CHAPTER2:FirstOrderDifferentialEquations
EXERCISES2.2:SeparableEquations
2. Thisequationisseparablebecausewecanseparatevariablesbymultiplyingbothsides by dx anddividingby4y 2 3y +1.
4. Thisequationisseparablebecause
6. Writingtheequationintheform
weseethattheright-handsidecannotberepresentedintheform g (t)p(s).Therefore, theequationisnotseparable.
8. Toseparatevariables,wedividetheequationby x andmultiplyby dt.Integratingyields
where C1 isanarbitraryconstantand,therefore, C2 := eC1 isanarbitrary positive constant, C = ±C2 isany nonzero constant.Separatingvariables,welostasolution x ≡ 0,whichcanbeincludedintheaboveformulabytaking C =0.Thus, x = Cet3 , C –arbitraryconstant,isageneralsolution.
10. Separatingvariablesyields
Thesecondintegralcanbeevaluatedusingasubstitution
12. Separatingvariables,weobtain
Usingthetrigonometricidentitiessec y =1/ cos y andcos2 y =(1+cos2y )/2and integrating,weget
Thelastequationdefinesimplicitsolutionstothegivendifferentialequation.
14. Writingthegivenequationintheform dx/dt = x3 + x,weseparatethevariablestoget
where C> 0isanarbitraryconstant.(Notethat,separatingvariables,welostasolution x ≡ 0,whichcanbeincludedintheaboveformulabyletting C =0.)Solvingfor x,we get
16. Toseparatevariables,wemovetheterm,containing dx,totheright-handsideofthe equationanddividebothsidesby y .Thisyields y 1 dy = yecos x sin xdx ⇒ y 2 dy = ecos x sin xdx.
Integratingthelastequation,weobtain y 2 dy = ( ecos x sin x) dx ⇒−y 1 + C = eu du (u =cos x)
⇒− 1 y + C = eu = ecos x ⇒ y = 1 C ecos x , where C isanarbitraryconstant.
18. Firstwefindageneralsolutiontotheequation.Separatingvariablesandintegrating, weget
Weusetheinitialcondition, y (0)=3,tofind C .Thus,substitution y =3and x =0 intothelastequationyields |1 3| =
Therefore, |1 y | =2e x4 /4 .Finally,since1 y (0)=1 3 < 0,onanintervalcontaining x =0onehas1 y (x) < 0sothat |1
1.Thus,theanswer
20. Separatingvariablesyields
Forintegratingtheright-handside,weusepartialfractions.
Solvingfor A, B ,and C ,weget A =1, B = 2,and C =3.Thus,integrating(2.1), weobtain 1 2 (y +1)2 =ln x + 2 x +3ln(x +1)+ C.
With y (1)=1, C = 3ln2sothattheansweris 1 2 (y +1)2 =ln x + 2 x +3ln(x +1) 3ln2 or,solvingfor y , y = 1+ ln(x2 )+ 4 x +ln[(x +1)6 ] 6ln2 .
22. Writing2ydy = x2 dx andintegrating,wefind
With y (0)=2, (2)2 = (0)3 3 +
Wenotethat,takingthesquareroot,wechosethepositivesignbecause y (0) > 0.
24. Forageneralsolution,weseparatevariablesandintegrate.
Wesubstitutenowtheinitialcondition, y (1)=0,andobtain
Hence,theanswerisgivenby
26. Weseparatevariablesandobtain
becauseatinitialpoint, x =0,1+ x> 0.Usingthefactthat y (0)=1,wefind C 2=0+ C ⇒ C =2, andso y =[2 ln(1+ x)] 2 /4istheanswer.
28. Wehave
where C1 =0isanyconstant.Separatingvariables,welostthesolution y ≡ 0,which canbeincludedintotheaboveformulabyletting C1 =0.So,ageneralsolutiontothe givenequationis y = C1 e (t 1)2 ,C1 isanarbitraryconstant .
Substituting t =0and y =3,wefind
ThegraphofthisfunctionisgiveninFig. 2–A attheendofthechapter.
Since y (t) > 0forany t,fromthegivenequationwehave y (t) > 0for t< 1and y (t) < 0 for t> 1.Thus t =1isthepointofabsolutemaximumwith ymax = y (1)=3e.
30.(a) Dividingby(y +1)2/3 ,multiplyingby dx,andintegrating,weobtain
(b) Substituting y ≡−1intotheoriginalequationyields
andsotheequationissatisfied.
(c) For y ≡−1forthesolutioninpart(a),wemusthave
whichisimpossiblesinceaquadraticpolynomialhasatmosttwozeros.
32.(a) ThedirectionfieldofthegivendifferentialequationisshowninFig. 2–B atthe endofthechapter.Usingthispicturewepredictthatlimx→∞ φ(x)=1.
