Second-OrderDifferential Equations
2.1TheLinearSecond-OrderEquation
1.Itisaroutineexerciseindifferentiationtoshowthat y1(x)and y2(x)are solutionsofthehomogeneousequation,while yp(x)isasolutionofthe nonhomogeneousequation.TheWronskianof y1(x)and y2(x)is
(x)= sin(6x)cos(6x) 6cos(6x) 6sin(6x)
andthisisnonzeroforall x,sothesesolutionsarelinearlyindependent ontherealline.Thegeneralsolutionofthenonhomogeneousdifferential equationis
Fortheinitialvalueproblem,weneed
so c2 = 179/36.And
(0)=2=6
so c1 =71/216.Theuniquesolutionoftheinitialvalueproblemis
2.TheWronskianof e4x and e 4x is
40 CHAPTER2.SECOND-ORDERDIFFERENTIALEQUATIONS
sothesesolutionsoftheassociatedhomogeneousequationareindependent.Withtheparticularsolution yp(x)ofthenonhomogeneousequation, thisequationhasgeneralsolution
Fromtheinitialconditionsweobtain
and
Solvethesetoobtain c1 =409/64and c2 =361/64toobtainthesolution
3.Theassociatedhomogeneousequationhassolutions e 2x and e x.Their Wronskianis
andthisisnonzeroforall x.Thegeneralsolutionofthenonhomogeneous differentialequationis
Fortheinitialvalueproblem,solve
(0)= 3= c
+ 15 2 and y (0)= 1= 2
c2 toget c1 =23/2,c2 = 22.Theinitialvalueproblemhassolution
(x)= 23
4.Theassociatedhomogeneousequationhassolutions
TheWronskianofthesesolutionsis
2.1.THELINEARSECOND-ORDEREQUATION
forall x.Thegeneralsolutionofthenonhomogeneousequationis
Tosatisfytheinitialconditions,itisrequiredthat
and
Solvethesetoobtain c1 = 7/8and c2 =15/8.Thesolutionoftheinitial valueproblemis
5.Theassociatedhomogeneousequationhassolutions
ThesehaveWronskian
sothesesolutionsareindependent.Thegeneralsolutionofthenonhomogeneousdifferentialequationis
Weneed
and
Solvethesetoget
6.Suppose y1 and y2 aresolutionsofthehomogeneousequation(2.2).Then
and
CHAPTER2.SECOND-ORDERDIFFERENTIALEQUATIONS
Multiplythefirstequationby y2 andthesecondby y1 andaddthe resultingequationstoobtain
WewanttorelatethisequationtotheWronskianofthesesolutions,which is
Thisisalinearfirst-orderdifferentialequationfor W .Multiplythisequationbytheintegratingfactor
toobtain We p(x) dx + pWe p(x) dx =0, whichwecanwriteas
We p(x) dx =0
Integratethistoobtain
We p(x) dx = k, with k constant.Then
Thisshowsthat W (x)=0forall x (if k =0),and W (x) =0forall x (if k =0).
Nowsupposethat y1 and y2 areindependentandobservethat
If k =0,then W (x)=0forall x andthequotient y2/y1 haszeroderivative andsoisconstant:
forsomeconstant c.Butthen y2(x)= cy1(x),contradictingtheassumptionthatthesesolutionsarelinearlyindependent.Therefore k =0and W (x) = forall x,aswastobeshown.
2.2.THECONSTANTCOEFFICIENTHOMOGENEOUSEQUATION 43
7.TheWronskianof x2 and x3 is W (x)= x2 x3 2x 3x2 = x 4
Then W (0)=0,while W (x) =0if x =0.Thisisimpossibleif x2 and x3 aresolutionsofequation(2.2)forsomefunctions p(x)and q(x).We concludethatthesefunctionsarenotsolutionsofequation(2.2).
8.Itisroutinetoverifythat y1(x)and y2(x)aresolutionsofthedifferential equation.Compute W (x)= xx2 12x = x 2 .
Then W (0)=0but W (x) > 0if x =0.However,towritethedifferential equationinthestandardformofequation(2.2),wemustdivideby x2 to obtain y 2 x y + 2 x2 y =0.
Thisisundefinedat x =0,whichisintheinterval 1 << 1,sothe theoremdoesnotapply.
9.If y2(x)and y2(x)bothhavearelativeextremum(maxormin)atsome x0 within(a,b),then y (x0)= y2(x0)=0
ButthentheWronskianofthesefunctionsvanishesat0,andthesesolutionsmustbeindependent.
10.Byassumption, ϕ(x)istheuniquesolutionoftheinitialvalueproblem y + py + qy =0; y(x0)=0.
Butthefunctionthatisidenticallyzeroon I isalsoasolutionofthisinitial valueproblem.Thereforethesesolutionsarethesame,and ϕ(x)=0for all x in I
11.If y1(x0)= y2(x0)=0,thentheWronskianof y1(x)and y2(x)iszeroat x0,andthesetwofunctionsmustbelinearlydependent.
2.2TheConstantCoefficientHomogeneousEquation
1.Fromthedifferentialequationwereadthecharacteristicequation λ2 λ 6=0, whichhasroots 2and3.Thegeneralsolutionis y(x)= c1e 2x + c2e 3x .
2.Thecharacteristicequationis
2 2λ +10=0
withroots1 ± 3i.Wecanwriteageneralsolution y(x)= c1ex cos(3x)+ c2ex sin(3x)
3.Thecharacteristicequationis
withrepeatedroots 3, 3.Then
isageneralsolution.
4.Thecharacteristicequationis
withroots0, 3,and y(x)= c1 +
isageneralsolution.
5.characteristicequation λ2 +10λ +26=0,withroots 5 ± i;general solution y(x)= c
6.characteristicequation λ2 +6λ 40=0,withroots4, 10;generalsolution
7.characteristicequation λ2 +3λ+18=0,withroots 3/2±3√7i/2;general solution
8.characteristicequation λ2 +16λ +64=0,withrepeatedroots 8, 8; generalsolution
9.characteristicequation λ2 14λ +49=0,withrepeatedroots7, 7;general solution y(x)= e 7x(c1 + c2x).
2.2.THECONSTANTCOEFFICIENTHOMOGENEOUSEQUATION 45
10.characteristicequation λ2 6λ+7=0,withroots3±√2i;generalsolution
(x)= c1e 3x cos(√2x)+ c2e 3x sin(√2x)
IneachofProblems11–20thesolutionisfoundbyfindingageneralsolution ofthedifferentialequationandthenusingtheinitialconditionstofindthe particularsolutionoftheinitialvalueproblem.
11.Thedifferentialequationhascharacteristicequation λ2 +3λ =0,with roots0, 3.Thegeneralsolutionis y(x)= c1 + c2e 3x
Choose c1 and c2 tosatisfy:
Then c2 = 2and c1 =5,sotheuniquesolutionoftheinitialvalue problemis
)=5
12.characteristicequation λ2 +2λ 3=0,withroots1, 3;generalsolution y(x)= c1ex
Solve
toget c1 =4and c2 =2.Thesolutionis
13.Theinitialvalueproblemhasthesolution y(x)=0forall x.Thiscan beseenbyinspectionorbyfindingthegeneralsolutionofthedifferential equationandthensolvingfortheconstantstosatisfytheinitialconditions.
14. y(x)= e2x(3 x)
15.characteristicequation λ2 + λ 12=0,withroots3, 4.Thegeneral solutionis
Weneed
and
Solvethesetoobtain
CHAPTER2.SECOND-ORDERDIFFERENTIALEQUATIONS
Thesolutionoftheinitialvalueproblemis
Thiscanalsobewritten
16.
17.
18.
19.
20.
21.(a)Thecharacteristicequationis
asarepeated root.Thegeneralsolutionis
(b)Thecharacteristicequationis
)=0,withroots α + ,α .Thegeneralsolutionis
Wecanalsowrite
Note,however,thatthecoefficientsinthedifferentialequationsin(a)and (b)canbemadearbitrarilyclosebychoosing sufficientlysmall.
2.2.THECONSTANTCOEFFICIENTHOMOGENEOUSEQUATION 47
22.With a2 =4b,onesolutionis y1(x)= e ax/2.Attemptasecondsolution y2(x)= u(x)e ax/2.Substitutethisintothedifferentialequationtoget
Because a2 4b =0,thisreducesto
(x)=0
Then u(x)= cx + d,with c and d arbitraryconstants,andthefunctions (cx + d)e ax/2 arealsosolutionsofthedifferentialequation.Ifwechoose c =1and d =0,weobtain y2(x)= xe ax/2 asasecondsolution.Further, thissolutionisindependentfrom y1(x),becausetheWronskianofthese solutionsis
andthisisnonzero.
