel futios of the first kid. Though Bessel futios of itegral order a be treated without the gamma futio, the Copyright 5 Pearso Eduatio, I.

43 Setio subsetio o the gamma futio is also eeded for Chapter 7 o Laplae trasforms. The fial subsetios o Bessel futio idetities ad the parametri Bessel equatio will ot be eeded util Setio.4, ad therefore may be osidered optioal at this poit i the ourse.. m m m m ( ) ( ) m J ( ) D m m m ( m!) m ( m!) m m m m ( ) ( ) m m ( m)!( m!) ( m)!( m!) m m ( ) m m m ( )!(!) m m m J ( ). (a) ()() ()!! (b) J ( ) m m m m ( ) ( ) / m m m mm! ( m ) mm! (m)!! m m ( ) (4 m)( (m)) m m m ( ) si (m)! m J /( ) os similarly (See the figure below for the graphs.) y.5 J J.5 Copyright 5 Pearso Eduatio, I.

44 56 BESSEL S EQUATION. (a) m m m4 m4 m m m (m) m (b) J / ( ) m m/ / m m m ( ) ( /) ( ) m! ( m / ) ( / ) m! 8 (m) m m 4. With p = / i Equatio (6) i the tet we have J/( ) J/( ) J /( ) si os si os si os The figure below shows the graphs of J/( ) ad J /( ). y.5 J J.5 5. Startig with p = i Equatio (6) we get 6 64 J4( ) J( ) J( ) J( ) J( ) J( ) 4 6 J ( ) J( ) J( ) 4 8(6 ) J ( ) J ( ) 8. Whe we arry out the differetiatios idiated i Equatios () ad () i the tet, we get p p J ( ) p J ( ) p J ( ), p p p p J ( ) J ( ) J ( ). p p p p p p Copyright 5 Pearso Eduatio, I.

45 Setio Whe we solve these two equatios for J ( ) we get Equatios (4) ad (5) i the tet.. Whe we add equatios (4) ad (5) we get Jp( ) J p( ) J p( ), so J p( ) Jp ( ) Jp ( ). Replaig p with p ad with p + i the first equatio, we get Jp ( ) J p( ) J p( ) ad Jp ( ) J p( ) J p( ). Whe we use these equatios to substitute for J ( ) ad J ( ) i the equatio for J ( ) p above, we fid that p J p( ) J p( ) J p( ) J p( ). 4. (p + m + ) = (p + m)(p + m ) (p + )(p + )(p + ), so p p J p ( ) m ( ) m m! ( pm) m p p m ( /) ( ) p m p p p m ( ) m!( )( ) ( ) m.. Substitutio of the power series of Problem yields y ( ) 5/ 5/ 5 A B A B / / ( ) ( ) ( ) ( ) ( C) ( D) ( C) ( D) where A = /( 5/ (7/)), B = /( 5/ (/),) C = /( / (/)), ad D = (/ / )(/). Hee ( A) ( B) B y( ). / (/) (/) ( C) ( D) D 5/ ( /) (4/) (/ The graph of y( ) show at the top of the et page orroborates this value. Copyright 5 Pearso Eduatio, I.

46 58 BESSEL S EQUATION y I Problems we use a ospiuous dot to idiate our hoie of u ad dv i the itegratio by parts formula udv uv vdu. We use repeatedly the fats (from Eample ) that J( ) d J( ) C ad J( ) d J( ) C J( ) d J( ) d J( ) J( ) d J( ) J( ) J( ) d J ( ) d J ( ) d J( ) J( ) J( ) d C J ( ) J ( ) d J( ) J( ) J( ) d J( ) J( ) 4 J( ) C ( 4 J ) ( ) J( ) C J ( ) d J ( ) d 4 J ( ) J ( ) d 4 J ( ) J ( ) J ( ) d 4 4 J ( ) J ( ) 9 J ( ) J ( ) d 4 J( ) J( ) 9 J( ) 9 J( ) J( ) d ( 9 ) J ( ) ( 9 ) J ( ) 9 J ( ) dc 4 J ( ) d J ( ) d J ( ) J ( ) dc 6. Copyright 5 Pearso Eduatio, I.

