- Information
- AI Chat
424580021 George F Simmons Differential Equations With Applications and Historical Notes Mc Graw Hill Science 1991 Solutions pdf
Differential equation (f211)
Related Studylists
MathsPreview text
1
Student’s Solutions Manual
for
Differential Equations:
Theory, Technique, and Practice
with Boundary Value Problems
Second Edition
by Steven G. Krantz (with the assistance of Yao Xie)
2 CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?
(h) Ify= arcsinxy, theny′= xy √ ′+y 1 −(xy) 2 soxy
′+y=y′√ 1 −x 2 y 2.
(i) Ify=xtanx, theny′=xsec 2 x+ tanx=x(tan 2 x+ 1) + tanx. Using tanx=y/xwe gety′=y 2 /x+x+y/xorxy′=x 2 +y 2 +y. (j) Ifx 2 = 2y 2 lny, then 2x= [2y 2 (1/y) + 4ylny]y′= 2yy′(1 + 2 lny). Consequently,y′=y+2xylny. Using lny= x 2 2 y 2 we gety
′= xy x 2 +y 2. (k) Ify 2 = x 2 −cx, then 2yy′ = 2x−c so 2xyy′ = 2x 2 −cx = x 2 +x 2 −cx=x 2 +y 2. (l) Ify=c 2 +c/x, theny′=−c/x 2 sox 4 (y′) 2 =c 2 =y−c/x. Use the fact that−c/x=xy′to obtainx 4 (y′) 2 =y+xy′.
(m) Ify=cey/x, theny′=xy
′−y x 2 ce
y/x=xyy′−y 2 x 2. Solve fory
′to obtain y′=y 2 /(xy−y 2 ). (n) Ify+siny=x, theny′+y′cosy= 1 ory′= 1/(1+cosy). Multiply the numerator and denominator of the right side byyto obtain y′=y/(y+ycosy). Now use the identityy=x−sinyto obtain y′= (x−siny+ycosy). (o) Ifx+y = arctany, then 1 +y′ = y′/(1 +y 2 ). Consequently, (1 +y′)(1 +y 2 ) =y′. This simplifies to 1 +y 2 +y 2 y′= 0.
- For each of the following differential equations, find the particular so- lution that satisfies the given initial condition.
(a) Ify′=xex, theny=
∫
xexdx+C= (x−1)ex+C(integrate by parts,u=x). Whenx= 1, y=Cso the particular solution is y(x) = (x−1)ex+ 3. (b) Ify′ = 2 sinxcosx, theny=
∫
2 sinxcosxdx+C= sin 2 x+C. Whenx= 0,y=Cso the particular solution isy(x) = sin 2 x+ 1. (c) Ify′= lnx, theny=
∫
lnxdx+C=xlnx−x+C(integrate by parts,u= lnx). Whenx=e,y=Cso the particular solution is y(x) =xlnx−x. (d) Ify′= 1/(x 2 −1), theny=
∫
1 /(x 2 −1)dx+C= 1/ 2
∫
1 /(x− 1)− 1 /(x+ 1)dx+C= 12 lnxx−+1 1 +C(method of partial fractions). Whenx= 2,y= 12 ln 13 +C=C−ln 3 2 so the particular solution isy(x) = 12 lnxx−+1 1 +ln 3 2.
1. THE NATURE OF SOLUTIONS 3
(e) Ify′=x(x 12 −4), theny=
∫ 1
x(x 2 −4)dx+C= 1/ 8
∫
1 /(x+2)+1/(x− 2)− 2 /xdx+C = 18 ln|x
2 − 4 | x 2 +C (method of partial fractions). Whenx= 1,y= 18 ln 3 +Cso the particular solution isy(x) = 1 8 ln
|x 2 − 4 | x 2 −
1 8 ln 3 =
1 8 ln
|x 2 − 4 | 3 x 2. (f) Ify′ = 2 x
2 +x (x+1)(x 2 +1), then y =
∫ 2 x 2 +x (x+1)(x 2 +1)dx+C =
1 2
∫ 1
x+1+ 3 x− 1 x 2 +1dx+C=
1 2 ln(x+1)+
3 4 ln(x
2 +1)− 1
2 arctanx+C(method of partial fractions). Whenx= 0,y=C so the particular solution isy(x) = 12 ln(x+ 1) + 34 ln(x 2 + 1)− 12 arctanx+ 1.