23.Therootsofthecharacteristicequationare
Because a2 4b<a2 byassumption, λ1 and λ2 arebothnegative(if a2 4b ≥ 0),orcomplexconjugates(if a2 4b< 0).Therearethree cases.
Case1-Suppose λ1 and λ2 arerealandunequal.Thenthegeneral solutionis
)=
andthishaslimitzeroas x →∞ because λ1 and λ2 arenegative.
Case2-Suppose λ1 = λ2.Nowthegeneralsolutionis y(x)=(c1 + c2x)e λ1 x , andthisalsohaslimitzeroas x →∞
Case3-Suppose λ1 and λ2 arecomplex.Nowthegeneralsolutionis y(x)= c1 cos( 4b a2x/2)+
e ax/2 , andthishaslimitzeroas x →∞ because a> 0. If,forexample, a =1and b = 1,thenonesolutionis e( 1+√5)x/2,and thistendsto ∞ as x →∞.
2.3ParticularSolutionsoftheNonhomogeneous Equation
1.Twoindependentsolutionsof y + y =0are y1(x)=cos(x)and y2(x)= sin(x),withWronskian W (x)= cos(x)sin(x) sin(x)cos(x) =1 Let f (x)=tan(x)anduseequations(2.7)and(2.8).First,
x)sin(x) dx
sin2(x) cos(x) dx
1 cos2(x) cos(x) dx = cos(x) dx sec(x) dx =sin(x) ln | sec(x)+tan(x)| Next,
Thegeneralsolutionis
2.Twoindependentsolutionsoftheassociatedhomogeneousequationare
2.3.PARTICULARSOLUTIONSOFTHENONHOMOGENEOUSEQUATION49 and
Thegeneralsolutionis
Morecompactly,thegeneralsolutionis
(
)=
ForProblems3–6,somedetailsofthecalculationsareomitted.
3.Theassociatedhomogeneousequationhasindependentsolutions y1(x)= cos(3x)and y2(x)=sin(3x),withWronskian3.Thegeneralsolutionis y(x)= c1 cos(3x)+ c2 sin(3x)+4x sin(3x)+ 4 3 cos(3x)ln | cos(3x)|
4. y1(x)= e3x and y2(x)= e x,with W (x)= 4e 2x.With f (x)=2sin2(x)=1 cos(2x)
wefindthegeneralsolution y(x)= c1e 3x + c2e x 1 3 + 7 65 cos(2x)+ 4 65 sin(2x)
5. y1(x)= ex and y2(x)= e2x,withWronskian W (x)= e3x.With f (x)= cos(e x),wefindthegeneralsolution
y(x)= c1ex + c2e 2x e 2x cos(e x)
6. y1(x)= e3x and y2(x)= e2x,withWronskian W (x)= e 5x.Usethe identity 8sin2(4x)=4cos(8x) 1
indetermining u1(x)and u2(x)towritethegeneralsolution
y(x)= c1e 3x + c2e 2x + 2 3 + 58 1241 cos(8x)+ 40 1241 sin(8x)
InProblems7–16themethodofundeterminedcoefficientsisusedtofind aparticularsolutionofthenonhomogeneousequation.Detailsareincluded forProblems7and8,andsolutionsareoutlinedfortheremainderofthese problems.
7.Theassociatedhomogeneousequationhasindependentsolutions y1(x)= e2x and e x.Because2x2 +5isapolynomialofdegree2,attemptasecond degreepolynomial
yp(x)= Ax2 + Bx + C
forthenonhomogeneousequation.Substitute yp(x)intothisnonhomogeneousequationtoobtain
2A (2Ax + B) 2(Ax2 + Bx + C)=2x 2 +5.
Equatingcoefficientsoflikepowersof x ontheleftandright,wehavethe equations
2A =2(coefficientsof x 2) 2A 2B 0(coefficientsof x 2A 2B 2C =5(constantterm.)
Then A = 1,B =1and C = 4.Then yp(x)= x 2 + x 4
andageneralsolutionofthe(nonhomogeneous)equationis
8. y1(x)= e3x and y2(x)= e 2x areindependentsolutionsoftheassociated homogeneousequation.Because e2x isnotasolutionofthehomogeneous equation,attemptaparticularsolution yp(x)= Ae2x ofthenonhomogeneousequation.Substitutethisintothedifferentialequationtoget
4A 2A 6A =8, so A = 2andageneralsolutionis y(x)= c1
9. y1(x)= ex cos(3x)and y2(x)= ex sin(3x)areindependentsolutionsof theassociatedhomogeneousequation.Tryaparticularsolution yp(x)= Ax2 + Bx + C toobtainthegeneralsolution y(x)= c1ex cos(3x)+ c2ex sin(3x)+2x 2 + x 1.
2.3.PARTICULARSOLUTIONSOFTHENONHOMOGENEOUSEQUATION51
10.Fortheassociatedhomogeneousequation, y1(x)= e2x cos(x)and y2(x)= e2x sin(x).Try yp(x)= Ae2x toget A =21andobtainthegeneralsolution
(x)=
11.Fortheassociatedhomogeneousequation, y1(x)= e2x and y2(x)= e4x
Because ex isnotasolutionofthehomogeneousequation,attempta particularsolutionofthenonhomogeneousequationoftheform yp(x)= Aex.Weget A =1,soageneralsolutionis
(x)= c
12. y1(x)= e 3x and y2(x)= e 3x.Because f (x)=9cos(3x)(whichisnot asolutionoftheassociatedhomogeneousequation),attemptaparticular solution
yp(x)= A cos(3x)+ B sin(3x)
Thisattemptincludesbothasineandcosinetermeventhough f (x)has onlyacosineterm,becausebothtermsmaybeneededtofindaparticular solution.Substitutethisintothenonhomogeneousequationtoobtain A =0and B =1/2,soageneralsolutionis
y(x)=(c1 + c2x)e 3x + 1 2 sin(3x)
Inthiscase yp(x)containsonlyasineterm,although f (x)hasonlythe cosineterm.
13. y1(x)= ex and y2(x)= e2x.Because f (x)=10sin(x),attempt
yp(x)= A cos(x)+ B sin(x).
Substitutethisintothe(nonhomogeneous)equationtofindthat A =3 and B =1.Ageneralsolutionis
(x)= c1ex +
2
2x +3cos(x)+sin(x).
14. y1(x)=1and y2(x)= e 4x.Findingaparticularsolution yp(x)forthis problemrequiressomecare.First, f (x)containsapolynomialtermand anexponentialterm,sowearetemptedtotry yp(x)asaseconddegree polynomial Ax2 + Bx + C plusanexponentialterm De3x toaccountfor theexponentialtermintheequation.However,notethat y1(x)=1,a constantsolution,isonetermoftheproposedpolynomialpart,somultiply thispartby x totry
yp(x)= Ax3 + Bx2 + Cx + De3x
Substitutethisintothenonhomogeneousdifferentialequationtoget
6Ax +2B +9De3x 4(3Ax2 +2Bx + C +3De3x)=8x 2 +2e 3x .
CHAPTER2.SECOND-ORDERDIFFERENTIALEQUATIONS
Matchingcoefficientsofliketerms,weconcludethat
2B 4C =0(fromtheconstantterms)
6A 8B =0(fromthe x terms), 12A =8(fromthe x 2 terms) 3D =2 Then D =
andageneralsolutionofthenonhomogeneousequationis
Thiswillworkbecauseneither e2x nor e3x isasolutionoftheassociated homogeneousequation.Substitute yp(x)intothedifferentialequationand obtain A =1/3,B = 1/2.Thedifferentialequationhasgeneralsolution
16. y1(x)= ex and y2(x)= xex.Try
p(x)= Ax + B + C cos(3
Thisleadstothegeneralsolution
InProblems17–24thestrategyistofirstfindageneralsolutionofthedifferentialequation,thensolvefortheconstantstofindasolutionsatisfyingthe initialconditions.Problems17–22arewellsuitedtotheuseofundeterminedcoefficients,whileProblems23and24canbesolvedfairlydirectlyusingvariation ofparameters.