47 Setio J ( ) d J ( ) d J ( ) J ( ) d J ( ) J ( ) C J ( ) d J ( ) d J( ) J( ) d J ( ) J ( ) J ( ) d J( ) J( ) J( ) J( ) d ( J ) ( ) J( ) J( d ) C J ( ) d J ( ) d 4 4 J ( ) 4 J ( ) d 4 J ( ) 4 J ( ) J ( ) d 4 4 J( ) 4 J( ) 8 J( ) J( ) d 4 ( 8 ) J ( ) (4 6 ) J ( ) C. With p =, Eq. () i the tet gives J ( ) d J ( ) d J( ) d J( ) C. Hee J ( ) J ( ) J ( ) J ( ) d. But Eq. (6) with p = gives J J J ( ) ( ) ( ), so J ( ) d J ( ) J ( ) d J ( ) d. Fially, we a solve this last equatio for J ( ) d J ( ) J ( ) d C.. With p =, Eq. () i the tet gives J ( ) d J ( ) d J ( ) J ( ) d J( ) d J( ) C. Hee J( ) J( ) C (by Eample ) Copyright 5 Pearso Eduatio, I.

48 5 BESSEL S EQUATION J( ) J( ) J( ) C 4 J( ) J( ) C. (By Eq. (6) with p =). Let us defie ad ote first that g ( ) os( si ) d g() os() d J (). Differetiatio uder the itegral sig yields g( ) si( si )si d. Whe we itegrate by parts with we get u = si( si ) dv = si d du = ( os )os( si ) d v = os ( ) os os( si ). g d But differetiatio of the first equatio for g() yields si os( si ). g( ) d Fially, beause os + si =, it follows that g( ) g( ) os( si ) d g( ). Thus y = g() satisfies Bessel's equatio of order zero i the form y'' + (/)y' + y =. Therefore the futio g takes the form g() = a J () + b Y (). Sie g() = is fiite ad J () =, we must have a = ad b =, so g() = J (), as desired.. This is a speial ase of the disussio below i Problem Give a iteger, let us defie Copyright 5 Pearso Eduatio, I.

49 Setio Differetiatio yields g ( ) os( si ) d. g ( ) si( si )si d. Itegratio by parts with u = si( si ) ad dv = si d yields ( ) os os( si ) os os( si ). g d d But differetiatio of the first equatio for g '() yields ( ) si os( si ). g d It follows that g ( ) g ( ) g( ) os os( si ) d g( ) ( os ) os( si ) d g( ) si( si g ( ) ( ). g Upo equatig the first ad last members of this otiued iequality ad multiplyig by, we see that y = g () satisfies Bessel's equatio of order. The iitial values of g () are g () os( ) d ad g () si( )si( ) d. If = the g () = /, whereas g () = if. I either ase the values of g () ad g () are times those of J () ad J (), respetively. Now we kow from the geeral solutio of Bessel's equatio that g () = J () for some ostat. If = the the fat that / = g () = J () = / implies that =, as desired. But if > the fat that does ot suffie to determie. = g () = J () = 6. The graph show at the top of the et page illustrates the iterlaed zeros of the Bessel futios J( ) ad J( ). Copyright 5 Pearso Eduatio, I.

50 5 APPLICATIONS OF BESSEL FUNCTIONS. J 4 J -. SECTION 8.6 APPLICATIONS OF BESSEL FUNCTIONS Problems are routie appliatios of the theorem i this setio. I eah ase it is eessary oly to idetify the oeffiiets A, B, C ad the epoet q i the differetial equatio y Ay ( BC q ) y. () The we a alulate the values ( A) 4 A q C B,, k, p () q q ad fially write the geeral solutio y( ) J p( k ) J p( k ) () speified i Theorem of this setio. This is a "template proedure" that we illustrate oly i a ouple of problems.. We have A, B, C, q so ( ( )) 4() ( ),, k, p, so our geeral solutio is y() = [ J () + Y ()], usig Y () beause p = is a iteger.. y() = [ J () + Y ()]. y() = [ J / ( ) + J / ( )] Copyright 5 Pearso Eduatio, I.

