Second-Order Linear Differential Equations

Thomson Brooks-Cole copyright 2007 SECOND-ORDER LINEARDIFFERENTIALEQUATIONS fteohr o ntne h ucin ad arelinearlydependent, and b isasolutionofEquation 2. thefunctions of theother. For instance, independent advanced courses.Itsaysthatthegeneralsolution isalinearcombinationoftwo Thus, wehave usingthebasicrulesfordifferentiation, Therefore, and Proof thenthe ofsuchanequation, Equation1is , and forsome that ifweknow two solutions T If homogeneous differential equationis tions arecalled arecontinuousfunctions.Equationsofthistypeariseinthestudy circuits. and , Equations , the motionofaspring.In , where A y pc:Nonhomogeneous LinearEquations opics: tad are ut and linearly independent. second-order lineardifferential equation G h te atw edi ie ytefloigterm whichisproved inmore The otherfact weneedisgiven bythefollowing theorem, Tw inEquation1.Suchequa- , for all , In thissectionwestudythecasewhere is alsoasolutionofEquation2. thenthefunction areany constants, and tion (2)and 3 2 1 f c P x 1 o basicfacts enableustosolve homogeneouslinearequations. The ﬁrstofthesesays x y ic ad r ouin fEuto ,wehave aresolutionsofEquation2, and Since P y Theorem x 1 Q y 0 c we willfurtherpursuethisapplicationaswellthetoelectric e 1 c R y y x 2 1 ouin ad hsmasta ete nr is aconstantmultiple nor Thismeans thatneither and solutions 1 y Q 2 c f n arebothsolutionsofthelinearhomogeneousequa- and If homogeneous 1 is alsoasolution. x c y t

G 2 y 1 c P P c y y 2 1 1 x x 2 x x c 0 P x 2 y x 1 c c P R 1 1 xe y y y P diinlTpc:ApplicationsofSecond-Order Differential Additional Topics: x x 1 1 1 c y x 2 P P x 2

y y 0 d y dx x

1 x x iereutos hs theformofasecond-orderlinear linear equations. Thus, 2 c 2 d y dx Q c . y 2 y y 2 2 2 y y 2 1 x y 2 x 2 2 0 y y Q 2 1 Q Q nonhomogeneous Q c f x Q 1 Q x x y x x R

G 1 . y y x dx dy x

x 2 1 x dx dy x has theform c c y 1 1 x 1 y y 2 R R 1 1 c R 0 2 x x R y x 2 y y c c c x 2 1 t y 2 x 2 2 y y y P x 2 2 y and isdiscussedin 0 0 x 1 G x y 0 2 x R 5 R x x 2 y x 2 Q c c 1 linear combination 1 y x y 1 1 y 2 c c 2 2 y R y 2 2 Additional x linearly y 2 1 Thomson Brooks-Cole copyright 2007 2 ■ SECOND-ORDER LINEARDIFFERENTIALEQUATIONS Equation 6iscalledthe isarootoftheequation isasolutionofEquation5if isnever 0. Thus, . But or isasolutionif .Furthermore, isaconstant)hastheprop- (where weseethat If wesubstitutetheseexpressions intoEquation5, isequal erty thatitsderivative is aconstantmultipleofitself: plusathirdconstanttimes to 0. We . know thattheexponential function plusanotherconstanttimes its secondderivative suchthataconstanttimes we statetheequationverbally. We areconstantsand arelookingforafunction and , , where areconstant and ifthedifferential equationhastheform , thatis, functions, , tion. Butitisalways possibletodosoifthecoefﬁcient functions We .Noticethatitisanalgebraicequationobtained . by and , by other casesthey arefound byusingthequadraticformula: , by from thedifferential equationbyreplacing tial equation thenweknow pendent solutions, n aetolnal needn ouin fEuto .(oeta isnot a aretwo linearly independentsolutionsofEquation5.(Notethat wehave thefollowing fact. by Theorem 4, .) Therefore, constant multipleof so oftheauxiliaryequationarerealanddistinct, and and In thiscasetheroots CASE I It’s nothardtothinkofsomelikely candidatesforparticularsolutionsofEquation5if itisnoteasytodiscover particularsolutionstoasecond-orderlinearequa- In general, Theorem 4isvery usefulbecauseitsaysthatifweknow where andare arbitrary constants. thenthegeneralsolutionisgiven by is never 0, oeie h ot ad oftheauxiliaryequationcanbefoundbyfactoring. In and Sometimes theroots nqa,te h eea ouino is thenthegeneralsolution of unequal,