17. y1(x)= e2x and y2(x)= e 2x.Because e2x isasolutionoftheassociatedhomogeneousequation,use xe2x inthemethodofundetermined coefficients,attempting
p(x)= Axe2x + Bx + C.
2.3.PARTICULARSOLUTIONSOFTHENONHOMOGENEOUSEQUATION53
Substitutethisintothenonhomogeneousdifferentialequationtoobtain
Then A = 7/4,B = 1/4and C =0,sothedifferentialequationhas thegeneralsolution
Weneed
and
Then c1 =7/4and c2 = 3/4.Theinitialvalueproblemhastheunique solution
18. y1 =1and y2(x)= e 4x areindependentsolutionsoftheassociated homogeneousequation.For yp(x),try
p(x)= Ax + B cos(x)+ C sin(x),
withtheterm Ax because1isasolutionofthehomogeneousequation. Thisleadstothegeneralsolution
Nowweneed
and
.
Then c1 =3and c2 =2,sotheinitialvalueproblemhasthesolution
19.Wefindthegeneralsolution
Thesolutionoftheinitialvalueproblemis
CHAPTER2.SECOND-ORDERDIFFERENTIALEQUATIONS
20.1and e3x areindependentsolutionsoftheassociatedhomogeneousequation.Attemptaparticularsolution
yp(x)= Ae2x cos(x)+ Be2x sin(x)
ofthenonhomogeneousequationtofindthegeneralsolution
))
Thesolutionoftheinitialvalueproblemis
21. e4x and e 2x areindependentsolutionsoftheassociatedhomogeneous equation.Thenonhomogeneousequationhasgeneralsolution
Thesolutionoftheinitialvalueproblemis
22.Thegeneralsolutionofthedifferentialequationis
Itiseasiertofittheinitialconditionsspecifiedat x =1ifwewritethis generalsolutionas
Now
Solvethesetoobtain
3.Thesolutionof theinitialvalueproblemis
23.Thedifferentialequationhasgeneralsolution y(x)= c1ex + c2e x sin2(x) 2
Thesolutionoftheinitialvalueproblemis y(x)=4e x sin2(x) 2.
2.4.THEEULERDIFFERENTIALEQUATION
24.Thegeneralsolutionis
y(x)= c1 cos(x)+ c2 sin(x) cos(x)ln | sec(x)+tan(x)|, andthesolutionoftheinitialvalueproblemis
y(x)=4cos(x)+4sin(x) cos(x)ln | sec(x)+tan(x)|.
2.4TheEulerDifferentialEquation
DetailsareincludedwithsolutionsforProblems1–2,whilejustthesolutions aregivenforProblems3–10.Thesesolutionsarefor x> 0.
1.Readfromthedifferentialequationthatthecharacteristicequationis r 2 + r 6=0
withroots2, 3.Thegeneralsolutionis
2.Thecharacteristicequationis
withrepeatedroot 1, 1.Thegeneralsolutionis
Wecanalsowrite
for x> 0.
3.
8. y(x)= x 2(c1 cos(7ln(x))+ c2 sin(7ln(x)))
9. y(x)= 1 x12 (c1 + c2 ln(x))
10. y(x)= c1x 7 + c2x 5
11.Thegeneralsolutionofthedifferentialequationis
(x)= c1x 3 + c2x 7 .
Fromtheinitialconditions,weneed
(2)=8c1 +2 7 c2 =1and y (2)=3c122 7c22 8 =0
Solvefor c1 and c2 toobtainthesolutionoftheinitialvalueproblem y(x)= 7 10 x 2 3 + 3 10 x 2 7
12.Theinitialvalueproblemhasthesolution y(x)= 3+2x 2 .
13. y(x)= x2(4 3ln(x))
14. y(x)= 4x 12(1+12ln(x))
15. y(x)=3x6 2x4
16. y(x)= 11 4 x 2 + 17 4 x 2
17.With Y (t)= y(et),usethechainruletoget
(x)= dY dt dt dx = 1 x Y (t) andthen
Then x 2 y (x)= Y (t) Y (t).
SubstitutetheseintoEuler’sequationtoget Y (t)+(A 1)Y (t)+ BY (t)=0
Thisisaconstantcoefficientsecond-orderhomogeneousdifferentialequationfor Y (t),whichweknowhowtosolve.
18.If x< 0,let t =ln( x)=ln |x|.Wecanalsowrite x = et.Notethat dt dx = 1 x ( 1)= 1 x
justasinthecase x> 0.Nowlet y(x)= y( et)= Y (t)andproceedwith chainruledifferentiationsasinthesolutionofProblem17.First,
and
Then x 2 y (x)= Y (t) Y (t)
justaswesawwith x> 0.NowEuler’sequationtransformsto Y +(A 1)Y + BY =0.
Weobtainthesolutioninallcasesbysolvingthislinearconstantcoefficient second-orderequation.Omittingallthedetails,weobtainthesolutionof Euler’sequationfornegative x byreplacing x with |x| inthesolutionfor positive x.Forexample,supposewewanttosolve x 2 y + xy + y =0 for x< 0.Solvethisfor x> 0toget
y(x)= c1 cos(ln(x))+ c2 sin(ln(x))
Thesolutionfor x< 0is
y(x)= c1 cos(ln |x|)+ c2 sin(ln |x|).
19.Theproblemtosolveis
x 2 y 5dxy +10y =0; y(1)=4,y (1)= 6
Weknowhowtosolvethisproblem.Hereisanalternativemethod,usingthetransformation x = et,or t =ln(x)for x> 0(sincetheinitial conditionsarespecifiedat x =1).Euler’sequationtransformsto
However,alsotransformtheinitialconditions:
Y (0)= y(1)=4,Y (0)=(1)y (1)= 6.
Thisdifferentialequationfor Y (t)hasgeneralsolution
Y (t)= c1e 3t cos(t)+ c2e 3t sin(t).
(0)= c2 =4 and Y (0)=3c1 + c2 = 6, so c2 = 18.Thesolutionofthetransformedinitialvalueproblemis
Y (t)=4e 3t cos(t) 18e 3t sin(t).
Theoriginalinitialvalueproblemthereforehasthesolution
y(x)=4x 3 cos(ln(x)) 19x 3 sin(ln(x))
for x> 0.Thenewtwisthereisthattheentireinitialvalueproblem (includinginitialconditions)wastransformedintermsof t andsolvedfor Y (t),thenthissolution Y (t)intermsof t wastransformedbacktothe solution y(x)intermsof x
20.Suppose
x 2 y + Axy + By =0
hasrepeatedroots.Thenthecharacteristicequation
r 2 +(A 1)r + B =0
has(1 A)/2asarepeatedroot,andwehaveonlyonesolution y1(x)= x(1 A)/2 sofar.Foranothersolution,independentfrom y1,lookfora solutionoftheform y2(x)= u(x)y1(x).Then y2 = u y1 + uy1 and y2 = u y1 +2u y1 + uy1 .
2.4.THEEULERDIFFERENTIALEQUATION
Substitute y2 intothedifferentialequationtoget x 2(u y1 +2u y1 + uy1 )+ Ax(u y1 + uy1)+ Buy1 =0
Threetermsinthisequationcancel,because u(x 2 y1 + Axy1 + By1)=0
byvirtueof y1 beingasolution.Thisleaves x 2 u y1 +2x 2 u y1 + Axu y1 =0.
Assumingthat x> 0,divideby x toget xu y1 +2xu y1 + Au y1 =0.
Substitute y1(x)= x(1 A)/2 intothistoobtain xu x(1 A)/2 +2xu 1 A 2 x( 1 A)/2 + Au x(1 A)/2 =0
Dividethisby x(1 A)/2 toget xu +(1 A)u + Au =0, andthisreducesto
xu + u =0.
Let z = u toobtain xz + z =0, or (xz) =0.
Then xz = c,constant,so
z = u = c x
Then u(x)= c ln(x)+ d.Weonlyneedonesecondsolution,solet c =1 and d =0toget u(x)=ln(x).Asecondsolution,independentfrom y1(x), is
y2(x)= y1(x)ln(x), asgivenwithoutderivationinthechapter.
