Math 201 Lecture 12: CauchyEuler Equations


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1 Mah 20 Lecure 2: CauchyEuler Equaions Feb., 202 Many examples here are aken from he exbook. The firs number in () refers o he problem number in he UA Cusom ediion, he second number in () refers o he problem number in he 8h ediion. To solve general 2nd order linear equaions, 0. Review a() y +b() y + c() y = f(). (). Guess one soluion y. (Popular guesses: consans; exponenials; simple polynomials; sin,cos) 2. Wrie he equaion ino sandard form y + p() y + q() y = 0 (2) and apply he reducion of order formula: e p()d y 2 ()= y () y () 2 d (). Apply variaion of parameers formula wih Quiz: Solve v = y p = v y + v 2 y 2 (4) f() y 2 () a()[y y 2 y y 2 ], v 2 = o obain y p. Noe ha a() is no a consan anymore. 4. The general soluion is hen given by Simplify if possible. 5. Check your soluion if ime allows. (As = 0 is a singular poin, le s jus consider > 0). Soluion.. Guess one soluion for I is clear ha any consan is a soluion. So le s ake y =. 2. Use reducion of order o ge y 2. a. Wrie he equaion in sandard form: f() y () a()[y y 2 y y 2 ]. (5) y =C y + C 2 y 2 + y p. (6) y y =. (7) y y =0. (8) b. Compue y y = p()=. (9) e p y 2 = y 2 = e = e ln = = 2 y 2. (0). Use variaion of parameers o ge y p.
2 2 Mah 20 Lecure 2: CauchyEuler Equaions Compue [ a()[y y 2 y y 2 ] = ( 2 2 ) ] 2 2 () = 2. () f() y v = 2 () a()[y y 2 y y 2 ] = = 2 ; (2) f() y v 2 = () a() [y y 2 y y 2 ] = 2 = () So ( y p = ) ( + ) 2 =. (4) The general soluion is The Equaion: which can be simplifiied o y =C + C (5) y = C +C 2 2. (6). Basic Informaion a 2 y + b y + c y = 0. (7) Such equaions are called CauchyEuler equaions and hey are as easy as equaions of consan coefficiens (In fac, CauchyEuler equaions is jus consancoefficien equaions in disguise, see Noes and Commens). How o ge general soluion Idea: From general heory of 2nd order linear equaions, we know ha as soon as we figure ou wo linearly independen soluions y, y 2, he general soluion is simply C y + C 2 y 2. (8) Recall ha in suing consancoefficien equaions a y + b y + c y = 0 we obain y, y 2 hrough guess y =e r. Here he idea is similar bu he guess is differen: y = r. Procedure:. Wrie down he characerisic equaion Solve i o ge r, r Three cases: Examples: r r 2, boh real: r = r 2 =r, real. r,2 = α ± iβ.. Wrie down general soluion. a r 2 + (b a) r + c = 0. (9) y = r, y 2 = r2 ; (20) y = r, y 2 = r ln. (2) y = α cos (β ln ), y 2 = α sin (β ln ). (22) Example. Solve 2 y +7y 7 y = 0. (2)
3 Feb., 202 Soluion. Guided by he heory, we only need o find wo linearly independen soluions. The key now is o realize he following propery of r : ( r ) (k) k = C r. Subsiue y = r ino he equaion, we have 0= 2 y +7y 7 y = [r (r )+7r 7] r = 0 r 2 + 6r 7=0 r = 7, r 2 =. (24) Thus he general soluion is y = c 7 + c 2. (25) Example 2. solve 2 y y + 4 y = 0. (26) Soluion. Subsiuing y = r we have r (r ) r + 4=0 r 2 4r +4=0 r = r 2 =2. (27) This is double roo so he general soluion is given by y = c 2 + c 2 2 ln. (28) Example. Solve y y + 5 y = 0. (29) 2 Soluion. Muliply boh sides by 2 : Subsiuing y = r gives 2 y y + 5 y = 0. (0) r (r ) r + 5=0 r 2 2r +5=0 r = +2i, r 2 = 2 i. () The general soluion is hen given by Relaions o consancoefficien equaions. The formulas r r 2, boh real: r = r 2 =r, real. r,2 = α ± iβ. y = c cos (2 ln )+c 2 sin (2 ln ). (2) y = r, y 2 = r2 ; () y = r, y 