(b) InnotationofSection1.4,wehave x0 =0, y0 =1 5, f (x,y )= y 2 3y +2,and h =0 1.Withthisstepsize,weneed(1 0)/0 1=10stepstoapproximate φ(1). TheresultsofcomputationaregiveninTable 2–A attheendofthechapter.From thistableweconcludethat φ(1) ≈ 1.26660. Copyright c 2018PearsonEducation,Inc.
(c) Separatingvariablesandintegrating,weobtain
wherewehaveusedapartialfractionsdecomposition
toevaluatetheintegral.Theinitialcondition, y (0)=1.5,impliesthat C =0,and so
(Wehavechosenthenegativesignbecauseoftheinitialcondition.)Solvingfor y yields
ThegraphofthissolutionisshowninFig. 2–B attheendofthechapter.
(d) Wefind
Thus,theapproximatevalue φ(1) ≈ 1 26660foundinpart(b)differsfromthe actualvaluebylessthan0 003.
(e) Wefindthelimitof φ(x)atinfinitywriting
, whichconfirmsourguessinpart(a).
34.(a) Separatingvariablesandintegrating,weget
where C isanynonzeroconstant.Wecanincludethelostsolution T ≡ M intothis formulabyletting C =0.Thus,ageneralsolutiontotheequationis
(b) Giventhat M =70◦ , T (0)=100◦ , T (6)=80◦ ,weformasystemtodetermine C and k .
100=70+ C
80=70+ Ce 6k ⇒ C =30 k = (1/6)ln[(80 70)/30]=(1/6)ln3. Therefore,
T =70+30e (t ln3)/6 =70+(30)3 t/6 , andafter20minthereadingis
T (20)=70+(30)3 20/6 ≈ 70 77◦
36. AgeneralsolutiontothecoolingequationfoundinProblem34,thatis, T = M + Ce kt . Since T (0)=100◦ , T (5)=80◦ ,and T (10)=65◦ ,wedetermine M , C ,and k fromthe system
M + C =100 M + Ce 5k =80 M + Ce 10k =65
C (1 e 5k )=20 Ce 5k (1 e 5k )=15
Tofind M ,wecannowusethefirsttwoequationsintheabovesystem.
M + C =100 M +(3/4)C =80 ⇒ M =20.
38. With m =100, g =9.8,and k =5,theequationbecomes
100 dv dt =100(9 8) 5v ⇒ 20 dv dt =196 v.
e 5k =3/4
Separatingvariablesandintegratingyields dv v 196 = 1 20 dt ⇒ ln |v 196| = t 20 + C1 ⇒ v =196+ Ce t/20 , where C isanarbitrarynonzeroconstant.With C =0,thisformulaalsogivesthe(lost) constantsolution v (t) ≡ 196.Fromtheinitialcondition, v (0)=10,wefind C .
196+ C =10 ⇒ C = 186 ⇒ v (t)=196 186e t/20
Theterminalvelocityoftheobjectcanbefoundbyletting t →∞. v∞ =lim t→∞ 196 186e t/20 =196(m/sec)
40.(a) Substituting ρ = Mp/RT into dp/dz = ρg yields
Separatingvariablesandintegrating,wefind
(b) If T = T (z )varies,thenintegrating(2.2)from z0 to z weobtain
(c) Withthegivenformulafor T (z ),theintegralinpart(b)gives
Substitutingthegivendataintothisequationandcomputing,wegettheheight
z0 ≈ 168(m)
2. Neither.
4. Linear.
6. Linear.
8. Writingtheequationinstandardform,
weseethat
Multiplyingthegivenequationby μ(x),weget
10. Fromthestandardformofthegivenequation,
wefindthat
12. Here, P (x)=4,
14. Instandardform,wehave
and
(Integrationbypartswasusedtointegratethefirstterm.)
16. Here, |x| < 1and
Therefore,
18. Since μ(x)=exp
,wehave
Substitutingtheinitialcondition, y =4/3at x =0,yields
20. Wehave
isthesolutiontothegiveninitialvalueproblem.
22. Fromthestandardformofthisequation,
wefind
(Alternatively,onecannoticethattheleft-handsideoftheoriginalequationisthe derivativeoftheproduct y sin x.)So,usingintegrationbyparts,weobtain
Wefind C usingtheinitialcondition y (π/2)=2:
andthesolutionisgivenby
24.(a) Theequation(12)ofthetextbecomes
Since y (0)=40,wehave
Theterm5e 10t willeventuallydominate.
(b) Thistime,theequation(12)hastheform
26. Here
Substitutingtheinitialconditionyields
Thus,
and y (1)=(1+sin 2 1) 1/2
WenowusetheSimpson’sRuletofindthat y (1) ≈ 0 860.
28.(a) Substituting y = e x intotheequation(16)yields
So, y = e x isasolutionto(16).