- For the differential equation
y′′− 5 y′+ 4y= 0,
carry out the detailed calculations required to verify these assertions.
(a) Ify=ex, theny′′− 5 y′+ 4y=ex− 5 ex+ 4ex≡0. Ify=e 4 x, theny′′− 5 y′+ 4y= 16e 4 x− 20 e 4 x+ 4e 4 x≡0. (b) If y= c 1 ex+c 2 e 4 x, theny′′− 5 y′+ 4y =c 1 (ex− 5 ex+ 4ex) + c 2 (16e 4 x− 20 e 4 x+ 4e 4 x≡0.
- For which values ofmwill the functiony=ym=emxbe a solution of the differential equation
2 y′′′+y′′− 5 y′+ 2y= 0?
Find three such valuesm. Use the ideas in Exercise 5 to find a solution containing three arbitrary constantsc 1 , c 2 , c 3. Substitutey=emxinto the differential equation to obtain
2 m 3 emx+m 2 emx− 5 memx+ 2emx= 0.
Cancelemxin each term (it is never 0) to obtain the equivalent equation
2 m 3 +m 2 − 5 m+ 2 = 0.
Observing thatm=m 1 = 1 is a solution (andy 1 =exis a solution to the differential equation). Using this we can factor the polynomial– divide bym−1–to obtain
2 m 3 +m 2 − 5 m+ 2 = (m−1)(2m 2 + 3m−2).
1. FIRST-ORDER LINEAR EQUATIONS 5
(g) The Leibnitz form dydxsiny=x 2 separates to sinydy=x 2 dx. In- tegrate to obtain−cosy=x 3 /3 +Cory= arccos(C−x 3 /3). (h) Write the equationy′−ytanx= 0 in Leibnitz formdydx−ytanx= 0 and separate the variables: dyy = tanxdx. Integrate
∫dy ∫ y = tanxdx, to obtain the solution: ln|y|=−ln|cosx|+C. This can also be written in the formy=C/cosxory=Csecx. (i) From Leibnitz form,xydydx =y−1 we obtain yydy− 1 = dxx. Write this in the formy−y−1+1 1 dy=dxx and integrate to obtain the implicit solutiony+ ln|y− 1 |= ln|x|+C. (j) Leibnitz formxy 2 −dydxx 2 = 0 separates todxx =y− 2 dy. Integrating yields the implicit solution ln|x|=− 1 /y+C. The solution can be expressed explicitly asy=C−ln 1 |x|.
- Substitutingy′=pyields p
′ p =x
- Separation of variables (or direct
integration) gives ln|p|=x 3 /3 +C. This implies thatp=Cex 3 / 3 and so y′ =Cex 3 / 3 . Consequently,y= C
∫
ex 3 / 3 dx+D. As we expect, the solution contains two arbitrary constants. The integral cannot be evaluated in terms of elementary functions.
1 First-Order Linear Equations
- Find the general solution of the following equations.
(a) The equation is linear and separable. The integrating factor is e−x 2 / 2 so it simplifies to (e−x 2 / 2 y)′= 0 ande−x 2 / 2 y=C. Therefore, y=Cex 2 / 2 . (b) This equation is also linear and separable. The integrating factor isex
2 / 2 so it simplifies to (ex
2 / 2 y y)′=xex
2 / 2 . Integrate to obtain ex 2 / 2 y=ex 2 / 2 +Candy= 1 +Ce−x 2 / 2 . (c) The integrating factor isexso the equation simplifies to (exy)′= ex 1+e 2 x. Integrating we obtaine
xy= arctanex+C. The general solution isy=e−xarctan(ex) +Ce−x. (d) The integrating factor isexso the equation simplifies to (exy)′= 2 x+x 2 ex. Integrate to getexy=x 2 +(2− 2 x+x 2 )ex+C(integrate