2.5SeriesSolutions
2.5.1PowerSeriesSolutions
1.Put y(x)= ∞ n
intothedifferentialequationtoobtain
Thisistherecurrencerelation.Ifweset
+1,weobtainthe coefficients
andsoon.Further,
andsoon.Thesolutioncanbewritten
2.Write
Therecurrencerelationis
with a0 arbitrary, a1 =4and a2 = a3 =0.Thisyieldsthesolution
3.Write
Therecurrencerelationis
4.Beginwith
Therecurrencerelationis
Taking a0 =1,a1 =0,weobtainthesolution
With a0 =0,a1 =1wegetasecond,linearlyindependentsolution
5.Write
Here a0 and a1 arearbitraryand a2 =(3 a0)/2.Therecurrencerelation is
Thisyieldsthegeneralsolution
6.Beginwith
Here a0 and a1 arearbitraryand a2 =0.Therecurrencerelationis
With a0 =0and a1 =1,weobtainasecond,linearlyindependentsolution
2.5.SERIESSOLUTIONS
7.Wehave
relationis
for n =4, 5, .Thegeneralsolutionhastheform
Notethat a0 = y(0)and a1 = y (0).Thethirdseriesrepresentsthe solutionobtainedsubjectto y(0)= y (0)=0.
8.UsingtheMaclaurinexpansionforcos(x),wehave
a0 isarbitraryand a1 =1.Therecurrencerelationis
CHAPTER2.SECOND-ORDERDIFFERENTIALEQUATIONS
for n =1, 2, .Thisyieldsthesolution
9.Wehave
for n =5, 6, .Thegeneralsolutionis
Here a0 = y(0)and a1 = y (0).
10.UsingtheMaclaurinexpansionof ex,wehave
Then a0 and a1 arearbitrary, a2 =0and an = (n 2)an 2 1/(n 2)! n(n 1)
2.5.SERIESSOLUTIONS
for n =3, 4, .Thisleadstothesolution
Notethat a0 = y(0)and a1 = y (0).
2.5.2FrobeniusSolutions
1.Substitute
intothedifferentialequationtoget xy +(1
=0
Because c0 isassumedtobenonzero, r mustsatisfytheindicialequation r2 =0,so r1 = r2 =0.Onesolutionhastheform
whileasecondsolutionhastheform
Forthefirstsolution,choosethecoefficientstosatisfy c0 =1and cn = n 2 n2 cn 1 for n =1, 2,
Thisyieldsthesolution y1(x)=1 x.Thesecondsolutionistherefore y2(x)=(1 x)ln(x)+
CHAPTER2.SECOND-ORDERDIFFERENTIALEQUATIONS
Substitutethisintothedifferentialequationtoobtain x 2 x 1 x x2 +(1 x) ln(x)+ 1 x x
x)ln(
)+
Thecoefficientsaredeterminedby c∗ 1 =3, c∗ 2 = 1/4,and c ∗ n = n 2 n2 for n =3, 4, .
Asecondsolutionis y2(x)=(1 x)ln(x)+3x ∞ n=2 1 n(n 1) xn .
2.Omittingsomeroutinedetails,theindicialequationis r(r 1)=0,so r1 =1and r2 =0.Therearesolutions y1(x)= ∞ n=0 cnxn+1 and y2(x)= ky1(x)ln(x) ∞ n=0 c ∗ nxn
For y1,therecurrencerelationis
cn = 2(n + r 2) (n + r)(n + r 1) cn 1 for n =1, 2, .With r =1and c0 =1,thisyields
y1(x)= x, asolutionthatcanbeseenbyinspectionfromthedifferentialequation. Forthesecondsolution,substitute y2(x)intothedifferentialequationto get (2c ∗ 0 + k)+2(c ∗ 2 k)x + ∞ n=1 (n(n 1)c ∗ n 2(n 2)c ∗ n 1)xn 1 =0.
Choose c∗ 0 =1toobtain k = 2. c∗ 1 isarbitrary,andwewilltake c∗ 1 =0. Finally, c∗ 2 = 2and c ∗ n = 2(n 2) n(n 1) c ∗ n 1 for n =3, 4,
Thisyieldsthesecondsolution
3.Theindicialequationis r2 4r =0,so r1 =4and r2 =0.Thereare solutionsoftheform y1(x)= ∞ n=0 cnxn+4 and y2(x)= ky1(x)ln(x)+ ∞ n=0 c ∗ nxn
With r =4therecurrencerelationis cn = n +1 n cn 1 for n =1, 2, ···
Then
Usingthegeometricseries,wecanobservethat y1(x)= x 4 d dx (1+ x + x 2 + x 3 + )
Thisgivesusthesecondsolution
4.Theindicialequationis4r2 9=0,withroots r1 =3/2and r2 = 3/2. Therearesolutions
Uponsubstitutingtheseintothedifferentialequation,weobtain
5.Theindicialequationis4r2 2r =0,withroots r1 =1/2and r2 =0. Therearesolutionsoftheform
1(x)=
Substitutetheseintothedifferentialequationtoget
and
6.Theindicialequationis4r2 1=0,withroots r1 =1/2and r2 = 1/2.
Therearesolutions y1(x)=
n=0 cnxn+1/2 and y2(x)= ky1(x)ln(x)+
Aftersubstitutingtheseintothedifferentialequation,weobtainthesimple solutions y1(x)= x 1/2 and y2(x)= x 1/2
Thesesolutionsareconsistentwiththeobservationthat,upondivisionby 4,thedifferentialequationisanEulerequation.
7.Theindicialequationis r2 3r +2=0,withroots r1 =2and r2 =1. Therearesolutions y(x)= ∞ n=0
Substitutetheseinturnintothedifferentialequationtoobtainthesolutions
and
Wecanrecognizetheseseriesas
y1(x)= x sinh(x)and y2(x)= xe x .
8.Theindicialequationis r2 2r =0,withroots r1 =2,r2 =0.Thereare solutions y1(x)=
Therecurrencerelationforthe cns is
, 4, ··· andweobtain,with c0 =1,
Forthesecondsolution,substitute y2(x)intothedifferentialequationto get
Setting c∗ 0 =1forsimplicity,weobtain c∗ 1 =2,k = 2, c∗ 2 arbitrary(we takethistobezero),andtherecurrencerelation
for n =3, 4, .Weobtainthesecondsolution
9.Theindicialequationis2r2 =0,withroots r1 = r2 =0.Thereare solutions
Uponsubstitutingtheseintothedifferentialequation,weobtaintheindependentsolutions
(
)=1
and
10.Theindicialequationis r2 1=0,withroots r1 =1and r2 = 1.There aresolutions y1(x)= ∞ n=0 cnxn+1 and y2(x)= ky1(x)ln(x)+ ∞ n=0 c ∗ nxn 1
Substituteeachoftheseintothedifferentialequationtoget
=1 ( 1)n(1 4 7 (3n 2)) 3nn!(5 8 11 (3n +2)) x 3n and
1(x)= x 1+
2(x)= 1 x 1+
n=1 ( 1)n+1(1 · 2 · 5 ··· (3n 1)) 3nn!(4 7 (3n 2)) x 3n .
Table1:Sumsofpairsofdice,Problem1,Section2.
3.Themeanis
Themedianisthesixteenthnumberfromtheleft,or1(thelast1tothe rightintheorderedlist).
Thestandarddeviationis
Section2RandomVariablesandProbabilityDistributions
1.Ifwerolltwodice,therearethirty-sixpossibleoutcomes.Thesumsof thenumbersthatcancomeuponthetwodicearelistedinTable1.
Forexample,if o istheoutcomethatonediecomesup2andtheother3, thenthesumofthediceis5,so X (o)=5.
Thetablegivesallofthevaluesthat X (o)cantakeon,overalloutcomes o oftheexperiment.Eachvalueislistedasoftenasitoccursasavalueof X .Forexample,4occursthreetimes,because X (o)=4forthreedifferent outcomes(namely(2, 2),(1, 3)and(3, 1)).
Defineaprobabilitydistribution P on X bybyletting P (x)betheprobabilityof x,foreachvalue x that X canassume.
Forexample,since2occursonceoutof36entriesinthistable,assignto thisvalueof X theprobability
P (2)= 1 36
Similarly,11occurstwice,sogivethevalue11of X theprobability
P (11)= 2 36 3
Since3occurstwiceinthetable, P (3)=2/36,andsoon.
Calculating P (n)foreach n inthetable,weobtain
Noticethat x P (x)=1,asrequiredforaprobabilityfunction.
Themeanof X is
Thisisinterpretedtomeanthat,onaverage,weexpecttocomeupwith asevenifwerolltwodice.Thisisareasonableexpectationinviewofthe factthattherearemorewaystoroll7thananyothersumwithtwodice.