Thefunction y = x 1 isasolutionto(17)because
(x 1 )
(b) Foranyconstant C , d(Ce x )
Thus y = Ce x isasolutionto(16).
Substituting y = Cx 1 into(17),weobtain
(Cx 1 )
andsowemusthave C (C 1)=0inorderthat y = Cx 1 isasolutionto(17). Thus,either C =0or C =1.
(c) Forthefunction y = C ˆ y ,onehas
ifˆ y isasolutionto y + P (x)y =0.
30.(a) Multiplyingbothsidesof(18)by y 2 ,weget y 2 dy dx +2y 3 = x.
If v = y 3 ,then v =3y 2 y .Thus, y 2 y = v /3,andwehave 1 3 dv dx +2v = x, whichisequivalentto(19).
(b) Theequation(19)islinearwith P (x)=6and Q(x)=3x.So,
where C = C1 /2isanarbitraryconstant.Thebacksubstitutionyields
.
32. Inthegivenequation, P (x)=2,whichimpliesthat μ(x)= e2x .Followingguidelines, firstwesolvetheequationon[0, 3].Onthisinterval, Q(x) ≡ 2.Therefore,
Since y1 (0)=0,weget
For x> 3, Q(x)= 2andso
Wenowchoose C2 sothat
Therefore, y2 (x)= 1+(2
,andthecontinuoussolutiontothegiveninitial valueproblemon[0, ∞)is
ThegraphofthisfunctionisshowninFig. 2–D attheendofthechapter.
34.(a) Since P (x)iscontinuouson(a,b),itsantiderivativesgivenby P (x)dx arecontinuouslydifferentiable,andthereforecontinuous,functionson(a,b).Sincethe function ex iscontinuouson(−∞, ∞),compositefunctions μ(x)= e P (x)dx are continuouson(a,b).Therangeoftheexponentialfunctionis(0, ∞).Thisimplies that μ(x)ispositivewithanychoiceoftheintegrationconstant.Usingthechain rule,weconcludethat
forany x on(a,b).
(b) Differentiating(8),weapplytheproductruleandobtain
andso
(c) Suggestedchoiceoftheantiderivativeandtheconstant C yields
(d) Weassumethat y (x)isasolutiontotheinitialvalueproblem(15).Since μ(x)isa continuouspositivefunctionon(a,b),theequation(5)isequivalentto(4).Since, fromthepart(a),theleft-handsideof(5)isthederivativeoftheproduct μ(x)y (x), thisfunctionmustbeanantiderivativeoftheright-handside,whichis μ(x)Q(x). Thus,wecomeupwith(8),wheretheintegralmeansoneoftheantiderivatives,for example,theonesuggestedinthepart(c)(whichhaszerovalueat x0 ).Substituting x = x0 into(8),weconcludethat y0 = y (x0 )= μ(x0 ) 1
+
=
(x
) 1 , andso C = y0 μ(x0 )isuniquelydefined.
36.(a) If μ(x)=exp Pdx and yh (x)= μ(x) 1 ,then dyh dx =( 1)μ(x) 2 dμ(x) dx = μ(x) 2 μ(x)P (x)= μ(x) 1 P (x)
andso
i.e., yh isasolutiontotheequation y + Py =0.Now,theformula(8)yields
(b) Separatingvariablesin(22)andintegrating,weobtain
Sinceweneedjustonesolution yh ,wetake C =0
andwechoose,say, y
(c) Substituting
Therefore, dv/dx = x
(d) Integratingyields
(Wehavechosenzerointegrationconstant.)
(e) Thefunction
isageneralsolutionto(21)because
38. Dividingbothsidesof(6)by μ andmultiplyingby dx yields
40.(a) Figure2-Eattheendofthischapterdepictsthedirectionfieldfor α = 1 4 ,β = 1 2 If u =1,u = β< 0; ifu =0,u = α> 0.Thus u(t)isconfinedto[0,1]ifitever entersthisinterval.
(b) Thegeneralsolution u(t)= α α+β + Ce (α+β )t approaches α α+β as t →∞.
EXERCISES2.4:ExactEquations
2. Thisequationisnotseparablebecausethecoefficient x10/3 2y cannotbewrittenasa product p(x)q (y ).Writingtheequationintheform dy dx 2x 1 y = x 7/3 , weseethattheequationislinear.Since M (x,y )= x10/3 2y , N (x,y )= x,
, andsotheequationisnotexact.
4. Here M (x,y )= yexy +2x, N (x,y )= xexy 2y .Thus
=
), Since ∂M/∂y = ∂N/∂x,theequationisexact.
Writingtheequationintheform
dy dx = yexy +2x xexy 2y , weconcludethatitisnotseparable,becausetheright-handsidecannotberepresented asaproductoftwofunctionsofsinglevariables x and y .Also,theright-handsideis notlinearwithrespectto y ,whichimpliesthattheequationisnotlinearwith y as thedependentvariable.Similarly,choosing x asthedependentvariable(takingthe reciprocalsofbothsides),weconcludethattheequationisnotlinearin x either.