6 CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?
x 2 exby parts, twice, or use an integral table). The general solution isy=x 2 e−x+ 2− 2 x+x 2 +Ce−x. (e) Write asxy′= 2y−x 3 and theny′−(2/x)y=−x 2. The integrating factor isx− 2 so the equation simplifies to (x− 2 y)′=−1. Integrate to obtainx− 2 y=−x+C. The general solution isy=−x 3 +Cx 2. (f) The integrating factor isex 2 so the equation simplifies to (ex 2 y)′= 0. Consequently,ex
2 y=Candy=Ce−x
2 . (g) Write asy′−(3/x)y=x 3. The integrating factor isx− 3 so the equation simplifies to (x− 3 y)′= 1. Integrating we obtainx− 3 y= x+Csoy=x 4 +Cx 3. (h) Express the equation in the formy′+1+ 2 xx 2 y=1+cotxx 2. The integrat- ing factor is 1 +x 2 and the equation simplifies to ((1 +x 2 )y)′= cotx. Consequently, (1 +x 2 )y= ln|sinx|+Candy=ln|sin1+xx 2 |+C. (i) The integrating factor is sinxand the equation simplifies to (ysinx)′= 2 x. Therefore,ysinx=x 2 +Candy=x
2 +C sinx. (j) Express the equation in the formy′+ (1/x+ cotx)y= 1. The in- tegrating factor isxsinxso the equation simplifies to (xysinx)′= xsinx. Integrate to obtainxysinx= sinx−xcosx+C (use a table of integrals or integratexsinxby parts,u=x). Therefore, y=sinx−xxsincosxx+C.
- Bernoulli Equations To verify the technique write the Bernoulli equation in the formy−ny′+P y 1 −n=Q. The substitutionz=y 1 −n andz′= (1−n)y−ny′yield 1 − 1 nz′+P z=Q.
(a) Bernoulli,n= 3. Substitutez=y− 2 ,z′=− 2 y− 3 y′intoxy− 3 y′+ y− 2 =x 4 to obtain (− 1 /2)xz′+z=x 4. This is linear,z′−(2/x)z= − 2 x 3 , with integrating factorx− 2. It simplifies to (x− 2 z)′=− 2 x. COnsequently,x− 2 z=−x 2 +Candz=−x 4 +Cx 2. This means thaty− 2 =Cx 2 −x 4 soy 2 =Cx 21 −x 4. (b) Write the equation in the form y′+ (1/x)y = y− 2 cosxto see that it is Bernoulli,n=−2. Substitutez =y 3 ,z′ = 3y 2 y′ into the equationy 2 y′+ (1/x)y 3 = cosxto obtain the linear equation (1/3)z′+ (1/x)z= cosx. This isz′+ (3/x)z= 3 cosx, with inte- grating factorx 3. It simplifies to (x 3 z)′= 3x 3 cosx. Consequently, x 3 z = 3F(x) +C whereF(x) is and antiderivative forx 3 cosx.
8 CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?
we first obtain (40−t) 2 = 16003 , thent= 40−√ 403 = 16. 905.. ..
1 Exact Equations
All references toMandNrefer to the equation in the formMdx+Ndy= 0.
M =y,N =x+ 2 y, so∂M∂y = 1 =∂N∂x. The equation is exact. ∂f∂x=y implies thatf(x, y) =xy+φ(y). ∂f∂y=x+ 2 y implies thatx+φ′(y) = x+ 2 ysoφ(y) = 2 ln|y|andf(x, y) =xy+2 ln|y|. The implicit solution isxy+ 2 ln|y|=C.
M =y−x 3 ,N =x+y 3 , so∂M∂y = 1 =∂N∂x. The equation is exact. ∂f ∂x=y−x
3 implies thatf(x, y) =xy−x 4 /4+φ(y).∂f ∂y=x+y
3 implies thatx+φ′(y) =x+y 3 soφ(y) =y 4 /4 andf(x, y) =xy−x 4 /4 +y 4 /4. The implicit solution isxy−x 4 /4 +y 4 /4 =C.
