# Math 201 Lecture 12: Cauchy-Euler Equations

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1 Mah 20 Lecure 2: Cauchy-Euler 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: Cauchy-Euler 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 Cauchy-Euler equaions and hey are as easy as equaions of consan coefficiens (In fac, Cauchy-Euler equaions is jus consan-coefficien 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 consan-coefficien 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 consan-coefficien 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 consan-coeffcien 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: Cauchy-Euler Equaions Thus we have ransformed he Euler-Cauchy equaion ino a consan-coefficien 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 Cauchy-Euler 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 Cauchy-Euler. (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 Cauchy-Euler 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 Cauchy-Euler, 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 Cauchy-Euler 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 Cauchy-Euler equaions. Examples.. Inroduce x = ln, ransform i o consan-coefficien 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: Cauchy-Euler 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 Cauchy-Euler. 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: Cauchy-Euler 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 Cauchy-Euler 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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