Thestandarddeviationis
Tocomputethis,firstcompute
2.Flipfourcoins,withsixteenpossibleoutcomes.If o isanoutcome, X (o) canhaveonlytwovalues,namely1iftwo,three,orfourtailsarein o,or 3otherwise(onetailornotailsin o).Therearefiveoutcomeswithone tailornotail,andelevenwithtwoormoretails,so P (1)= 11 16 and P (3)= 5 16 .
Themeanis
Forthestandarddeviationof X ,compute
3.Wehave X (1)=0, X (2)= X (3)= X (5)= X (7)= X (11)= X (13)= X (17)= X (19)=1, X (4)= X (6)= X (9)= X (10)= X (14)= X (15)=2, X (8)= X (12)= X (18)= X (20)=3, X (16)=4
Thevaluesassumedby X are0, 1, 2, 3, 4.Fromthelistofvalues,weget P (0)= 1
Thesearetheprobabilitiesofthevaluesoftherandomvariable X . Themeanof X is
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Table2:Outcomesofrollingtwodice,Problem4,Section2.
Forthestandarddeviationof X ,firstcompute
σ = (0 9600)=0 9798
4.TheoutcomesoftworollsofthedicearedisplayedinTable2. Then
X (n,n)= n for n =1, 2, 3, 4, 5, 6, X (1, 2)= X (2, 1)= X (2, 4)= X (4, 2)= X (3, 6)= X (6, 3)=2, X (1, 3)= X (3, 1)= X (2, 6)= X (6, 2)=3, X (1, 4)= X (4, 1)=4,X (1, 5)= X (5, 1)=5,X (1, 6)= X (6, 1)=6, X (2, 3)= X (3, 2)= X (4, 6)= X (6, 4)=3/2, X (2, 5)= X (5, 2)=5/2,X (3, 4)= X (4, 3)=4/3, X (5, 3)= X (3, 5)=5/3,X (4, 5)= X (5, 4)=5/4, X (5, 6)= X (6, 5)=6/5.
Usingthislisttocomputeprobabilities,wefindthat P (1)= 1 36 ,P (2)= 7 36 ,P (3)= 5 36 , P (4)= 3 36 ,P (5)= 3 36 ,P (6)= 3 36 6
Themeanof X is
Using μ,wecancomputethestandarddeviationof X :
5.Drawcardsfroma(fifty-twocard)deck.Thereare 52 C2 waystodothis, disregardingorder.If o isanoutcomeinwhichbothcardsarenumbered, then X (o)equalsthesumofthenumbersonthecards.Ifexactlyone ofthecardsisafacecardorace,then X (o)=11,andifbothcards arechosenfromthefacecardsoraces,then X (o)=12.Thereforethe valuesof X (o)forallpossibleoutcomesare4, 5, ··· , 20.Byaroutine buttediouscountingofthewaysthenumberedcardscantakeonvarious possibletotals,weobtain P (4)= 6 1326 ,P (5)= 16 1326 ,P (6)= 22 1326 ,P (7)= 32 1326 , P (8)= 38 1326 ,P (9)= 48 1326 ,P (10)= 54 1326 ,P (11)= 640 1326 , P (12)= 190 1326 ,P (13)= 64 1326 ,P (14)= 54 1326 ,P (15)= 48 1326 , P (16)= 38 1326 ,P (17)= 32 1326 ,P (18)= 22 1326 ,P (19)= 16 1326 , P (20)= 6 1326 . 7
Usingthese,computethemeanof X : μ = x xP (x)=11.566
andthestandarddeviationof X :
= x (x μ)2 P (x)= √5.4289=2.33.
6. X takesonvalues1, 2,π .Inparticular, X (1)= X (5)= X (7)= X (11)= X (13)= X (17) = X (19)= X (23)= X (25)= X (29)= π, X (2)= X (4)= ··· = X (eveninteger)=1, X (3)= X (9)= X (15)= X (21)= X (27)=2
Theseenableustowritetheprobabilitydistribution
(1)= 15 30 = 1 2 ,P (2)= 5 30 = 1 6 ,P (π )= 10 30 = 1 3 .
Themeanof X is
andthisisapproximately1.8805.
Thestandarddeviationof X isthesquarerootof
whichisapproximately0.92014.Then
σ = √92014=0.95924.
Section3TheBinomialandPoissonDistributions
Since x is H or T ineach,thereare8+6 +4+2=20outcomesofthis experiment.Then
Pr(dicetotalatleast9)= 20 72 = 5 18
(b)Supposethedicetotalatleast9andthecoincomesupheads.Now (from(a))thereare10outcomes,so
Pr(dicetotalatleast9,coinis H )= 10 72 = 5 36 .
(c)Nowsupposethecoinisatailandbothdicecomeupthesame.Now therearesixoutcomes,namely
(1, 1,T ), (2, 2,T ), , (6, 6,T ) so
Pr(tail,bothdicethesame)= 6 72 = 1 12 .
(d)Supposethedicebothcomeupevenandthecoinisatail.Nowthe outcomeshavetheappearance(x,y,T ),where x and y canindependently be2,4or6.Thereare9outcomesofthisform,so
Pr(tail,bothdiceeven)= 9 72 = 1 8
Section2FourCountingPrinciples
1.Thefactthatthesearelettersofthealphabetthatwearearrangingin orderisirrelevant.Theissueisthatthereareninedistinctobjects.The numberofarrangementsis9!,whichis362, 880.
2.Thereare26lettersintheEnglishslphabet.Theproblemisoneofdeterminingthenumberofwaysofchoosing17objectsfrom26objects,with ordertakenintoaccount.Thisis
3.Sinceanyofthenineintegerscanbeusedinanyofthenineplacesof theIDnumber,thereare9waysthefirstdigitcanbechosen,9waysthe seconddigitcanbechosen,andsoon.Thetotalnumberofcodesis99 , or 3.87420489(10)8 . 5
4.Thefirstplan(alldifferentsymbolsinthedifferentplaces)allowsfor 5!=120passwords.Thesecondplan(choosingwithreplacement)allows for55 =3, 125passwords.
5.These7symbolshave7!=5, 040permutationsorarrangements. If a isfixedasthefirstsymbol,thentherearesixsymbolstochoosein anyorderfortheothersixplaces,andthenumberofchoicesis6!=720.
If a isfixedasthefirstsymbol,and g asthefifth,thenwehavefivesymbols lefttochooseinanyorderfortheremainingfiveplaces.Thenumberof waysofdoingthisis5!=120.
6.Wewanttofind n sothat n! ≥ 20, 000.Clearlythereareinfinitelymany suchintegers n,butwewouldlikethesmallestpossible n.Withalittle experimentation,wefindthat8!=40, 320,morethanenough,while7!= 5040istoosmall,sochoose n =8.
For n! ≥ 100, 000,trysomevaluesof n.Wefindthat10!=3, 628, 800, morethanenough,while9!=362, 880istoosmall.Themostefficient choiceinthiscaseis n =10.
7.Thereare12!=479, 001, 600outcomes.
8.Thereare10lettersfrom a through l ,inclusive.Ifthreeoftheplacesare fixed(itdoesnotmatterwhichthree),therearesevenlettersavailableto putinanyorderintotheremainingsevenplaces.Thereare7!=5, 040 waystodothis.
9.Wewanttopick7objectsfrom25,takingorderintoaccount.Thenumber ofwaystodothisis
25 P7 = 25!
18! =(19)(20)(21)(22)(23)(24)(25)=2, 422, 728, 000
10.Thenumberofballotsis
16 P5 = 16! 11! =12(13)(14)(15)(16)=524, 160
11.Becauseorderisimportant,thenumberofpossibilitiesis
22 P6 = 22! 16! =53, 721, 360
12.(a)Thenumberofchoicesis
20 P3 = 20! 17! =6, 840
(b)Ifthelistbeginswith4,thereareonlytwonumberstochoosefrom theremainingnineteennumbers.Thereare
19 P 2 = 19! 17! =(18)(19)=342
waystodothis.Thisresultdoesnotdependonwhichnumberisfixedin thefirstposition.Thepercentageofchoicesbeginningwith4is100(342/6, 840), or5percent.Thisisreasonablefromacommonsensepointofview,since, with20numbertochoosefrom,wewouldexpect5percenttobeginwith anyparticularoneofthenumbers.
(c)Thisquestionisreallythesameasthequestionin(b),sinceitdoes notmatterwhichnumberisfixed.Theansweristhat5percentofthe choicesendsin9.