M=y+ycosxy,N=x+xcosxy, so∂M∂y = 1−xysinxy+cosxy=∂N∂x. The equation is exact. ∂f∂x=y+ycosxyimplies thatf(x, y) =xy+ sinxy+φ(y). ∂f∂y =x+xcosxy implies thatx+xcosxy+φ′(y) = x+xcosxy, soφ(y) = 0 andf(x, y) =xy+sinxy. The implicit solution isxy+ sinxy=C. For a given value ofC, the equationt+ sint=C has exactly one solution,t 0. Therefore, the implicit solution for the differential equation also be expressed simply asxy=C.
M=ey+cosxcosy,N=xey−sinxsiny, so∂M∂y =ey−cosxsiny=∂N∂x. The equation is exact. ∂f∂x=ey+cosxcosyimplies thatf(x, y) =xey+ sinxcosy+φ(y). ∂f∂y=xey−sinxsinyimplies thatxey−sinxsiny+ φ′(y) =xey−sinxsinysoφ(y) = 0 andf(x, y) =xey+sinxcosy. The implicit solution isxey+ sinxcosy=C.
M = 1 +y,N = 1−x, so∂M∂y = 1 6 =−1 = ∂N∂x. The equation is not exact.
M= 1 −xy 2 y 2 −1,N= 1 −xx 2 y 2 so ∂M∂y = 1+x
2 y 2 (1−x 2 y 2 ) 2 =
∂N ∂x. The equation is exact. ∂f∂x= 1 −xy 2 y 2 −1 implies that
f(x, y) =
1
2
ln(xy+ 1)−
1
2
ln(xy−1)−x+φ(y).
1. EXACT EQUATIONS 9
∂f ∂y=
x 1 −x 2 y 2 implies that
1 2 (
x xy+1−
x xy− 1 )+φ
′(y) = x 1 −x 2 y 2. This simplifies to 1 −xx 2 y 2 +φ′(y) = 1 −xx 2 y 2 , soφ(y) = 0 andf(x, y) = 12 ln(xy+ 1)− 1 2 ln(xy−1)−x. The implicit solution is
1 2 ln(xy+1)−
1 2 ln(xy−1)−x= C. This equation can be solved fory(exercise) to produce an explicit solutiony= Ce
2 x+ x(Ce 2 x−1).
- M= 1 −xy 2 y 2 +x,N= 1 −xx 2 y 2 , so∂M∂y = 1+x
2 y 2 (1−x 2 y 2 ) 2 =
∂N ∂x. The equation is exact. ∂f∂x= 1 −xy 2 y 2 +ximplies that
f(x, y) =
1
2
ln(xy+ 1)−
1
2
ln(xy−1) +
x 2 2
+φ(y).
∂f ∂y=
x 1 −x 2 y 2 implies that
1 2 (
x xy+1−
x xy− 1 +φ
′(y) = x 1 −x 2 y 2. This simplifies to 1 −xx 2 y 2 +φ′(y) = 1 −xx 2 y 2 , soφ(y) = 0 andf(x, y) = 12 ln(xy+ 1)− 1 2 ln(xy−1)+
x 2 2. The implicit solution is
1 2 ln(xy+1)−
1 2 ln(xy−1)+
x 2 2 = C. This equation can be solved fory(exercise) to produce an explicit solutiony= Ce
−x 2 + x(Ce−x 2 −1).
M=xlny+xy,N=ylnx+xy, so∂M∂y =xy+x 6 =yx+y=∂N∂x. The equation is not exact.
M= 1+y 2 sin 2x,N=− 2 ycos 2 x, so∂M∂y = 2ysin 2x= 4ycosxsinx= ∂N ∂x. The equation is exact.