itnus he ae codn otesg ftedsrmnn . distinguish threecasesaccordingtothesignofdiscriminant 4 8 7 6 5 e y 2 rx ■ a Theorem fteros n o h uiir qain arerealand oftheauxiliaryequation and If theroots b 2 b e c 1 r 2 x ay 4 ac f n aelnal needn ouin fEuto ,and arelinearlyindependentsolutionsofEquation2, and If c c r 2 1 y r 1 by 1 e y 0 r r 1 1 x b auxiliary equation r y y 1 r 2 cy 2 r s 2 e 2 rx b r ar a y 2 2 0 every 2 ay x e y ar a rx ar 4 2 ac 2 solution. c c ay 0 bre 1 1 by e y br br r 1 y 1 rx x (or y x by c cy characteristic equation r c c r ce 2 e 2 c 2 e rx e y 2 rx rx r y y 0 2 2 x ar 0 cy y x b 2 0 0 r re 0 r br s two y rx 2 r y b a 2 y P particular linearlyinde- c e Q b 4 rx 2 1 ac 0 ) ofthedifferen- 4 R ac e y r 2 y y x P 1 x r 2 e e r rx 1 x Thomson Brooks-Cole copyright 2007 f ■■ iercmiain f n ,areshown in blue. and linear combinationsof red, respectively. Someoftheothersolutions, equation inExample1areshownblackand FIGURE 1 11 _1 x In Figure1thegraphsofbasicsolutions fg e f-g 2 x n ofthedifferential and 5f+g t x _5 8 f f+g e f+5g 3 x t g-f We by(8)thegeneralsolutionofgiven differen- . Therefore, , tial equationis whose rootsare ic h ot r eladdsic,thegeneralsolutionis Since therootsarerealanddistinct, formula: qain ic ad r ieryidpnetsltos Theorem4pro- arelinearlyindependentsolutions, and vides uswiththegeneralsolution. equation. Since isarootoftheauxiliary The ﬁrsttermis0byEquations9;thesecondbecause a solution: SOLUTION EXAMPLE 1 SOLUTION EXAMPLE 3 We wehave fromEquations7, Then, therootsofauxiliaryequationarerealandequal.Let’s ;thatis, and thecommonvalue of denote by In thiscase CASE II SOLUTION EXAMPLE 2 differential equation. eea ouino is general solutionof

10 nwta i n ouino qain5 enwvrf ht isalso isonesolutionofEquation5. We now verify that know that could verify thatthisisindeedasolutionby differentiating andsubstitutingintothe 9 ■ fteaxlayeuto hsol n elro ,thenthe , hasonly onerealroot If theauxiliaryequation b h uiir qain canbe factored as The auxiliaryequation To The auxiliaryequationis 2 r ov h qain . Solve theequation Solve . ov h qain . Solve theequation

ov h uiir qain we usethequadratic solve theauxiliaryequation ay r 1 4 y 2 1 ac y r 1 3 r 2 by e d dx 2 rx e 2 0 2 ay y rx 2 3 cy r y dy dx r 2 2 y by

2 4 y y 4 c r 0 a ar 1 r 2 r e xe y 2 1 2 e ar y y 2 2 ( b cy y re r rx a rx 1 6 12 2 rx 12 0 s r br c c y 13 r 3 b 1 1 r 6 2 e e r 0 ) . 0 x e y r 2 rx 2 so 1 r x 3 rx 6 2 xe xe c 9 9 2 6 y rx r 2 0 rx 2 c s SECOND-ORDER LINEARDIFFERENTIALEQUATIONS c c 2 ar 2 2 13 ar 0 xe e r 0 e 0 0 1 ( 2 3 rx 0 x 1 b 3 b s e 0 br 13 rx ) x 0 0 6 rxe c r xe rx rx cxe y 2 r rx xe rx ■ 3 4 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