6. Thedifferentialequationisnotseparablebecause(2xy +cos y )cannotbefactored.This equationcanbeputinstandardformbydefining x asthedependentvariableand y as theindependentvariable.Thisgives
dx dy + 2 y x = cos y y 2 ,
andsoweseethatitislinear.
Ifweset M (x,y )= y 2 and N (x,y )=2xy +cos y ,weseethatthedifferentialequation isalsoexact,because My (x,y )=2y = Nx (x,y ).
8. Inthisproblem,thevariablesare r and θ , M (r,θ )= θ ,and N (r,θ )=3r θ 1.Since ∂M ∂θ =1 =3= ∂N ∂r , theequationisnotexact.With r asthedependentvariable,theequationtakestheform dr dθ = 3r θ 1 θ = 3 θ r + θ +1 θ , andsoitislinear.Sincetheright-handsideintheaboveequationcannotbefactored as p(θ )q (r ),theequationisnotseparable.
10. Inthisproblem, M (x,y )=2x + y , N (x,y )= x 2y .Thus, My = Nx =1,andthe equationisexact.Wefind F (x,y )= (2x + y )dx = x 2 + xy + g (y ),
∂y = x + g (y )= N (x,y )= x 2y
⇒ g (y )= 2y ⇒ g (y )= ( 2y )dy = y 2
⇒ F (x,y )= x 2 + xy y 2 , andso x2 + xy y 2 = C isageneralsolution.
12. Takingpartialderivativesof M (x,y )=cos x cos y +2x and N (x,y )= (sin x sin y +2y ), weobtain
∂M ∂y = ∂ ∂y (cos x cos y +2x)= cos x sin y, ∂N ∂x = ∂ ∂x [ (sin x sin y +2y )]= cos x sin y, andsotheequationisexact.
Integrating M (x,y )withrespectto x yields
F (x,y )= M (x,y )dx = (cos x cos y +2x) dx
=cos y cos xdx + 2xdx =sin x cos y + x 2 + g (y ).
Tofind g (y ),wecomputethepartialderivativeof F (x,y )withrespectto y andcompare theresultwith N (x,y ).
=
⇒ g (y )= 2y ⇒ g (y )= ( 2y )dy = y 2 .
(Wehavetakenzerointegrationconstant.)Therefore, F (x,y )=sin x cos y + x 2 y 2 = C isageneralsolutiontothegivenequation.
14. Inthisequation,thevariablesare y and t, M (y,t)= t/y , N (y,t)=1+ln y .Since
∂t = ∂ ∂t t y = 1 y and ∂N ∂y = ∂ ∂y (1+ln y )= 1 y , theequationisexact.
Integrating M (y,t)withrespectto y ,weget
F (y,t)= t y dy = t ln |y | + g (t)= t ln y + g (t).
(From N (y,t)=1+ln y weconcludethat y> 0.)Therefore,
∂F
∂t = ∂ ∂t [t ln y + g (t)]=ln y + g (t)=1+ln y
⇒ g (t)=1 ⇒ g (t)= t
⇒ F (y,t)= t ln y + t, andageneralsolutionisgivenby t ln y + t = C (or,explicitly, t = C/(ln y +1)).
16. Computing
∂M
∂y = ∂ ∂y yexy y 1 = exy + xyexy + y 2 ,
∂N
∂x = ∂ ∂x xexy + xy 2 = exy + xyexy + y 2 , weseethattheequationisexact.Therefore,
F (x,y )= yexy y 1 dx = exy xy 1 + g (y )
So,
= xexy + xy 2 + g (y )= N (x,y ) ⇒ g (y )=0.
Thus, g (y )=0,andtheansweris exy xy 1 = C .
18. Since
∂y = ∂N ∂x =2y 2 +sin(x + y ), theequationisexact.Wefind F (x,y )= 2x + y 2 cos(x + y ) dx = x 2 + xy 2 sin(x + y )+ g (y ),
∂y =2xy cos(x + y )+ g (y )=2xy cos(x + y ) ey
g (y )= ey
Therefore,
givesageneralsolution.
20. Wefind ∂M ∂y = ∂ ∂y [y cos(xy )]=cos(xy ) xy sin(xy ), ∂N ∂x = ∂ ∂x [ x cos(xy )]=cos(xy ) xy
Therefore,theequationisexactand
andageneralsolutionisgivenby
22. InProblem16,wefoundthatageneralsolutiontothisequationis
Substitutingtheinitialcondition, y (1)=1,yields e 1= C .So,theansweris exy xy 1 = e 1.