(d)Iftwoplacesarefixedat3and15,thenthereareeighteennumbers left,fromwhichwewanttochooseone.Thereare18suchchoices.
13.Withoutorder(and,weassume,withoutreplacement),thenumberoften cardhandsis
14.Disregardingorder,thenumberofninemanlineupsthatcanbeformed fromaseventeenpersonrosteris
Iftheordermakesadifference,thenthenumberis
15.Thenumberofcombinationsis
16.Thenumberis
C12 =
12!28! =5, 586, 853, 480
17.Thenumberofoutcomesofflippingfivecoinsis25 =32.
(a)Thenumberofwaysofgettingexactlytwoheadsfromthefiveflips is 5 C2 =5!/(2!3!)=10.Theprobabilityofgettingexactlytwoheads(or exactlytwotails)is
Pr(exactlytwoheads)= 10 32 = 5 16 .
(b)Wegetatleasttwoheadsifwegetexactlytwo,orexactlythree,or exactlyfour,orexactlyfiveheads.Thesumofthenumberofwaysof doingeachoftheseis
Pr(atleasttwoheads)= 26 32 = 13 16 .
18.Iffourdicearerolled,thenumberofoutcomesis64 =1, 296.
(a)Weneedthenumberofwaysthefourdicecancomeupwithexactly twoofthediceshowing4.Wegetexactlytwodiceshowing4bychoosing anytwoofthefourdicetocomeup4,andallowingtheothertwodiceto comeupanyof1, 2, 3, 5, 6,fivepossibilitiesfortheothertwo.Thenumber ofwaysthiscanhappen,bythemultiplicationprinciple,is
Theprobabilityofgettingexactlytwofour’sis
Pr(exactlytwofour’s)= 150 1296 = 23 216 .
(b)Thenumberofwaysofgettingexactlythreefour’sis5(4 C1 )=20,so
Pr(exactlythreefour’s)= 20 1296 = 5 324
(c)Wegetatleasttwofour’sifwegetexactlytwofour’s,exactlythree four’s,orallfourtossescomingup4.Thereare150waysofgetting exactlytwo,and20waysofgettingexactlythreefour’s.Clearlyinfour tossesthereisonewayofgettingfourfour’s.Therefore
Pr(atleasttwofour’s)= 150+20+1 1296 = 19 144
(d)Tototal22,thedicecouldcomeup6, 6, 6, 4,inanyorder(fourways), or6, 6, 5, 5inanyorder(sixways).Therefore
Pr(dicetotal22)= 10 1296 = 5 648
19.Thenumberofwaysofdrawingtwocardsoutof52cards,withoutregard toorder,is 52 C2 =1, 326.
(a)Wegettwokingsifwehappentogettwoofthefourkings,andthere are 4 C2 =6waysofdoingthis.Then
Pr(twokingsaredrawn)= 6 1326 = 1 221
(b)Theacesandfacecardsconstitute16ofthe52cards.Ifnoneofthe twocardsisdrawnfromthesesixteencards,thenthetwocardsaredrawn fromtheremaining36cards.Disregardingorder,thereare 36 C2 =630 waystodothis.Therefore
Pr(noaceorfacecard)= 630 1326 = 105 221 . 8
20.Thenumberofwaysofchoosingfourlettersfrom26letters,withorder,is
26 P4 = 26! (26 4)! = 26! 22! =(23)(24)(25)(26)=14, 950
(a)Ifthefirstletterissetat q ,thenwearelefttochoose,withorder, threelettersfromtheremaining25letters.Thereare 25 P3 =2, 300ways todothis,so
Pr(firstletterisq)= 2300 14950 = 2 13 .
(b)Suppose a and b aretwooftheletters.Howmanywayscanthisoccur? Imagineastringoffourboxes.Pickanytwo,andput a and b inthese boxes.Thereare 4 P2 =12waystodothis.Fortheothertwoboxes,put (keepingtrackoftheorder)anytwooftheremaining24letters.There are 24 P2 waystodothis.Thenumberoforderedstringsoffourletters with a and b twooftheletters,is (4 P2 )(24 P2 )= 4! 2! 24! 22! =564
Then
Pr(a and b werechosen)= 560 14950 = 7 187
(c)Theprobabilityof abdz occurringis1/14950,sincethisstringcan occurinexactlyoneway.
21.Thenumberofwaysofchoosing,withoutorder,threeoftheeightbowling ballsis 8 C3 =56.
(a)Fornoneoftheballstobedefective,theyhadtocomefromthesix nondefectiveones.Thereare 6 C3 =20waystodothis.Then
Pr(nonedefective)= 20 56 = 5 14
(b)Therearetwowaystotakeonedefectiveball.Theothertwowould havetobetakenfromthesixnondefectiveones,whichcanbedonein 6 C2 =15ways.Then
Pr(exactlyonedefectiveball)= 30 56 = 15 28
(c)Inchoosingthreebowlingballs,thereare3waysofpickingthetwo defectiveones.Thentherearesixwaysofchoosingthethirdballas nondefective.Therefore
Pr(bothdefectiveballsarechosen)= 6(3) 56 = 9 28
22.Thisprobleminvolvestossingdice,exceptnoweachdiehasonlyfour faces,numbered1, 2, 3, 4.Thenumberofoutcomesis47 =16, 384,since eachoftheseventosseshasfourpossibilities.
(a)Thereisonewayallsevendicecancomeup3,so
Pr(allcomeup3)= 1 16384
(b)Thereare 7 C5 =21waysofpickingfiveofthediceandimaginingthey comeup1,andimaginingtheothertwocomeup4.Therefore
Pr(fiveones,twofours)= 21 16384
(c)Theonlywaystorollatotalof26aretoroll(inanyorder),
4, 4, 4, 4, 4, 4, 2
andtherearesevenwaystodothis(sevenpossiblelocationsforthe2),or toroll
4, 4, 4, 4, 4, 3, 3
andthereare 7 C2 =21waystodothis.Therefore
Pr(total26)= 7+21 16384 = 7 4096 .
(d)Thesumisatleast26ifthesumis26,27,or,thelargestpossible,28. Wealreadyknowfrom(c)thatthereare28waystototalexactly26.To rolla27,wemustget(inanyorder),
4, 4, 4, 4, 4, 4, 3
andthereare7waystodothis.Toroll28,wemustgetallfour’s,and thereisonewaytodothis.Therefore
Pr(totalatleast26)= 28+7+1 16384 = 9 4096
23.Takingorderintoaccount,thereare
20 P5 = 20! 15! =(16)(17)(18)(19)(20)=1, 860, 480 waystochoose5oftheballs.
(a)Thereisonlyonewaytochoosetheballsnumbered1, 2, 3, 4, 5inthis order.theprobabilityis
Pr(select1, 2, 3, 4, 5inthisorder)= 1 1860480 .
(b)Wewanttheprobabilitythatballnumber3wasdrawn(somewherein thefivedrawings).Wethereforeneedthenumberoforderedchoicesoffive ofthetwentynumbers,thatincludethenumber3.Wecanthinkofthis aschoosing,inorder,fourofthenineteennumbers1, 2, 4, 5, , 18, 19, 20, andtheninserting3inanyofthepositionsfromthefirstthroughfifth numbersofthedrawing.Thiswillresultinallorderedsequencesoflength fivefromthetwentynumbers,andcontainingthenumber3insomeposition.Therearetherefore5(19 P4 )=465, 120suchsequences.Therefore
Pr(selectinga3)= 465120 1860480 = 1 4
(c)Wemustcountallthedrawings(sequences)thatcontainatleastone evennumber.Thiscouldmeanthesequencecontainsone,two,three, fourorfiveevennumbers.Thisisacomplicatedcountingproblemif approacheddirectly.Itiseasiertocountthesequencesthathavenoeven number,henceareformedjustfromthetenevenintegersfrom1through 20.Ifthisnumberis N0 ,thenthenumber X ofsequenceshavinganeven numberis
X =1, 860, 480 N0 .
Now N0 iseasytocompute,sincethisisthenumberoforderedfive-term sequencesofthetenoddnumbers.Thus
N0 =10 P5 =30, 240
Then X =1860480-30240=1830240.
Then
Pr(anevennumberwasdrawn)= 1830240 1860480 . Thisisapproximately0.984,soitisverylikelythatanevennumbered ballwasdrawn.