51 Setio y() = [ J ( / ) + Y ( / )] 5. To math the give equatio with Eq. () above, we first divide through by the leadig oeffiiet 6 to obtai the equatio y y y with A5/, B 5/ 6, C / 4, ad q. The (5/) 4( 5/6) 5/ /4,, k, p, so our geeral solutio is y() = / [ J / ( / /) + J / ( / /)]. 6. y() = /4 [ J ( / ) + Y ( / )] 7. y() = [ J () + Y ()] 8. y() = [ J (4 / ) + Y (4 / )] 9. y() = / [ J / ( / ) + J / ( / )]. y() = /4 [ J / ( 5/ /5) + J / ( 5/ /5)]. y() = / [ J /6 ( /) + J /6 ( /)]. y() = / [ J /5 (4 5/ /5) + J /5 (4 5/ /5)]. We wat to solve the equatio y'' + y' + y =. If we rewrite it as y'' + y' + y = the we have the form i Equatio () with A =, B =, C =, ad q =. The Equatio () gives = /, =, k =, ad p = /, so by Equatio () the geeral solutio is y J J / ( ) /( ) /( ) / os si aos asi, (with a / ) usig Equatios (9) i Setio.5. i i Copyright 5 Pearso Eduatio, I.

52 54 APPLICATIONS OF BESSEL FUNCTIONS 5. The substitutio u ( ), u y y u u u u immediately trasforms y' = + y to u'' + u =. The equivalet equatio u'' + 4 u = is of the form i () with A = B =, C =, ad q = 4. Equatios () give = /, =, k = /, ad p = /4, so the geeral solutio is u() = / [ J /4 ( /) + J /4 ( /)]. To ompute u'(), let z = / so = / z /. The Equatio () i Setio 8.5 with p = /4 yields d d dz / J /4 /4 /4( /) z J/4( z) d dz d /4 /4 dz z J /4( z) d / /4 / J /4 /4( /) J /4( /). Similarly, Equatio () i Setio 8.5 with p = /4 yields d / d /4 /4 dz / J /4( /) z J /4( z) J/4( /). d dz d Therefore u'() = / [ J /4 ( /) J /4 ( /)]. It follows fially that the geeral solutio of the Riati equatio y = + y is y ( ) u J ( ) ( ) J u J ( ) J ( ) /4 /4 /4 /4 where the arbitrary ostat is = /. 6. Substitutio of the series epressios for the Bessel futios i the formula for y() i Problem 5 yields y ( ) /4 /4 /4 /4 A B C D where eah pair of paretheses eloses a power series i with ostat term, ad Copyright 5 Pearso Eduatio, I.

53 Setio A = /4 /(7/4) C = /4 /(5/4) B = /4 /(/4) D = /4 /(/4). Multipliatio of umerator ad deomiator by / ad a bit of simplifiatio gives B D /4 /4 A y ( ). /4 /4 C It ow follows that /4 / /4 B / (/4) (/4) /4 /4 D / (/4) (/4) y(). (*) (a) If y() = the (*) gives = i the geeral solutio formula of Problem 5. (b) If y() = the (*) gives = (/4)/(/4). More geerally, (*) yields the formula 4J/4( ) y 4J/4( ) y ( ) J ( ) y J ( ) 4 /4 4 /4 for the solutio of the iitial value problem y = + y, y() = y. 7. If we write the equatio 4 y + y = i the form y + y =, the we see that it is of the form i Equatio () of this setio with A = B =, C =, ad q =. The Equatios (5) give = /, =, k =, ad p = /, so the theorem yields the geeral solutio y() = / [ J / (/) + J / (/)] = [A os(/) + B si(/)], usig Equatios (9) i Setio 8.5 for J / () ad J / (). With a ad b both ozero, the iitial oditios y(a) = y(b) = yield the equatios A os(/a) + B si(/a) = A os(/b) + B si(/b) =. These equatios have a otrivial solutio for A ad B oly if the oeffiiet determiat Copyright 5 Pearso Eduatio, I.

54 56 APPLICATIONS OF BESSEL FUNCTIONS = si(/b) os(/a) si(/a) os(/b) = si(/b /a) = si(l/ab) is ozero. Hee L/ab must be a itegral multiple of, ad the the th buklig fore is EI EI ab a P EI 4 4. b b L L b 8. The substitutio L = a + bt i L'' + L'' + g = yields the trasformed equatio L ''(L) + L'(L) + (g/b )L = with idepedet variable L that is of the form i () with A =, B =, q =, ad C = g/b. Hee so = /, = /, k = g / /b, ad p =, ( L) AJ gl BY gl. L b b Copyright 5 Pearso Eduatio, I.