24. First,wecheckthegivenequationforexactness.
dx = e t = ∂N ∂t . So,itisexact.Wefind
isageneralsolution.With x(1)=1,weget (1)(e 1)+1= C
C = e, andthesolutionisgivenby x = e t et 1
26. Takingpartialderivatives My and Nx ,wefindthattheequationisexact.So,
and x(tan y 2)+ln |y | = C
isageneralsolution.Substituting y (0)=1yields C =0.Therefore,theansweris x(tan y 2)+ln y =0. (Weremovedtheabsolutevaluesigninthelogarithmicfunctionbecause y (0) > 0.)
28.(a) Computing ∂M ∂y =cos(xy ) xy sin(xy ), whichmustbeequalto ∂N/∂x,wefindthat N (x,y )= [cos(xy ) xy sin(xy )] dx = [x cos(xy )]x dx = x cos(xy )+ g (y ).
(b) Since
weconcludethat
30.(a) Differentiating,wefindthat
Since My = Nx ,theequationisnotexact.
(b) Multiplyinggivenequationby xn y m andtakingpartialderivativesofnewcoefficientsyields
Inorderthatthesepolynomialsareequal, wemusthaveequalcoefficientsatsimilar monomials.Thus, n and m mustsatisfythesystem
5(m +1)=2(n +3) 6(m +2)=3(n +4) 4(m +2)=3(n +2)
Solving,weobtain n =2and m =1.Therefore,multiplyingthegivenequationby x2 y yieldsanexactequation.
(c) Wefind
Therefore,
andageneralsolutiontothegivenequationis
32.(a) Theslopeoftheorthogonalcurves,say m⊥ ,mustbe 1/m,where m istheslope oftheoriginalcurves.Therefore,wehave
(b) Let F (x,y )= x2 + y 2 .Thenwehave Fx (x,y )=2x and Fy (x,y )=2y .Substituting theseexpressionsintotheresultofpart(a)givesus
ydx 2xdy =0
Tofindtheorthogonaltrajectories,wemustsolvethisdifferentialequation.Tothis end,notethatthisequationisseparableandthus
Therefore,theorthogonaltrajectoriesarelinesthroughtheorigin.
(c) Let F (x,y )= xy .Thenwehave Fx (x,y )= y and Fy (x,y )= x.Substitutingthese expressionsintotheresultofpart(a)yields xdx ydy =0
Tofindtheorthogonaltrajectories,wemustsolvethisdifferentialequation.Tothis end,notethatthisequationisseparableandthus
where k :=2C .Therefore,theorthogonaltrajectoriesarehyperbolas.
34. TousethemethoddescribedinProblem32,werewritetheequation x2 + y 2 = kx in theform x + x 1 y 2 = k .Thus, F (x,y )= x + x 1 y 2 ,
=1 x 2 y 2 ,
SubstitutingthesederivativesintheequationgiveninProblem32(b),wegettherequired.Multiplyingtheequationby xn y m ,weobtain
Therefore,
Thus,tohaveanexactequation, n and m mustsatisfy n 2=0 2(m +1)= n.
Solving,weobtain n =2, m = 2.Withthischoice,theequationbecomes
, andso
Therefore, g (y )= y ,andthefamilyoforthogonaltrajectoriesisgivenby x2 y 1 + y = C
Writingthisequationintheform x2 + y 2 Cy =0,weseethat,given C ,thetrajectory isthecirclecenteredat(0,C/2)andofradius C/2.
SeveralgivencurvesandtheirorthogonaltrajectoriesareshowninFig. 2–E attheend ofthechapter.
36. Thefirstequationin(4)followsfrom(9)andtheFundamentalTheoremofCalculus.
Forthesecondequationin(4),
x x0 M (t,y )dt + g (y )
)dt + ⎡ ⎣N (x,y )
EXERCISES2.5:SpecialIntegratingFactors
x x
M (t,y )dt⎤ ⎦ = N (x,y ).
2. Here, M (x,y )=2y 3 +2y 2 and N (x,y )=3y 2 x +2xy .Computing ∂M ∂y =6y 2 +4y and ∂N ∂x =3y 2 +2y,
weconcludethatthisequationisnotexact.Notethatthesederivatives,aswellas M itself,dependon y only.Then,clearly,sodoestheexpression(∂N/∂x ∂M/∂y )/M , andthegivenequationhasanintegratingfactordependingon y alone.Also,since ∂M/∂y ∂N/∂x N = (6y 2 +4y ) (3y 2 +2y ) 3y 2 x +2xy = 3y 2 +2y x(3y 2 +2y ) = 1 x ,
theequationhasanintegratingfactordependingon x only.
Writingtheequationintheform dx dy = 3y 2 x +2xy 2y 3 +
(y +1) x,
weconcludethatitisseparableandlinearwith x asthedependentvariable.
