Solutions Manual Advanced Engineering Mathematics 10th edition by Erwin Kreyszig

PART A

Ordinary Differential Equations(ODEs)

Chap.1First-OrderODEs

Sec.1.1BasicConcepts.Modeling

Togetagoodstartintothischapterandthissection,quickly reviewyourbasiccalculus.Takealookat thefrontmatterofthetextbookandseeareviewofthemaindifferentiationandintegrationformulas.Also, Appendix3,pp.A63–A66,hasusefulformulasforsuchfunctionsasexponentialfunction,logarithm,sine andcosine,etc.Thebeautyofordinarydifferentialequationsisthatthesubjectisquitesystematicandhas differentmethodsfordifferenttypesofordinarydifferentialequations,asyoushalllearn.Letusdiscuss someExamplesofSec.1.1,pp.4–7.

Example2,p.5.SolutionbyCalculus.SolutionCurves. Tosolvethefirst-orderordinary differentialequation(ODE)

y = cos x

meansthatwearelookingforafunctionwhosederivativeiscos x .Yourfirstanswermightbethatthe desiredfunctionissin x ,because(sin x ) = cos x .Butyouranswerwouldbeincompletebecausealso (sin x + 2) = cos x ,sincethederivativeof2andofanyconstantis0.Hencethecompleteansweris y = cos x + c ,where c isanarbitraryconstant.Asyouvarytheconstantsyougetaninfinitefamily ofsolutions.Someofthesesolutionsareshownin Fig.3.Thelessonhereisthatyoushouldnever forgetyourconstants!

Example4,pp.6–7.InitialValueProblem. Inaninitialvalueproblem(IVP)forafirst-orderODE wearegivenanODE,here y = 3y ,andaninitialvaluecondition y (0) = 5 7.Forsuchaproblem,the firststepistosolvetheODE.Hereweobtain y (x ) = ce 3x asshownin Example3,p.5.Sincewealso haveaninitialcondition,wemustsubstitutethatconditionintooursolutionandget y (0) = ce 3 0 = ce 0 = c · 1 = c = 5 7 Hencethecompletesolutionis y (x ) = 5 7e 3x .Thelessonhereisthatforan initialvalueproblemyougetauniquesolution,alsoknownasaparticularsolution.

Modeling meansthatyouinterpretaphysicalproblem,setupanappropriatemathematicalmodel, andthentrytosolvethemathematicalformula.Finally,youhavetointerpretyouranswer. Examples3(exponentialgrowth,exponentialdecay)and5(radioactivity)areexamplesofmodeling problems.Takeacloselookat Example5,p.7,becauseitoutlinesallthestepsofmodeling.

ProblemSet1.1.Page8

3.Calculus. FromExample3,replacingtheindependentvariable t by x weknowthat y = 0 2y hasa solution y = 0 2ce 0 2x .Thusbyanalogy, y = y hasasolution

where c isanarbitraryconstant.

Anotherapproach(tobediscussedindetailsinSec.1.3)istowritetheODEas

andthenbyalgebraobtain

Integratebothsides,andthenapplyexponentialfunctionsonbothsidestoobtainthesame solutionasabove

Thetechniqueusediscalled separationofvariables becauseweseparatedthevariables,sothat y appearedononesideoftheequationand x ontheothersidebeforeweintegrated.

7.Solvebyintegration. Integrating y = cosh5.13x weobtain(chainrule!) y = cosh5.13xdx = 1 5 13 (sinh5.13x ) + c .Check:Differentiateyouranswer:

11.Initialvalueproblem(IVP). (a) Differentiationof y = (x + c )e x byproductruleanddefinitionof y gives

ButthislookspreciselylikethegivenODE y = e x + y .Hencewehaveshownthatindeed y = (x + c )e x isasolutionofthegivenODE. (b) Substitutetheinitialvalueconditioninto thesolutiontogive y (0) = (0 +

(c) Thegraphintersectsthe x -axisat x = 0 5andshootsexponentiallyupward.

19.Modeling:FreeFall. y = g = constisthemodeloftheproblem,anODEofsecondorder. IntegrateonbothsidesoftheODEwithrespectto t andobtainthevelocity v = y = gt + c1 (c1 arbitrary).Integrateoncemoretoobtainthedistancefallen y = 1 2 gt 2 + c1 t + c2 (c2 arbitrary). Todothesesteps,weusedcalculus.Fromthelastequationweobtain y = 1 2 gt 2 byimposingthe initialconditions y (0) = 0and y (0) = 0,arisingfromthestonestartingatrestatourchoiceoforigin, thatistheinitialpositionis y = 0withinitialvelocity0.Fromthiswehave y (0) = c2 = 0and v(0) = y (0) = c1 = 0.