■■ Figure 2 shows the basic solutions 3 so the only root is r 2 . By (10) the general solution is f x e3x2 and tx xe3x2 in Example 3 and some other members of the y c e3x2 c xe3x2 family of solutions. Notice that all of them 1 2 approach 0 as x l . 2 CASE III ■ b 4ac 0 f-g 8 In this case the roots r1 and r2 of the auxiliary equation are complex numbers. (See f Additional Topics: Complex Numbers for information about complex numbers.) We can 5f+g write f+5g r1 i r2 i _2 2 f+g g-f g where and are real numbers. [In fact, b2a , s4ac b 22a .] Then, using Euler’s equation _5

i FIGURE 2 e cos i sin from Additional Topics: Complex Numbers, we write the solution of the differential equation as

r1x r2 x ix ix y C1e C2e C1e C2e

x x C1e cos x i sin x C2e cos x i sin x

x e C1 C2 cos x iC1 C2 sin x

x e c1 cos x c2 sin x

where c1 C1 C2 ,c2 iC1 C2 . This gives all solutions (real or complex) of the differential equation. The solutions are real when the constants c1 and c2 are real. We sum- marize the discussion as follows.

11 If the roots of the auxiliary equation ar 2 br c 0 are the complex num-

bers r1 i ,r2 i , then the general solution of aybycy 0 is

x y e c1 cos x c2 sin x

■■ Figure 3 shows the graphs of the solu- EXAMPLE 4 Solve the equation y6y13y 0 . tions in Example 4, f x e 3x cos 2x and 2 tx e 3x sin 2x, together with some linear SOLUTION The auxiliary equation is r 6r 13 0 . By the quadratic formula, the combinations. All solutions approach 0 roots are l as x . 6 s36 52 6 s16 r 3 2i 3 2 2 f+g g By (11) the general solution of the differential equation is f-g f 3x _3 2 y e c1 cos 2x c2 sin 2x

INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS _3 An initial-value problem for the second-order Equation 1 or 2 consists of ﬁnding a solu- FIGURE 3 tion y of the differential equation that also satisﬁes initial conditions of the form

yx0 y0 yx0 y1

where y0 and y1 are given constants. If P ,Q ,R , and G are continuous on an interval and Px 0 there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem. Examples 5 and 6 illustrate the technique for solving such a problem. Thomson Brooks-Cole copyright 2007 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 5

EXAMPLE 5 Solve the initial-value problem yy6y 0 y0 1 y0 0

■■ Figure 4 shows the graph of the solution of SOLUTION From Example 1 we know that the general solution of the differential equa- the initial-value problem in Example 5. Compare tion is with Figure 1. yx c e 2x c e3x 20 1 2 Differentiating this solution, we get

2x 3x yx 2c1e 3c2 e To satisfy the initial conditions we require that _2 2 12 y 0 c1 c2 1 0

13 y0 2c1 3c2 0 FIGURE 4 2 From (13) we have c2 3 c1 and so (12) gives 2 3 2 c1 3 c1 1 c1 5 c2 5 Thus, the required solution of the initial-value problem is

3 2x 2 3x y 5 e 5 e

EXAMPLE 6 Solve the initial-value problem

■■ The solution to Example 6 is graphed in yy y y Figure 5. It appears to be a shifted sine curve 0 0 2 0 3 and, indeed, you can verify that another way of SOLUTION The auxiliary equation is r 2 1 0 , or r 2 1 , whose roots are i . Thus writing the solution is 0,, 1 and since e 0x 1 , the general solution is s 2 y 13 sin x where tan 3

yx c1 cos x c2 sin x 5

Since yx c1 sin x c2 cos x the initial conditions become _2π 2π y0 c1 2 y0 c2 3 Therefore, the solution of the initial-value problem is _5 yx 2cos x 3sin x FIGURE 5 A boundary-value problem for Equation 1 consists of ﬁnding a solution y of the differential equation that also satisﬁes boundary conditions of the form

yx0 y0 yx1 y1 In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution.

EXAMPLE 7 Solve the boundary-value problem y2yy 0 y0 1 y1 3

SOLUTION The auxiliary equation is r 2 2r 1 0 or r 12 0 whose only root is r 1 . Therefore, the general solution is

■■ Figure 6 shows the graph of the solution of The boundary conditions are satisﬁed if the boundary-value problem in Example 7. y0 c1 1 5 1 1 y1 c1e c2 e 3

The ﬁrst condition gives c1 1 , so the second condition becomes

_1 5 1 1 e c2 e 3

Solving this equation for c2 by ﬁrst multiplying through by e , we get _5 1 c2 3e so c2 3e 1 FIGURE 6 Thus, the solution of the boundary-value problem is y ex 3e 1xex

Summary: Solutions of ay byc 0

Roots ofar 2 br c 0 General solution

r1 x r2 x r1, r2 real and distinct y c1e c2e rx rx r1 r2 r y c1e c2 xe x r1, r2 complex: i y e c1 cos x c2 sin x

EXERCISES

20. 2y5y3y 0,,y0 1 y0 4 A Click here for answers. S Click here for solutions. 21. y16y 0,,y4 3 y4 4 1–13 Solve the differential equation. 22. y2y5y 0,,y 0 y 2

1.y6y8y 0 2. y4y8y 0 23. y2y2y 0,,y0 2 y0 1 3.y8y41y 0 4. 2yyy 0 24. y12y36y 0,,y1 0 y1 1 5.y2yy 0 6. 3y 5y ■■■■■■■■■■■■■ 25–32 Solve the boundary-value problem, if possible. 7.4yy 0 8. 16y24y9y 0 25. 4yy 0,,y0 3 y 4 9.4yy 0 10. 9y4y 0 d 2y dy d 2y dy 26. y 2y 0,,y 0 1 y 1 2 11. 2 y 0 12. 6 4y 0 dt 2 dt dt 2 dt 27. y3y2y 0,,y0 1 y3 0 d 2y dy 28. y100y 0,,y0 2 y 5 13. y 0 dt 2 dt 29. y6y25y 0,,y0 1 y 2

■■■■■■■■■■■■■ 30. y6y9y 0,,y0 1 y1 0 ; 14–16 Graph the two basic solutions of the differential equation and several other solutions. What features do the solutions have in 31. y 4y 13y 0,,y 0 2 y 2 1 common? 32. 9y18y10y 0,,y0 0 y 1 2 2 d y dy d y dy ■■■■■■■■■■■■■ 14.6 2 2y 0 15. 2 8 16y 0 dx dx dx dx 33. Let L be a nonzero real number. d 2y dy (a) Show that the boundary-value problem y y 0 , 16. 2 5y 0 dx2 dx y 0 0, y L 0 has only the trivial solution y 0 for the cases 0 and 0 . ■■■■■■■■■■■■■ (b) For the case 0 , ﬁnd the values of for which this prob- 17–24 Solve the initial-value problem. lem has a nontrivial solution and give the corresponding 17. 2y5y3y 0,,y0 3 y0 4 solution. a b c yx 18. y3y 0,,y0 1 y0 3 34. If , , and are all positive constants and is a solution of the differential equation aybycy 0 , show that 19. 4y 4y y 0,,y 0 1 y 0 1.5 lim x l yx 0. Thomson Brooks-Cole copyright 2007 Thomson Brooks-Cole copyright 2007 l ouin prah0a adapoc a . as andapproach All solutionsapproach0as 11. 7. 3. 15. 13. 9. 5. 1. S y y y y 021 _0.2 Click hereforsolutions. c c c 1 1 1 e ANSWERS e e x 4 t _40 x 40 2 c [ c 2 e 1 c cos c 2 2 xe x e 4 2 ( x x s 3 t 2 ) x y y y g l c 2 sin c c e f 1 1 cos e 4 ( ( x 1 s c 3 s 1 x cos5 2 t ) t 2 2 )] c x 2 c e 2 ( sin 1 c s 2 sin5 2 ) x t 2 x x l 33. 31. 29. 27. 25. 23. 21. 19. 17. b , (b) No solution y y y y y y y