4. Thisequationisnotseparable,becauseofthefactor(y 2 +2xy ).Itisnotlinearineither variablebecauseoftheterms y 2 and x2 .Toseeifitisexact,wecompute My (x,y )and Nx (x,y ),andfindthat My (x,y )=2y +2x = 2x = Nx (x,y ).
Therefore,theequationisnotexact.Toseeifwecanfindanintegratingfactorofthe form μ(x),wecompute
∂M/∂y ∂N/∂x N = 2y +4x x2 , whichisnotafunctionof x alone.Toseeifwecanfindanintegratingfactoroftheform μ(y ),wecompute
∂N/∂x ∂M/∂y M = 4x 2y y 2 +2xy = 2(2x + y ) y (y +2x) = 2 y
Thustheequationhasanintegratingfactorthatisafunctionof y alone.
6. Inthisproblem, M (x,y )=2y 2 x y and N (x,y )= x.Therefore,
=4yx 1and
=2 4yx.
Theequationisnotexact,because ∂M/∂y = ∂N/∂x,butithasanintegratingfactor, whichdependson y because
.
Isolating dy/dx,weobtain
dx = y 2y 2 x x = y x 2y 2 .
Theright-handsidecannotbewrittenas p(x)q (y ),sotheequationisnotseparable.
Also,itisnotlinearwith y asthedependentvariable(becauseof2y 2 term).Takingthe reciprocals,weconcludethatitisnotlinearwiththedependentvariable x.
8. Theequation(3x2 + y ) dx +(x2 y x) dy =0isnotseparableorlinear.Toseeifitis exact,wecompute
Thus,theequationisnotexact.Toseeifwe canfindanintegratingfactor,wecompute
Thus,anintegratingfactor μ(x)isgivenby μ(x)=exp 2 x dx =exp( 2ln |x|)= x 2 .
Tosolvetheequation,wemultiplyitbytheintegratingfactor x 2 andget (3+ yx 2 )
Thisnewequationisnowexact.Thus,wefind F (x,y )integrating M (x,y )=3+ yx 2 withrespectto x.
Therefore, F (x,y )=3x yx 1 + y 2 2 , andanimplicitsolutiontothegivenequationis y 2 2 y x +3x = C.
Since μ(x)= x 2 wemustcheckifthesolution x ≡ 0waslost.Thefunction x ≡ 0isa solutiontotheoriginalequation,butisnotgivenbytheaboveimplicitsolutionforany choiceof C .Hence, y 2 2 y x +3x = C and x ≡ 0 giveageneralsolution.
10. Wecomputethepartialderivativesof M (x,y )=2y 2 +2y +4x2 and N (x,y )=2xy + x.
∂y = ∂
2y 2 +2y +4x 2 =4y +2,
= ∂ ∂x (2xy + x)=2y +1.
Althoughtheequationisnotexact(∂M/∂y = ∂N/∂x),thequotient ∂M/∂y ∂N/∂x N = (4y +2) (2y +1) 2xy + x = 2
y +1) = 1 x dependson x only,andsotheequationhasanintegratingfactor,whichcanbefound byapplyingformula(8).Namely, μ(x)=exp 1 x dx =exp(ln |x|)= |x|.
Notethatif μ isanintegratingfactor,then μ isanintegratingfactoraswell.This observationallowsustotake μ(x)= x.Multiplyingthegivendifferentialequationby x yieldsanexactequation
Therefore,
+
=
y
+
y + x 4 , and x2 y 2 + x2 y + x4 = C isageneralsolution.
12. Here, M (x,y )=2xy 3 +1, N (x,y )=3x2 y 2 y 1 .Since ∂M
∂y =6xy 2 = ∂N ∂x ,
theequationisexact.So,wefindthat
andthegivenequationhasageneralsolution
14. Multiplyingthegivenequationby xn y m yields
Therefore,
Matchingthecoefficients,wegetasystem 12m =6(n +1) 5(m +1)=3(n +2)
todetermine n and m.Thissystemhasthesolution n =3, m =2.Thus,thegiven equationmultipliedby x3 y 2 ,thatis,
isexact.Wecompute
⇒ g (y )=0 ⇒ g (y )=0, andso3x4 y 2 + x5 y 3 = C isageneralsolutiontothegivenequation. Copyright c 2018PearsonEducation,Inc.
16.(a) Anequation Mdx + Ndy =0hasanintegratingfactor μ(x + y )ifandonlyifthe equation
(x + y )M (x,y )dx + μ(x + y )N (x,y )dy =0 isexact.AccordingtoTheorem2,Section2.4,thismeansthat
Applyingtheproductandchainrulesyields
Collectingsimilartermsyields
Theright-handsideof(2.4)dependson x + y onlysotheleft-handsidedoes.
Tofindanintegratingfactor,welet s = x + y anddenote G(s)= ∂N/∂x ∂M/∂y M N .
Then(2.4)impliesthat
Inthisformula,wecanchooseeithersignandanyintegrationconstant.