∂f ∂x = 1 +y
2 sin 2ximplies thatf(x, y) = x− 12 y 2 cos 2x+φ(y). ∂f∂y=− 2 ycos 2 ximplies that−ycos 2x+φ′(y) = − 2 ycos 2 x. Since cos 2 x= 12 + 12 cos 2x, the last equation is equivalent to −ycos 2x+φ′(y) = −y−ycos 2xso φ′(y) = −y, φ(y) = − 12 y 2 , andf(x, y) =x− 12 y 2 cos 2x− 12 y 2. Therefore, the implicit function is x− 12 y 2 cos 2x− 12 y 2 =C.
- M = 3x 2 (1 + lny),N= x 3 y − 2 yso
∂M ∂y =
3 x 2 y =
∂N ∂x. The equation is exact. ∂f∂x= 3x 2 (1+lny) implies thatf(x, y) =x 3 (1+lny)+φ(y).∂f∂y= x 3 y − 2 yimplies that
x 3 y +φ
′(y) =x 3 y − 2 y, soφ
′(y) =− 2 y, φ(y) =−y 2 , andf(x, y) =x 3 (1+lny)−y 2. The implicit solution isx 3 (1+lny)−y 2 = C.
(a) ∂f∂x = (x+yy) 2 −1 implies thatf(x, y) =−x+yy−x+φ(y). ∂f∂y = 1 −(x+xy) 2 implies that 1−(x+xy) 2 =phi′(y)−(x+xy) 2 soφ′(y) = 1,
1. ORTHOGONAL TRAJECTORIES AND FAMILIES OF CURVES 11
!"# !$%& !& !'%& # '%& & $%& "#
!&
!'%&
'%&
&
The familyxy=cand orthogonal trajectories.
!"# !$%& !& !'%& # '%& & $%& "#
!&
!'%&
'%&
&
More on the familyxy=cand orthogonal trajectories.
!"#$ !" !%#& !%#" !'#( !'#$ ' '#$ '#( %#" %#& " "#$
!%#&
!%#"
!'#(
!'#$
'#$
'#(
%#"
%#&
12 CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?
!"#$ !" !%#& !%#" !'#( !'#$ ' '#$ '#( %#" %#& " "#$
!%#&
!%#"
!'#(
!'#$
'#$
'#(
%#"
%#&
The familyy=cx 2 and orthogonal trajectories.
!" !# !$ !% !& ' & % $ # "
!$
!%
!&
&
%
$
!" !# !$ !% !& ' & % $ # "
!$
!%
!&
&
%
$
The familyx+y=cand orthogonal trajectories.
14 CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?
!"# !$% !%" !&' !( # ( &' %" $% "#
!%"
!&'
!(
(
&'
%"
The familyr=c(1 + cosθ), c= 2, 4 , 8 ,16 and orthogonal trajectories.
!" !# !$ !% !& ' & % $ # "
!$
!%
!&
&
%
$
!" !# !$ !% !& ' & % $ # "
!$
!%
!&
&
%
$
The familyy=cexand orthogonal trajectories.
1. ORTHOGONAL TRAJECTORIES AND FAMILIES OF CURVES 15
!" !# !$ !% !& ' & % $ # "
!$
!%
!&
&
%
$
!" !# !$ !% !& ' & % $ # "
!$
!%
!&
&
%
$
The familyy=cx 4 and orthogonal trajectories.
1. HOMOGENEOUS EQUATIONS 17
1 Homogeneous Equations
The lettersMandNwill always refer to the equationMdx+Ndy= 0. Note that this equation is equivalent toy′=−MN.
- Verify the equation is homogeneous, solve.