e 3 cos e 2 3 cos4 e e e x 3 e /2 2 x x x 3 2 cos n x 2 cos3 3 ( 2 2 1

2 1 2 x x ) xe 2 x L e x 1 2 sin 4 x 2 x 4 sin SECOND-ORDER LINEARDIFFERENTIALEQUATIONS e n 3 sin 2 a positive integer; x e e x

( 3 2 1 sin3 x x )

x y C sin

n x L ■ 7 Thomson Brooks-Cole copyright 2007 8 ■ SECOND-ORDER LINEARDIFFERENTIALEQUATIONS 13. 15. 17. 11. 21. 27. 19. 25. 23. SOLUTIONS 7. 1. 3. 9. 5. solut y bo T T − solution The 2 r T r and r y ⇒ T the The 4 y y 1= 4 r and T T Then 2 2 2 2 r r r (0 h h h h h h 4= = = = undary 2 2 2 e e e e e e − − +2 +1 ) i +5 1= a +1 − nitial- a a a a g a a a y e e c i c s = 8 on 3 ux ux ux ux uxiliar uxi uxiliary rap y 1 x/ − (0) 1 y r r r 4 6 x e 4 r =1 ( 3 r ili ili ili ili 2 x x +1 +2 π +2 is to 4 = = li h y π →∞ -v +3 ( +1 − = + s a a a a a 0 v )= c y (0) the r r r r r alue alue 0 0 a ⇒ 1 y y y y = y y 6 2 and c c re = = = cos xe 2 1 eq eq eq eq e e eq = = = − xe i c = qu q ⇒ ⇒ all the + nitial-v c 0 ( 2 prob p u u u u u uation x/ ( 1 3 y c r ( ( , 5 r x a a a a atio a c 2 e 1 r c 2 0 oblem s a 2 tio tio tio tio x tio − 2 . (0) r 4 s 2 r o ⇒ sy + . − x olution r r − ⇒− + lem +3 and − 2)( n n n n n t n mp h + c = = 4) alue = c i i i i i i is e c 2 1) s s s s s s 1 2 ) r c r 2 ± ± solution totic i is =3 0= r 4 r 4 r r ( r − 2 s si 2 − r = =0 2 2 2 r r 2 2 e 4 s 2 1 y y prob 1 c n ⇒ =0 2 2 2 i +1 + − − − +8 i 1 tend . 1) x − = 5 =2 5 + +1 and and y t and . = o x 6 r 2 2 so 1 (3) = ) lem r r r ) r r e t +1 c ± . he e − ⇒ e of 3 to = t 2 − +8 y y +1 +4 = = t ⇒ x he 0 − he = 3 − 0 +3 i =3 = (0) ± 3 x the i 0 1=0 r 0 s and = x/ general ,s -axis 1 c 1 g (4 ∞ ⇒ = y = 2 1 c 1 ⇒ e r 2 1 o 0 bo = c = + e r n and =3 . e 1 + 0 ⇒ = t 3 ( e r he +1 4 r und r 0 − + 1 + a x r a = ⇒ e c − s 2 1 e t l 4 − − ⇒ ⇒ general + c 1 he =1 c s c 2 x s ) ary o − an o ⇒ x 2 1) x 2 olution =3 r e s c = l . e →− solution = ution ⇒− 3 2 3 d 2 r 4 = 2 6 - , xe , r . ( v x 0 t r = so =0 r − r he a r ± an =1 − lue 4 =2 − s 1 = ∞ − x = c olution 2 1 d ⇒ i general . i si . 2 s 5 s 4)( i 2 1 , − 2 3 y prob ,s − ,s y n ± =1 y to c an 0 ⇒ ± 1 4 1 4 o = o r = √ r the π d x 4 and − ± y c − √ l / . =0 2 e 2 em t 2 i c (1 he 3 s s = 0 ,s r c 5 1 2) =4 i olution = x i 2 nitial-v y t i co ,s =1 he o ( − . is , c ge = c = = − 1 r T y o s 1 y e ne general cos hen 2 = − cos . y 3 = e 0 2 1 =3 ⇒− ) r − T and 4 = x a i − alu c and he x ,s s l b 4 1 ⇒ ( s y 2 1 e 4 1 y x e c o c + n e ol t − x ,s ( 1 o h = ( s + c c b pro 1+ u t/ 1 e s c co olu 4 1 1 o y t 1 2 r + 2 c solu i c c √ = ) =2 o y 2 1 ( s 2 k si blem 1 2 =4 t 10 n 2 t x e c c x ion he n = si =4 ) e 1 2 x/ i t tion ), + s 3 n co si and − , c + / 2 2 1 general y i th 4 1 i n r s c ( x s s + x 4s e 2 c = e + y to =2 y 3 )= 2 ⇒ c si .T c g 2 1 √ e = i 2 − c c = 2 2 the e n n x ( 3 2 1 xe n =1 1 h . x e e c t 1) e − c e s e . T 2 1 1 − x c ) r 1 olution i x/ − n √ e a n x 2 hen .T . x/ + + cos . x l − i 2 3= 2 Th tial-v = (2 ) T s 4 . .T 3 h c t c o x/ . hu . e b e 2 2 lu 4 − cos n e y h x 2 si y s s tio 1 2 e i o alue 2= (0) ( + s n x + t ,s n 8) lu he x .T n y c ot ti c +3 the √ i 2 (0) = 2 on solution s 2 y pr h e 3 h si e (0) − e oblem t c n g of n l x s 1 = eneral i .T 4 n and = t . 1 x h x h e . c ) to e is 1 . n SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 9