(b) Wecompute ∂N/∂x ∂M/∂y M N = (1+ y ) (1+ x) (3+ y + xy ) (3+ x + xy ) =1 .
Applyingformula(2.5),weobtain
(s)=exp (1)ds = es ⇒
(x + y )= ex+y ,
Multiplyingthegivenequationby μ(x + y ),wegetanexactequation e x+y (3+ y + xy )dx + e x+y (3+ x + xy )dy =0
andfollowtheprocedureofsolvingexactequations,Section2.4.
Takingthepartialderivativeof F withrespectto y ,wefind h(y ).
h (y )=0
Thus,ageneralsolutionis
18. Thegivencondition, xM (x,y )+ yN (x,y ) ≡ 0,isequivalentto yN (x,y ) ≡−xM (x,y ).
Inparticular,substituting x =0,weobtain yN (0,y ) ≡−(0)M (0,y ) ≡ 0
Thisimpliesthat x ≡ 0isasolutiontothegivenequation.
Toobtainothersolutions,wemultiplytheequationby x 1 y .Thisgives
Therefore, x 1 y = C or y = Cx
Thus,ageneralsolutionis y = Cx and x ≡ 0
20. Fortheequation
wecompute
Therefore, ∂M/∂y = ∂N/∂x,andtheequationisexact.
EXERCISES2.6:SubstitutionsandTransformations
2. Wecanwritetheequationintheform
dy dx =(y 4x 1)2 =[(y 4x) 1]2 = G(y 4x), where G(t)=(t 1)2 .Thus,itisoftheform dy/dx = G(ax + by ).
4. Inthisequation,thevariablesare x and t.Itscoefficients, t + x +2and3t x 6, arelinearfunctionsof x and t.Therefore,thegivenequationisanequationwithlinear coefficients.
6. Thegivendifferentialequationisnothomogeneousduetothe e 2x terms.Sinceitcan bewrittenintheform dy/dx + P (x)y = Q(x)y n ,namely,
3 . itisaBernoulliequation.Thedifferentialequationdoesnothavelinearcoefficients,and itisnotoftheform y = G(ax + by )either.
8. Here,thevariablesare y and θ .Writing
weseethattheright-handsideisafunctionof y/θ alone.Hence,theequationis homogeneous.
10. First,wewritetheequationintheform
Therefore,itishomogeneous,andwewemakeasubstitution y/x = u or y = xu.Then y = u + xu ,andtheequationbecomes
Separatingvariablesandintegratingyields
Substitutingback y/x for u andsimplifying,wefinallyget
12. From
makingthesubstitution v = y/x,weobtain
where C1 = e3C2 isanypositiveconstant.Makingthebacksubstitution,wefinallyget
where C = ±C1 isanynonzeroconstant.
14. Substituting v = y/θ yields
16. Werewritetheequationintheform
andsubstitute v = y/x toget
where C =0isanyconstant.Notethat,separatingvariables,welostasolution, v ≡ 1, whichcanbeincludedintheaboveformulabyletting C =0.Thuswehave v = eCx where C isanyconstant.Substitutingback y = xv yieldsageneralsolution
tothegivenequation.
18. With z = x + y +2and z =1+ y ,wehave
20. Substitution z = x
Theleft-handsideintegralcanbefoundasfollows.
Thus,ageneralsolutionisgivenimplicitlyby
22. Dividingtheequationby y 3 yields
Wenowmakeasubstitution
Thisisalinearequation.So,
Dividingtheequationby y 3 ,welostaconstantsolution y ≡ 0.
24. WedividethisBernoulliequationby y 1/2 andmakeasubstitution v = y 1/2 .
Anintegratingfactorforthislinearequationis
Since y = v 2 ,wefinallyget
Inaddition, y ≡ 0isa(lost)solution.
26. Multiplyingtheequationby y 2 ,weget
With
,theequationbecomes
Therefore,
28. First,wenotethat y ≡ 0isasolution,whichwillbelostwhenwedividetheequation by y 3 andmakeasubstitution v = y 2 togetalinearequation
Anintegratingfactorforthisequationis
Thus,
So,
30. Wemakeasubstitution
thetext:
and k satisfysystem(14)in
Solvingyields
, thesubstitutionleadsto
Thereforewith
where C1 and C2 areconstants.Individingby4 z 2 ,welosttheconstantsolutions z ≡ 2and z ≡−2,buttheseareincludedinouranswerifwelet C2 =0.
32. Wemakeasubstitution x = u + h,y = v + k, where h and k satisfysystem(14)in thetext:
Solvingyields h =
, thesubstitutionleadsto
Thereforewith z = v u , v = z + uz wehave
+
where C1 and C2 areconstants.Individingby z 2 1,welosttheconstantsolutions z ≡ 1and z ≡−1,buttheseareincludedinouranswerifwelet C2 =0.