2 = r ln. (4) y = α cos (β ln ), y 2 = α sin (β ln ). (5) above looks similar o our heory for linear consancoeffcien equaions. This similariy becomes more sriking if we inroduce a new variable x=ln : The hree cases become Two disinc roos e rx, e r2x ; One double roo e rx, xe rx ; Complex roos e αx cos β x, e αx sin β x. This is no coincidence! In fac, seing x=ln gives y = d = dx dx d = dx, y = d2 y d 2 = d [ ] dx d 2 y dx dx d = 2 dx 2 2 dx. (6) Subsiuing ino he equaion 0=a 2 y + b y + cy = a d2 y + (b a) + c y. (7) dx2 dx
4 4 Mah 20 Lecure 2: CauchyEuler Equaions Thus we have ransformed he EulerCauchy equaion ino a consancoefficien equaion. Furhermore, he auxiliary equaion for his equaion is a r 2 + (b a)r + c = a r (r )+br + c (8) which is exacly he characerisic equaion of he CauchyEuler equaion! How o check soluions Noe: Check soluions, especially in he rd case, may involve so much calculaion ha i becomes no worhwhile. Insead, make sure you wrie down he correc characerisic equaion and solve i correcly. See Common Misakes for examples. 2. Things o be Careful/Tricky Issues Fail o see ha he equaion is CauchyEuler. (This is he hoes misake in 20!!) Afer finding r,2, wrie e r, e r2 insead of r, r2. (This misake is also very popular.) Iniial value problem.. More Examples Example 4. Solve he following iniial value problem for he CauchyEuler equaion Soluion. Subsiuing y = r gives 2 y 4y + 4 y = 0; y()= 2, y () =. (9) r (r ) 4r + 4=0 r 2 5r +4=0 r = 4, r 2 =. (40) Thus he general soluion is given by Using he iniial values, we have Solving his we reach y() =c 4 +c 2. (4) 2 = y() =c +c 2 ; = y ()=4c + c 2. (42) Thus he soluion o he iniial value problem is given by Wha happens if <0? Example 5. (NA; 4.7 5) Solve for <0 Soluion. Le x =. Then x > 0. We have d = dx dx d = dx ; d 2 y d 2 = d ( d d c =, c 2 =. (4) y() = 4 +. (44) y y + 5 = 0. (45) 2 ) = d d ( dx ) = d dx ( dx ) dx d = d2 y dx2. (46) So if we use x insead of as he variable, he equaion (wih unknown y and variable x) reads d 2 y dx 2 x dx + 5 = 0. (47) x2 I is sill CauchyEuler, wih a =, b =, c = 5. We wrie down characerisic equaion r 2 2 r +5=0 r,2 = ± 2i. (48)
5 Feb., So he soluion reads Back o : Which can be wrien as y(x) =C x cos (2 ln (x))+c 2 x sin (2 ln x). (49) y()=c ( ) cos(2 ln ( )) +C 2 ( ) sin (2 ln ( )). (50) y()=c cos (2 ln ) +C 2 sin (2 ln ). (5) Remark 6. In fac, we can solve CauchyEuler for 0 as r r 2, boh real: r = r 2 =r, real. r,2 = α ± iβ. Wha if i s no? Example 7. (4.7 2; 4.7 2) Solve y = r, y 2 = r2 ; (52) y = r, y 2 = r ln. (5) y = α cos (β ln ), y 2 = α sin (β ln ). (54) ( 2) 2 y 7( 2) y + 7 y = 0, >2. (55) Soluion. I is clear ha we should inroduce x = 2. Now chain rule gives The equaion becomes This can be easily solved: Back o : Nonhomogeneous problem. d = dx ; d 2 y d 2 = d2 y dx2. (56) x 2 d 2 y 7x +7 y =0, x > 0. (57) dx2 dx y(x)=c x+c 2 x 7. (58) y()=c ( 2)+C 2 ( 2) 7. (59) There are wo ways o aack nonhomogeneous problem for CauchyEuler equaions. Examples.. Inroduce x = ln, ransform i o consancoefficien case, hen apply undeermined coefficiens, or variaion of parameers; 2. Apply variaion of parameers direcly. Example 8. Solve Soluion (x = ln ). We know ha seing x=ln ransforms 2 y 4y + 4 y = 2. (60) a 2 y +by + c y o a d2 y + (b a) + cy. (6) dx2 dx So he equaion for x is (x=ln = e x ): d 2 y 5 dx2 dx + 4 y = e2x. (62)