Sec.1.2GeometricMeaningof y = f ( x,y ).DirectionFields,Euler’sMethod

ProblemSet1.2.Page11

1.Directionfield,verificationofsolution. Youmayverifybydifferentiationthatthegeneral solutionis y = tan(x + c )andtheparticularsolutionsatisfying

1is y = tan x .Indeed,forthe particularsolutionyouobtain

15.Initialvalueproblem.Parachutist. Inthissectiontheusualnotationis(1),thatis, y = f (x , y ), andthedirectionfieldliesinthe xy -plane.InProb.15theODEis v = f (t, v ) = g bv2 /m,where v suggestsvelocity.Hencethedirectionfieldliesinthe t v -plane.With m = 1and b = 1theODE becomes v = g v2 .Tofindthelimitingvelocitywefindthevelocityforwhichtheacceleration equalszero.Thisoccurswhen g v2 = 9.80 v2 = 0or v = 3.13(approximately).For v < 3.13 youhave v > 0(increasingcurves)andfor v > 3.13youhave v < 0(decreasingcurves).Notethat theisoclinesarethehorizontalparallelstraightlines g v2 = const,thus v = const.

Sec.1.2Prob.1. DirectionField

Sec.1.3SeparableODEs.Modeling ProblemSet1.3.Page18

1.CAUTION!Constantofintegration. Itisimportanttointroducetheconstantofintegration immediately,inordertoavoidgettingthewronganswer.Forinstance,let

y = y Thenln |y |= x + c , y = c ∗ e x (c ∗ = e c ), whichisthecorrectwaytodoit(thesameasinProb.3ofSec.1.1above)whereasintroducingthe constantofintegrationlateryields

y = y ,ln |y |= x , y = e x + C whichisnotasolutionof y = y when C = 0.

5.Generalsolution. Separatingvariables,wehave ydy =−36xdx .Byintegration,

Withtheplussignofthesquarerootwegettheupperhalfandwiththeminussignthelowerhalfof theellipsesintheansweronp.A4inAppendix2ofthetextbook. For y = 0(the x -axis)theseellipseshaveaverticaltangent,sothatatpointsofthe x -axisthe derivative y doesnotexist(isinfinite).

17.Initialvalueproblem. Usingtheextendedmethod(8)–(10),let u = y /x .Thenbyproductrule y = u + xu .Now

sothat u = 3x 2 cos2 u .

Separatingvariables,thelastequationbecomes du cos2 u = 3x 2 dx .

Integratebothsides,ontheleftwithrespectto u andontherightwithrespectto x ,asjustifiedinthe textthensolvefor u andexpresstheintermediateresultintermsof x and y

tan u = x 3 + c , u = y x = arctan(x 3 + c ), y = xu = x arctan(x 3 + c ).

Substitutingtheinitialconditionintothelastequation,wehave y (1) = 1arctan(13 + c ) = 0,hence c =−1

Togetherweobtaintheanswer

y = x arctan(x 3 1).

23.Modeling.Boyle–Mariotte’slawforidealgases. Fromthegiveninformationontherateof changeofthevolume dV dP =− V P .

https://buklibry.com/download/solutions-manual-advanced-engineering-mathematics-10th-edition-by-erwin-kreyszig/

Sec.1.4ExactODEs.IntegratingFactors

Use(6)or(6∗ ),onp.22,onlyifinspectionfails.Useonlyoneofthetwoformulas,namely,thatinwhich theintegrationissimpler.ForintegratingfactorstrybothTheorems1and2,onp.25.Usuallyonlyoneof them(orsometimesneither)willwork.Thereisnocompletelysystematicmethodforintegratingfactors, butthesetwotheoremswillhelpinmanycases.Thusthissectionisslightlymoredifficult.

ProblemSet1.4.Page26

1.ExactODE. WeproceedasinExample1ofSec.1.4.WecanwritethegivenODEas Mdx + Ndy = 0where M = 2xy and N = x 2 .

Nextwecompute ∂ M ∂ y = 2x (where,whentakingthispartialderivative,wetreat x asifitwerea constant)and ∂ N ∂ x = 2x (wetreat y asifitwereaconstant).(SeeAppendixA3.2forareviewofpartial derivatives.)ThisshowsthattheODEisexactby(5)ofSec.1.4.From(6)weobtainbyintegration

u = Mdx + k (y ) = 2xydx + k (y ) = x 2 y + k (y )

Tofind k (y )wedifferentiatethisformulawithrespectto y anduse(4b)toobtain

Thelastequationwasobtainedbyintegration.Insertthisintotheequationfor u ,comparewith(3)of Sec.1.4,andobtain u = x 2 y + c ∗ .Because u isaconstant,wehave x 2 y = c ,hence y = c /x 2