24.Weneedtobeclearwhatanoutcomeofthisexperimentlookslike.An outcomecanbewrittenasastring abc,witheachletterrepresenting(in someway)achosendraweroutofthenineavailabledrawers.Thenumber ofoutcomes,withoutregardtoorder,is 9 C3 =84.
(a)Apersongetsatleast1,000dollarsbypickingexactlyonedrawerwith thethousanddollarbillandtwowithout.Thereare2(7 C2 )waystodo this.Or,wecouldpickbothdrawerswiththethousanddollarbilland oneoftheothers.Thereare7waystodothis.Thenumberofwaysof gettingatleastathousanddollarsistherefore42+7,and
Pr(gettingatleastonethousanddollars)= 49 84 = 7 12
(b)Apersoncannotendupwithlessthanonedollar.Thisisnotan outcomeinthesamplespace.Wemayalsoassignaprobabilityofzeroto thisproposedoutcome.
(c)Thepayoffis1.50exactlywhenthepersonchoosesthreedrawers,each containingfiftycents.Thenumberofwaysthiscanhappenis 7 C3 =35, so
Pr(1.50payoff)= 35 84 = 5 12
Noticethatthesumoftheprobabilitiesof(a)and(c)is1.Thisisbecause, inchoosingthreedrawers,thepersonmusteitherchoosealldrawershaving fiftycents,ormustchooseatleastoneofthedrawershavingathousand dollarbill.
25.Thenumberoffive-cardhands(disregardingtheorderofthedeal)is
52 C5 =2, 598, 960.
(a)Thenumberofhandscontainingexactlyonejackandexactlyoneking is
4(4)(44 C3 )=211, 904
Thisisbecausetherearefourwaysofgettingonejack,fourwaysofgettingoneking,andthenwechoose(withoutorder)threecardsfromthe remaining44cards.Thus
Pr(exactlyonejackandexactlyoneking)= 211904 2598960
Thisisapproximately0 082,sothiseventisquitelikely.
(b)Thehandwillcontainatleasttwoacesifithasexactlytwoaces, exactlythreeaces,orexactlyfouraces.Thenumberofsuchhandsis
)=108, 336
Forthefirstterm,choosetwoacesoutoffour,thenthreecardsfrom theremainingforty-eightcards,andsimilarlyfortheothertwotermsfor drawingthreeacesordrawingfouraces.Then
Pr(drawatleasttwoaces)= 108336 2598960 .
Thisisapproximately0 042.Aswemightexpect,ahandwithatleast twoacesisnotverylikely.
26.Thereare101numberstochoosefrom,and 101 C2 =668, 324, 943, 343, 021, 950, 370
waystodothis(withoutregardtoorder).
(a)Therearetwenty-onenumbersinthe80to100range,inclusive.The numberofwaysofchoosingtwentyofthese(withoutorder)is 21 C20 =21. Therefore
Pr(allnumbersarelargerthan79)= 21 668324943343021950370 . 12
Thisisveryclosetozero.Intherealworldnosanepersonwouldbeton drawingallthetwentynumbersfromtheeightytoonehundredrange. (b)Thereare
100 C19 =132, 341, 52, 939, 212, 267, 400 waystochooseafive:chooseonefive,thennineteennumbersfromthe remainingonehundrednumbers.Therefore
Pr(chooseafive)= 132341572939212267400 668324943343021950370
Thisisapproximately0 190.Sincetwentynumbersisnearly1/5the101 numbers,itisnotsurprisingthattheprobabilityofgettinganyparticular numberiscloseto1/5.
Section3ComplementaryEvents
1.Therearemanywayssevendicecancomeupwithatleasttwofours(call thisevent E ).Wecancountthese,butitmaybeeasiertolookatthe complementaryevent E C ,whichisthatfewerthantwodicecomeup4. Thisistheeventthatexactlyonediecomesup4,ornoneofthemdo. Ifno4comesup,theneachofthesevendicehasfivepossiblenumbers showing,for57 possibilities.Ifexactlyonediecomesupwithafour,then theotherdicehavefivepossiblenumbersshowing,andthiscanoccurin 7(56 )ways.Thismeansthat E C has
57 +7(56 )=187, 500
outcomes.Then,sincethetotalnumberofoutcomesofsevenrollsis 67 =279, 939,wehave
Pr(E C )= 187500 279936 .
Theprobabilityweareinterestedinis
Pr(E )=Pr(atleastone4)=1 187500 279936
Thisisapproximately0.330.
2.Infourteencointosses,thereare214 =16, 384outcomes. Nowlet E betheeventthatatleastthreeofthefourteencoinscomeup heads. E hasmanyoutcomesinit.Itmaybeeasiertodealwith E C , whichistheeventthatfewerthanthreeofthefourteencoinscomeup heads. E C consistsoftheevents:noheadcomesup,oneheadcomesup, ortwoheadscomeup.
CompetingSpeciesPopulationModels
Inacompetingspeciesmodel,twopopulationscompetewitheachotherfor thesameresource.Wewilldeveloptwomodelsforanalyzingthebehaviorof thesepopulations.
ASimpleCompetingSpeciesModel
Arelativelysimplecompetingspeciesmodelcanbeobtainedbyassuming thatanincreaseineitherpopulationcausesareductionintheresourceforboth populations,henceshouldcontributetoadeclineofbothpopulations.The effectofthisinteractionismodeledastheproductofthepopulations,leading tothesystem
inwhich x(t)and y(t)arethepopulationsattime t andthecoefficientsare positiveconstants.Inmatrixform,
Thecriticalpointsare(0, 0)and(k/c,a/b).
Example1 Wewillanalyzethecompetingspeciesmodel
Thecriticalpointsare(0, 0)and( 40 7 , 20 3 ).
Firstlookat(0, 0).Thematrixofthelinearizedsystemis
A(0,0) = 20 04
witheigenvalues2, 4.Theoriginisanunstablenodalsourceforthissystem. Figure1isaphaseportraitshowingtrajectoriesmovingoutofandawayfrom theorigin.
Attheothercriticalpoint,thematrixofthelinearizedsystemis
(40/7,20/3) = 0 12 7 14 3 0 ,
witheigenvalues ±2√2.Thiscriticalpointisanunstablesaddlepointofthe linearizedmodel,hencealsooftheoriginalcompetingspeciesmodel.Figure2 isaphaseportraitofthiscompetingspeciessystem,showingsometrajectories
Figure1:TrajectoriesneartheorigininExample1.
Figure2:Trajectoriesnear(40/7, 20/3)inExample1.
nearthecriticalpoint( 40 7 , 20 3 ).Iftheinitialconditionsare“near”thiscritical point,say x(0)=15,y(0)=5
thenthe x populationprospersandgrowsindefinitelyas t increases,whilethe y populationdiesout.
Thisfirstcompetingspeciesmodelcanbereducedtoasingledifferential equationintermsof x and y.ForthesystemofExample1,firstwrite
andthevariablesseparatetoyield
Integrateandrearrangetermstoobtaintheimplicitlydefinedgeneralsolution
Initialconditions x(0)and y(0)determine k andthetrajectorythroughthis point.
Ingeneral,forthiscompetingspeciesmodel,theasymptotesofthetrajectoriesseparatethefirstquadrantintofourregions,labeled 1,2,3,4 inFigure3. Iftheinitialpopulationpoint(x(0),y(0))isinregion 1 or 4,thenthe x populationwillincreaseintimewhilethe y populationshrinkstozeroas t →∞.If theinitialpopulationpointisinregion 2 or 3,thenthe y populationsurvives andthe x populationbecomesextinct.
AnExtendedCompetingSpeciesModel
Thecompetingspeciesmodeljustconsideredleavesnoroomforcompromise ordiplomacy-onespeciessurvives,theotherdies.Wewouldlikeamore sophisticatedmodelallowingagreatervarietyofoutcomes(asoccurinthereal world).
Onewaytodothisistoaddatermtoeachequationthataccountsforfactors withineachpopulationthatmightlimititsgrowth,independentofinteractions withtheotherpopulation,whicharealreadytakenintoaccountbythe xy terms. Assumingthatsuchfactorsareproportionaltothesquareofthepopulation,we havethecompetingspeciesmodel X = Gx Dx2 Cxy, gy dy2 cxy
Unlikethepreviousmodel,theseequationsdonotadmitexponentialgrowthor decayintheabsenceoftheotherpopulation.Thinkof D and d astheinternal growth-limitingfactors,while C and c arethecompetitionfactors.
Figure3:Phaseportraitofasimplecompetingspeciesmodel.