34. InProblem2,wefoundthatthegivenequationisoftheform dy/dx = G(y 4x)with G(u)=(u 1)2 .Thuswemakeasubstitution u = y 4x toget
dx =(y 4x 1)2
=(
1)2 ⇒ du dx =(u 1)2 4=(u 3)(u +1) ⇒ du (u 3)(u +1) = dx.
Tointegratetheleft-handside,weusepartialfractionsdecomposition,
Thus,
where C =0isanarbitraryconstant.Separatingvariables,welosttheconstantsolutions u ≡ 3and u ≡−1,thatis, y =4x +3and y =4x 1.While y =4x +3canbeobtained from(2.6)bysetting C =0,thesolution y =4x 1isnotincludedin(2.6).Therefore, ageneralsolutiontothegivenequationis
36. Thisequationhaslinearcoefficients.Thuswemakeasubstitution t = u + h and x = v + k with h and k satisfying
As dt = du and dx = dv ,thesubstitutionyields (u + v )dv +(3u
With z = u/v ,wehave u = vz , u = z + vz ,andtheequationbecomes
Makingbacksubstitution,aftersomealgebraweget
38. InProblem6,wehavewrittentheequationintheform
Makingasubstitution u = y 2 (andso u = 2y 3 y )inthisBernoulliequation,weget
Also,welosttheconstantsolution y ≡ 0whendividedtheequationby y 3
40. Sincetheequationishomogeneous,wemakeasubstitution u = y/θ .Thus,weget
whichgives
Backsubstitution u = y/θ yields y 2 (θ + y )2 = C θ ⇒ θy 2 = C (θ + y )2 ,C =0. When C =0,theaboveformulagives θ ≡ 0or y ≡ 0,whichwerelostinseparating variables.Also,welostanothersolution, u +1 ≡ 0or y = θ .Thus,theansweris θy 2 = C (θ + y )2 and y = θ, where C isanarbitraryconstant.
42. Suggestedsubstitution, y = vx2 (sothat y =2xv + x2 v ),yields 2xv + x 2 dv dx =2vx +cos v ⇒ x 2 dv dx =cos v.
Solvingthisseparableequation,weobtain dv cos v = dx x2 ⇒ ln | sec v +tan v | = x 1 + C1
⇒ sec v +tan v = ±e C1 e 1/x = Ce 1/x ⇒ sec y x2 +tan y x2 = Ce 1/x , where C = ±eC1 isanarbitrarynonzeroconstant.With C =0,thisformulaalso includeslostsolutions
So,togetherwiththeothersetoflostsolutions,
wegetageneralsolutiontothegivenequation.
44. From
usingthat a2 = ka1 and b2 = kb1 ,weobtain
48. Applyformula(8),page50ofSection2.3,tothetransformedBernoulliequationbelow equation(10)ofSection2.6,page74.(If n =1,theBernoulliequationislinearand formula(8)canbeapplieddirectly.)
REVIEWPROBLEMS
2. y = 8x2 4x 1+ Ce4x 4. x3 6 4x2 5 + 3x 4 Cx 3
6. y 2 =2ln |1 x2 | + C and y ≡ 0
8. y =(Cx2 2x3 ) 1 and y ≡ 0
10. x + y +2y 1/2 +arctan(x + y )= C
12. 2ye2x + y 3 ex = C
14. x = t2 (t 1) 2 + t(t 1)+3(t 1)ln |t 1| + C (t 1)
16. y =cos x ln | cos x| + C cos x
18. y =1 2x + √2tan √2 x + C
20. y = Cθ 3 12θ 2 5 1/3
22. (3y 2x +9)(y + x 2)4 = C
24. 2√xy +sin x cos y = C
26. y = Ce x2 /2
28. (y +3)2 +2(y +3)(x +2) (x +2)2 = C
30. y = Ce4x x 1 4
32. y 2 = x2 ln(x2 )+16x2
34. y = x2 sin x + 2x2 π 2
36. sin(2x + y ) x3 3 + ey =sin2+ 2 3
38. y = 2 1 4 arctan x 2 2
40. y = 8 1 3e 4x 4x
TABLES
10.11.47560.61.353368921
20.21.450062570.71.330518988
30.31.42531187580.81.308391369
40.41.40086970790.91.287062756
50.51.376852388101.01.266596983
Table2–A:Euler’sapproximationsto y = x y , y (0)=0,on[0, 1]with h =0 1.
FIGURES
Figure2–A:ThegraphofthesolutioninProblem28.
Figure2–B:ThedirectionfieldandsolutioncurveinProblem32.
Figure2–C:ThegraphofthesolutioninProblem32.
Figure2–D:ThedirectionfieldforProblem40(α = 1 4 ,β = 1 2 ).
Figure2–E:CurvesandtheirorthogonaltrajectoriesinProblem34. Copyright c