2 3x 29. r 6r +25=0 r =3 4i and the general solution is y = e (c1 cos 4x + c2 sin 4x).But1=y(0) = c1 − ⇒ ± 3π 3π and 2=y(π)=c1e c1 =2/e , so there is no solution. ⇒ 2 2x 31. r +4r +13=0 r = 2 3i and the general solution is y = e− (c1 cos 3x + c2 sin 3x).But ⇒ − ± π π 2=y(0) = c1 and 1=y = e− ( c2), so the solution to the boundary-value problem is 2 − 2x π y = e− (2 cos 3x e sin 3x). − 2 33. (a) Case 1 (λ =0): y00 + λy =0 y00 =0which has an auxiliary equation r =0 r =0 ⇒ ⇒ ⇒ y = c1 + c2x where y(0) = 0 and y(L)=0. Thus, 0=y(0) = c1 and 0=y(L)=c2L c1 = c2 =0. ⇒ Thus, y =0.

2 Case 2 (λ<0): y00 + λy =0has auxiliary equation r = λ r = √ λ (distinct and real since − ⇒ ± − √ λx √ λx λ<0) y = c1e − + c2e− − where y(0) = 0 and y(L)=0.Thus,0=y(0) = c1 + c2 ( )and ⇒ ∗ √ λL √ λL 0=y(L)=c1e − + c2e− − ( ). † √ λL √ λL √ λL Multiplying ( )bye − andsubtracting( )givesc2 e − e− − =0 c2 =0and thus ∗ † − ⇒ c1 =0from ( ). Thus, y =0for the cases λ =0and λ<0. ∗ 2 (b) y00 + λy =0has an auxiliary equation r + λ =0 r = i √λ y = c1 cos √λx+ c2 sin √λx ⇒ ± ⇒ where y(0) = 0 and y(L)=0.Thus,0=y(0) = c1 and 0=y(L)=c2 sin √λL since c1 =0.Sincewe