Wewilllookattwoexamplesbeforemakingageneralanalysisofthismodel.
Example2 Themodel
= f (x,y) g(x,y) =
hasfourcriticalpointsinthefirstquadrant: (0, 0), (0, 4), ( 18 5 , 0), and( 6 5 , 8 5 )
Forbehavioroftrajectoriesnear(0, 0),lookat A(0,0) = fx(0, 0) fy (0, 0) gx(0, 0) gy (0, 0) = 3 5 0 01
Thishaseigenvalues1and 3 5 sotheoriginisanunstablenodalsource. For(0, 4),form A(0,4) = 2 5 0 2 1
Thishaseigenvalues 2 5 , 1,so(0, 4)isanasymptoticallystablenodalsink. Solutionsbeginning“near”(0, 4)tendtoward(0, 4)as t increases.Becausewe areonlyinterestedinintegervaluesof x(t)and y(t)aspopulationcounts,this
Figure4:PhaseportraitinExample2.
istheobviousresultthat,if x(0)=0,and y(0)=4,thenthepopulationstend toward x =0,y =4as t increases.
For( 18 5 , 0),compute
Thishaseigenvalues 3 5 , 4 5 andthiscriticalpointisalsoanasymptotically stablenodalsink.
Finally,forthecriticalpoint( 6 5 , 8 5 ),wefindthat
witheigenvalues 4 5 , 1 5 .Thiscriticalpointisanunstablesaddlepoint.
Figure4showsaphaseportraitforthissystem.Trajectoriesflowoutward fromtheoriginand,dependingonwheretheystartattimezero,flowtoward thecriticalpoint(0, 4)(so y survivesand x becomesextinct),ortoward( 18 5 , 0) (so x survivesand y becomesextinct).Someofthesetrajectoriesalsosuggest thebehaviorofthesystemnearthesaddlepoint( 6 5 , 8 5 ).Becausethispoint isunstable,itispossibletofindinitialpointsclosetothiscriticalpointfrom whichthe x speciessurvivesandthe y speciesdoesnot,orfromwhichthe y speciessurvivesandthe x speciesdiesout.
Example3 ContrasttheoutcomesofExample2withthoseofthemodel
Thecriticalpointsare (0, 0), (0, 4), (3, 0)and( 24 11 , 36 11 )
Analyzeeachcriticalpointasfollows.
For(0, 0),compute
(0,0) =
02 , witheigenvalues3, 2.Theoriginisanunstablenodalsource. For(0, 4),
witheigenvalues 2, 2.Then(0, 4)isanunstablesaddlepoint. For(3, 0),wehave
witheigenvalues 3, 3 2 .Then(3, 0)isalsoanunstablesaddlepoint.
Finally,at( 24 11 , 36 11 ),
witheigenvalues 21±3√5 11 .Thesenumbersarebothnegative,sothiscritical pointisanasymptoticallystablenodalsink.Figure5near(6/5, 8/5)showsa phaseportraitforthissystemwithsometrajectoriesnearthiscriticalpoint.
Itispossibletocarryoutageneralanalysisforthecompetingspeciesmodel X = Gx Dx2 Cxy, gy dy2 cxy
Firstlookatthecriticalpointsfromageometricpointofview.Thesecritical pointsaresimultaneoussolutionsof x(G Dx Cy)=0, y(g cx dy)=0,
Solutionsfor x and y arecoordinatesofpointsofintersectionofpairsoflines, namelythe x ,axis,the y axis,andthelines Dx + Cy = G cx + dy = g.
Figure5:PhaseportraitinExample3.
Figure6showsthefourpossiblerelativepositionsofthelasttwolinesinthe firstoctant.Thecriticalpointsare
(0, 0), ( G D , 0), (0, g d ), and
Nowconsiderthecriticalpointsinturn.
(0, 0)-Writethesystemas A(0,0) = G 0 0 K X + Dx2 Cxy dy2 cxy
.
Thematrixofthelinearparthaseigenvalues G,g,bothpositive.Iftheseare unequalthentheoriginisanunstablenode.If G = g thentheoriginisan unstablenode.
( G D , 0)-Nowcompute A(G/D,0) = G CG/D 0 cG/D
witheigenvalues g,g cG D .Certainly G< 0.If g c > G D thenthesecond eigenvalueispositiveandthiscriticalpointisanunstablenode.If g c < G D then botheigenvaluesarenegative.Iftheyaredistinct,thenthecriticalpointisan
asymptoticallystablenode.Iftheyareequal,thenthealmostlinearsystemhas anasymptoticallystablenodeorspiralpointat( G D , 0).
(0, g d )-Nowwefindthat
(0,g/d) = G Cg
,
witheigenvalues g,G Cg d .If G/C>g/d thenthesecondeigenvalueis positiveand(0,g/d)isanunstablenode.If G/C<g/d thenbotheigenvalues arenegative.Iftheseeigenvaluesaredistinct,then(0,g/d)isanasymptotically stablenode.Iftheseeigenvaluesareequal,thenthealmostlinearsystemasan asymptoticallystablenodeorspiralpointat(0,g/d).
((Gd Cg)/(dD cC), (Dg cG)/(dD cD))-Denotethispoint(˜x, y).
Thisisthepointofintersectionofthelines Dx + Cy = G,cx + dy = g.
Inthepresentcontext,lookatthecasethatthepointofintersectionfallsinthe firstquadrant(Figures6(3)andFigure6(4)).InFigure6(3),
whileinFigure6(4),
Nowcompute
Butrecallthat G = Dx + Cy and
+ dy. Then A(˜x,y) = Dx Cx cy dy , witheigenvalues 1 2 (Dx + dy) ± (Dx + dy)2 4(Dd Cc)˜xy .
Theseeigenvaluescanbewritten 1 2 (Dx + dy) ± (Dx dy)2 +4Ccxy .
Thisformulationmakesitclearsthattheseeigenvaluesarereal.Twocases occur.
Figure6:Relativepositionsoflines(1 ) Dx + Cy =0and(2 ) cx + dy =0.
If Dd = Cc< 0,thenoneeigenvalueispositiveandtheothernegative.In thiscase(˜x, y)isanunstablesaddlepoint.Ifthepopulationpoint(x(0),y(0)) startsnearenoughtothiscriticalpoint,onepopulationwilldieoutwithtime andtheotherwillsurvive.ThiscasecorrespondstoFigure6(3).Thecondition Dd<Cd)canbeinterpretedtomeanthattheproductoftheinternallimiting factorsislessthantheproductofthecompetitionfactors.Thecompetition factorstendingtoincreasethepopulationsaredominantinthemodelandonly onepopulationsurvives.
If Dd Cd> 0thenbotheigenvaluesarenegativeand(˜x, y)isanasymptoticallystablenode(Figure6(4)).If(x(0),y(0)issufficientlyclosetothisnode, trajectoriesthroughthisinitialpointapproachthenodeinthelimitandboth speciessurvive(coexistence).Now Cd<Dd andthecompetitionfactorsare lessimportantthaninternallimitingfactors.Withcompetitionplayinglessof aroleinthepopulation,mutualsurvivalcanoccur.
Problems
EachofProblems1–6dealswiththesimplecompetingspeciesmodeltreated firstinthissection.Ineach,(a)determinethecriticalpointsinthefirstoctant andclassifytheirtypeandstabilityproperties,(b)drawaphaseportraitfor
thesystem,(c)interpretthesurvivalprospectsforeachspecies,asdictatedby themodel.
1. x = x 4xy,yprime =3y 6xy
2. x =3x xy,y =4y 10xy
3. x =3x 2xy,y =6y 2xy
4. x =8x 3xy,y =2y 7xy
5. x =3x 7xy,y = y 4xy
6. x =4x 9xy,y =5y 2xy
Problems7–12dealwiththeextendedcompetingspeciesmodel.Ineach, (a)determinethecriticalpointsinthefirstoctantandthetypesandstability properties,(b)drawaphaseportraitforthesystem,(c)interpretthesurvival prospectsforeachspecies,asdictatedbythemodel.
7. x =7x 5x2 2xy,y =3y 4y2 xy
8. x =4x 7x2 xy,y = y 2y2 3xy
9. x =2x 3x2 xy,y =3y y2 2xy
10. x = x 9x2 3xy,y = y y2 xy
11. x =3x x2 2xy,y = y 2y2 xy
12. x = x 6x2 3xy,y =5y 2y2 8xy