6 6 Mah 20 Lecure 2: CauchyEuler Equaions We see ha his equaion is eligible for undeermined coefficiens (keep in mind ha whenever undeermined coefficiens applys, i is more efficien han variaion of parameers). To solve i we firs ge y, y 2 by solving which gives Now guess d 2 y 5 dx2 dx + 4 y = 0 (6) r = 4, r 2 = ; y = e 4x, y 2 =e x. (64) y p = A x s e 2x. (65) We have s = 0 because 2 does no appear in he roo lis r = 4, r 2 =. Subsiue y p =Ae 2x ino he equaion we ge d 2 y dx 2 =4Ae2x, dx =2Ae2x 4 A 0 A+4A= A = 2. (66) So y p = e2x. The general soluion (in x) is hen 2 Back o : (Replace every x by ln ): y = C e 4x + C 2 e x 2 e2x. (67) y()=c 4 +C (68) Soluion 2 (Direc applicaion of variaion of parameers). The equaion is Cauchy Euler so we can solve he homogeneous equaion as follows: So Now calculae: Thus Therefore 2 y 4 y +4 y =0 (69) r (r ) 4r + 4=0 r 2 5r + 4=0 r =4, r 2 =. (70) v = y = 4, y 2 =. (7) a()[y y 2 y y 2 ] = 2 [ 4 4 ] = 6. (72) f() y 2 () a()[y y 2 y y 2 ] = 2 6 = f() y v 2 = () a()[y y 2 y y 2 ] = y p =v y + v 2 y 2 = The general soluion is hen given by Example 9. (4.7 4; 4.7 4) Solve Soluion. Firs solve he homogeneous equaion = 6 2. (7) =. (74) ( 6 ) ( ) = 2 2. (75) y()=c 4 +C (76) 2 z +z + 9z = an ( ln ). (77) 2 z + z +9z = 0. (78)
7 Feb., I is CauchyEuler. So firs solve he characerisic equaion: So we have r (r )+r+9=0 r,2 = ±. (79) y = cos ( ln ), y 2 = sin ( ln ). (80) Nex we use variaion of parameers o obain y p. Firs calculae: ( ) a [y y 2 y y 2 ] = [cos 2 ( ln ) sin ( ln ) Now we have ( cos( ln ) ) sin ( ln ) = 2 =. (8) an ( ln ) sin ( ln ) v = d = sin 2 ( ln ) dln. (82) cos ( ln ) Leing x = ln, we compue sin 2 (x) cos (x) dx = dx cos ( x) cos(x)dx = cos 2 (x) cos ( x) dx sin (x) = sin 2 ( x) dsin (x) sin (x) = [ ] dsin ( x) 6 sin ( x) + dsin ( x) sin ( x) + sin ( x) = 6 [ln + sin (x) ln sin ( x) ] sin (x). (8) ] Back o : v = { 6 [ln + sin ( ln ) ln sin ( ln ) ] } sin ( ln ). (84) On he oher hand ( an ( ln )) cos ( ln ) v 2 = d sin ( ln ) = d = sin ( ln )dln Puing hings ogeher we have = cos ( ln ); (85) 9 y p = [ln + sin ( ln ) ln sin ( ln ) ] cos( ln ) 8 sin ( ln ) cos ( ln ) 9 + cos ( ln ) sin ( ln ) 9 = [ln + sin ( ln ) ln sin ( ln ) ] cos( ln ) (86) 8 Thus he soluion is y = C cos ( ln )+C 2 sin ( ln )+ [ln +sin ( ln ) ln sin ( ln ) ] cos ( ln ). (87) 8
8 8 Mah 20 Lecure 2: CauchyEuler Equaions Remark 0. For such problems someimes i is more efficien o le x=ln and ransform he equaion (See Noes and Commens). For example for he above problem, if we le x=ln, hen he equaion becomes d 2 y +9 y = an (x). (88) dx2 This can be solved by variaion of parameers, o obain y = C cos (x)+c 2 sin (x)+ [ln +sin ( x) ln sin (x) ] cos ( x). (89) 8 Replace x by ln we ge our soluion. 4. Noes and Commens Why b a? To remember i, jus remember ha our guess is y= r. Subsiuing his ino he equaion we reach which is exacly a r (r )+br + c =0 (90) a r 2 + (b a) r + c = 0. (9) Why do we need > 0? Noe ha, unlike e r, r is singular (meaning: eiher i s infiniy, or is cerain order of derivaive is infiniy) a = 0. More accuraely, when wriing he CauchyEuler equaion in sandard form y + b/a y + c/a y = 0. (92) 2 We see ha p() = b/a and q() = c/a are singular a 0. This is an indicaion ha good heories 2 (soluion exiss, soluion is unique, soluion is smooh...) break down when he inerval for conains 0. Therefore o make hings simple we eiher work in > 0 or in < 0, o avoid conaining = 0.
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