(a) M =x 2 − 2 y 2 andN =xyare homogeneous of degree 2; y′= −x
2 − 2 y 2 xy. The substitutions z =
y x andy
′ = xz′+z yield the separable equation xz′+z = 2z− 1 z. This equation has so- lutionsz = ±
√
1 +Cx 2 so the original equation has solutions y=±x
√
1 +Cx 2. (b) M= 3xy+ 2y 2 andN=−x. M has degree 2 andN has degree
- This equation is not homogeneous. Note that it is Bernoulli. (c) Divide the equation byx 2 to obtainy′= 3(1+(yx) 2 arctanxy+yx, a homogeneous equation. The substitutionsz=xyandy′=xz′+z yield the separable equationxz′+z= 3(1 +z 2 ) arctanz+z. This equation has the solutionz= tan(Cx 3 ) so the original equation has solutiony=xtan(Cx 3 ). (d) Divide the equation by x to make it homogeneous. The sub- stitutions z = yx and y′ = xz′+z yield the separable equa- tion (xz′+z)sinzz = sinz+ 1 z. This equation has the solution cosz+lncx= 0 so the original equation has solution cosyx+lncx=
- (e) Divide byxto make the equation homogeneous. The substitutions z = yx andy′ =xz′+z yield the separable equationxz′+z = z+ 2e−z. This equation has the solutionz= ln(2 lnx+C) so the original equation has solutiony=xln(2 lnx+C). (f) M = x−yand N = −(x+y) are homogeneous of degree 1; y′ = xx−+yy. The substitutionsz = yx andy′ =xz′+z yield the separable equationxz′+z= 1 1+−zz. This equation has the solution z 2 + 2z−1 =Cx− 2 and the original equation has solutiony 2 + 2 xy−x 2 =C. (g) This equation is linear and homogeneous. To solve it as a ho- mogeneous equation, divide byx: y′ = 2− 6 xy, then substitute z = yx to obtain the separable equationxz′+z = 2− 6 z. The
18 CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?
solution isz= 27 +Cx− 7 so the original equation has the solution y= 27 x+Cx− 6.
(h) Divide by x: y′ =
√
1 +y 2 x 2 , then substitutez =
y x to obtain the separable equationxz′+z =
√
1 +z 2. The solution isz 2 + z
√
1 +z 2 + ln(z+
√
1 +z 2 ) = 2 lnx+C. The solution to the original equation can be simplified toy 2 +y
√
√ x 2 +y 2 +x 2 ln(y+ x 2 +y 2 ) =x 2 (3 lnx+C). (i) The equation is Bernoulli and homogeneous. To solve it asho- mogeneous, divide byx 2 : y′= y
2 x 2 +
2 y x, and substitutez =
y x to obtain the separable equationxz′+z=z 2 + 2z. The solution is z=Cx−x. The solution to the original equation isy= x 2 C−x. (j) M =x 3 +y 3 andN =−xy 2 are homogeneous of degree 3;y′= x 3 +y 3 xy 2. The substitutionz=
y xyields the separable equationxz
′+
z=1+z 3 z 2. This equation has the solutionz=
√ 3 3 lnx+Cand the
original equation has the solutiony=x 3
√
3 lnx+C.
- Solvingdxdy=F(dxax++eyby++fc)
(a) The substitutionsx=z−handy=w−kyield the equation dw dz =F(
az+bw−(ah+bk−c) dz+ew−(dh+ek−f)). Note that
dy dx=
dy dw·
dw dz·
dz dx=
dw dz. Since ae 6 =bdthere are unique numbershandksuch thatah+bk=c anddh+ek=f. Using these values the new equation is dwdz = F(azdz++bwew) which is homogeneous. (b) Ifae=bd, then there is a constantksuch thatcx+dy=k(ax+by) for allxandy. We are assuming thataandbare not both 0. The equation then has the formdydx=F(k(axax++byby+)+cf). The substitution z=ax+byreduces it toz′=a+bF(kzz++cf) which is separable.
- Observe that ifz=xyntheny′=xnz′+nxn− 1 z.
(a) The substitution z = xyn yields z′ = x
−(2n+1)−z 2 (2n+1) 2 xz. Setting n=− 12 this becomesz′= 21 xz, a separable equation with solution z 2 = lnx+C. Therefore,y 2 =lnxx+C.
424580021 George F Simmons Differential Equations With Applications and Historical Notes Mc Graw Hill Science 1991 Solutions pdf
Course: Differential equation (f211)
- Discover more from: