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-
Solutions Manual
for
A Course in Ordinary DifferentialEquations
by
Randall J. SwiftStephen A. Wirkus
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2
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Preface
This solutions manual is a guide for instructors using A Course in Ordinary Differential Equations.Many problems have their solution presented in its entirety while some merely have an answer andfew are skipped. This should provide sufficient guidance through the problems posed in the text.
As with the book, code for Matlab, Maple, or Mathematica is not given. It is our experience thatthe syntax given in the book is sufficient to learn the relevant commands used to obtain solutions tothe various problems in the book. Please give Appendix A a chance, if you have not done so already.
This solutions manual was put together by many people and we note a few of them here. We owea big thanks to our former students David Monarres, for help in preparing portions of this book,and Walter Sosa and Moore Chung, for their help in preparing solutions. More recently Scott Wildehas helped tremendously in shaping this manual. Jenny Switkes and other colleagues and studentshave also given feedback on various drafts of this manual and all have been helpful.
This book has evolved over the last few years and we have tried to make this solution manualstay in step. However, we realize that there are probably many typos throughout and we encourageyou to contact us with your corrections. Hopefully future printings of this manual will have anexponentially decreasing number of such errors.
We would appreciate any comments that you might have regarding the book and the manual.
Randall J. Swift (e-mail: [email protected])Stephen A. Wirkus (e-mail: [email protected])
URL for typos and errata: http://www.csupomona.edu/swirkus/ACourseInODEs
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Contents
1 Traditional First-Order Differential Equations 11.1 Some Basic Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Physical Problems with Separable Equations . . . . . . . . . . . . . . . . . . . . . . 241.4 Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.5 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.6 Chapter 1: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2 Geometrical and Numerical Methods for First-Order Equations 512.1 Direction Fieldsthe Geometry of Differential Equations . . . . . . . . . . . . . . . 512.2 Existence and Uniqueness for First-Order Equations . . . . . . . . . . . . . . . . . . 562.3 First-Order Autonomous EquationsGeometrical Insight . . . . . . . . . . . . . . . 592.4 Population Modeling: An Application of Autonomous Equations . . . . . . . . . . . 822.5 Numerical Approximation with the Euler Method . . . . . . . . . . . . . . . . . . . . 832.6 Numerical Approximation with the Runge-Kutta Method . . . . . . . . . . . . . . . 872.7 An Introduction to Autonomous Second-Order Equations . . . . . . . . . . . . . . . 912.8 Chapter 2: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
3 Elements of Higher-Order Linear Equations 933.1 Some Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 933.2 Essential Topics from Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 943.3 Reduction of OrderThe Case of n = 2 . . . . . . . . . . . . . . . . . . . . . . . . . 1013.4 Operator Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1043.5 Numerical Consideration for nth Order Equations . . . . . . . . . . . . . . . . . . . 1063.6 Chapter 3: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
4 Techniques of Higher-Order Linear Equations 1154.1 Homogeneous Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . 1154.2 A Mass on a Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1234.3 Cauchy-Euler (Equidimensional) Equation . . . . . . . . . . . . . . . . . . . . . . . . 1294.4 Nonhomogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.5 Method of Undetermined Coefficients via Tables . . . . . . . . . . . . . . . . . . . . 1384.6 Method of Undetermined Coefficients via the Annihilator Method . . . . . . . . . . . 1464.7 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1544.8 Chapter 4: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
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5 Fundamentals of Systems of Differential Equations 1675.1 Systems of Two EquationsMotivational Examples . . . . . . . . . . . . . . . . . . 1675.2 Useful Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1715.3 Linear Transformations and the Fundamental Subspaces . . . . . . . . . . . . . . . . 1745.4 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1775.5 Matrix Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1805.6 Chapter 5: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
6 Techniques of Systems of Differential Equations 1856.1 A General Method, Part I: Solving Systems with Real, Distinct Eigenvalues . . . . . 1856.2 A General Method, Part II: Solving Systems with Repeated Real or Complex Eigen-
values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1886.3 Solving Linear Homogeneous and Nonhomogeneous Systems of Equations . . . . . . 1936.4 Nonlinear Equations and Phase Plane Analysis . . . . . . . . . . . . . . . . . . . . . 1956.5 Epidemiological Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2046.6 Chapter 6: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
7 Laplace Transforms 2137.1 Fundamentals of the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . 2137.2 Properties of the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 2147.3 Step Functions, Translated Functions, and Periodic Functions . . . . . . . . . . . . . 2147.4 The Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2157.5 Laplace Transform Solution of Linear Differential Equations . . . . . . . . . . . . . . 2167.6 Solving Linear Systems using Laplace Transforms . . . . . . . . . . . . . . . . . . . . 2177.7 The Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2177.8 Chapter 7: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
8 Series Methods 2218.1 Power Series Representations of Functions . . . . . . . . . . . . . . . . . . . . . . . . 2218.2 The Power Series Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2238.3 Ordinary and Singular Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2248.4 The Method of Frobenius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2258.5 Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2268.6 Chapter 8: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
B Graphing Factored Polynomials 231
C Selected Topics from Linear Algebra 233C.1 A Primer on Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233C.2 Gaussian Elimination, Matrix Inverses, and Cramers Rule . . . . . . . . . . . . . . . 235C.3 Coordinates and Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
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Chapter 1
Traditional First-Order DifferentialEquations
1.1 Some Basic Terminology
1. With y(x) = 2x3, we have y(x) = 6x2. Substituting into the ODE gives x(6x2) = 3(2x3),which is true for x (,).
2. y = 2,dy
dx=
d
dx(2) = 0. Substituting y and
dy
dxinto the ODE, we get 0 = x3(2 2)2, which
is true for all x. Thus y = 2 is a solution tody
dx= x3(y 2)2 on (,).
3. y(x) =1
5x + 4= (5x + 4)1, dy
dx= (5x + 4)2(5) =
5(5x + 4)2
. y anddy
dxdo not exist when
5x + 4 = 0 x = 45. Substituting y and
dy
dxinto the ODE, we get
5(5x + 4)2
= 5( 1
5x + 4
)2=
5(1)2(5x + 4)2
=5
(5x + 4)2
which is true for {x | x = 45}. Thus y =1
5x + 4is a solution to the ODE on ( 45 ,).
4. Since y(x) = ex x, we calculate y(x) = ex 1. Substituting into the ODE gives(ex 1) + (ex x)2 = e2x + (1 2x)ex + x2 1
ex 1 + e2x 2xex + x2 = e2x + ex 2xex + x2 1which is true for all x (,).
5. y(x) = x3,dy
dx= 3x2. Substituting y and
dy
dxinto the ODE, we get
3x2 = 3(x3)2/3
= 3x2
1
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2 Section 1.1
which is true for all x. Thus y = x3 is a solution to the ODE on (,).
6. y(x) =1
x 3 ,dy
dx= (x 3)2 = 1
(x 3)2 . y anddy
dxdo not exist when x 3 = 0 x = 3.
Substituting y anddy
dxinto the ODE, we get
1(x 3)2 =
( 1x 3
)2=
(1)2(x 3)2
=1
(x 3)
which is true for {x | x = 3}. Thus y = 1x 3 is a solution to the ODE on (, 3).
7. Taking the derivative of y(x) = x2 x1 gives us y(x) = 2x x2 and y(x) = 2 + x3.Substitution gives
x2(2 + x3) = 2(x2 x1) 2x2 + x1 = 2x2 x1
which is true where both sides are defined, which occurs when x = 0.
8. y(x) = sin x + 2 cosx,dy
dx= cosx 2 sinx, and d
2y
dx2= sinx 2 cosx. Substituting y and
d2y
dx2into the ODE, we get
( sinx 2 cosx) + (sin x + 2 cosx) = 0
which is true for all x. Thus y = sinx + 2 cosx is a solution to the ODE on (,).
9. y(x) = x,dy
dx= 1, and
d2y
dx2= 0. Substituting y and
d2y
dx2into the ODE, we get
(0) + (x) = x
which is true for all x. Thus y = x is a solution to the ODE on (,).
10. y(x) = x + C sin x,dy
dx= 1 + C cosx, and
d2y
dx2= C sin x. Substituting y and d
2y
dx2into the
ODE, we get
(C sin x) + (x + C sinx) = xx = x
which is true for all x. Thus y = x + C sin x is a solution to the ODE on (,) for anyconstant C.
11. (a) y = ex, y = ex, y = ex y 3y + 2y = ex 3ex + 2ex = 0(b) y = e2x, y = 2e2x, y = 4e2x
y 3y + 2y = 4e2x 6e2x + 2e2x = 0
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Section 1.1 3
12. (a) y = ex, y = ex, y = ex y 2y + y = ex 2ex + ex = 0(b) y = xex, y = xex + ex, y = xex + 2ex
y 2y + y = xex + 2ex 2xex 2ex + ex = 013. (a) y = sin 3x, y = 3 cos 3x, y = 9 sin 3x
y + 9y = 9 sin3x + 9 sin 3x = 0 YES(b) y = sinx, y = cosx, y = sinx
y + 9y = sinx + 9 sinx = 0 NO(c) y = cos 3x, y = 3 sin3x, y = 9 cos 3x
y + 9y = 9 cos 3x + 9 cos 3x = 0 YES(d) y = e3x, y = 3e3x, y = 9e3x
y + 9y = 9e3x + 9e3x = 0 NO(e) y = x3, y = 3x2, y = 6x
y + 9y = 6x + 9x3 = 0(unless x = 0)NO14. y + 6y + 9y = 0
(a) y = ex, y = ex, y = ex ex + 6ex + 9ex = 0 NO
(b) y = e3x, y = 3e3x y = 9e3x 9e3x 18e3x + 9e3x = 0 YES
(c) y = xe3x, y = 3xe3x + e3x y = 9e3x 6e3x 9xe3x 6e3x 18xe3x + 6e3x + 9xe3x = 0 YES
(d) y = 4e3x, y = 12e3x, y = 36e3x
36e3x + 72e3x + 36e3x = 0 NO(e) y = e3x(x + 2), y = e3x(3x 5), ye3x(9x + 12)
e3x(81x + 108 18x 30 + 18 + 9x) = 0 NO15. y 7y + 12y = 0
(a) y = e2x, y = 2e2x, y = 4e2x
4e2x 14e2x + 12e2x = 0 NO(b) y = e3x, y = 3e3x, y = 9e3x
9e3x 21e3x + 12e3x = 0 YES(c) y = e4x, y = 4e4x, y = 16e4x
16e4x 28e4x + 12e4x = 0 YES(d) y = e5x, y = 5e5x, y = 25e5x
25e5x 35e5x + 12e5x = 0 NO(e) y = e3x + 2e4x, y = 3e3x + 8e4x, y = 9e3x + 32e4x
32e4x + 9e3x 21e3x 56e3x + 12e3x + 24e4x = 0 YES16. y + 4y + 5y = 0
(a) y = e2x, y = 2e2x, y = 4e2x 4e2x 8e2x + 5e2x = 0 NO
(b) y = e2x sin 2x, y = 2e2x cos 2x 2e2x sin 2x, y = 8e2x cos 2x 8e2x cos 2x + 8e2x cos 2x 8e2x sin 2x + 5e2x sin 2x = 0NO
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4 Section 1.1
(c) y = e2x cos 2x, y = 2e2x cos 2x 2e2x sin 2x, y = 8e2x sin 2x 8e2x sin 2x 8e2x cos 2x 8e2x sin 2x + 5e2x cos 2x = 0NO
(d) y = cos 2x, y = 2 sin2x, y = 4 cos 2x 4 cos 2x 8 sin 2x + 5 cos 2x = 0 NO
17. y = erx, y = rerx, and y = r2erx. The ODE gives
r2erx + 3rerx + 2erx = 0 erx(r2 + 3r + 2) = 0 erx(r + 2)(r + 1) = 0 r = 2,1
18. y + 4y + 4y = 0y = xerx, y = rxerx + erx, y = r2xerx + 2rerx
erx(r2x + 2r + 4rx + 4 + 4x
)= 0
xr2 + (2 + 4x)r + (4 + 4x) = 0
r = 2 4x
42x
=4x2x
or4 4x
2x
= 2 or 2 2xx
We can disregard the value involving x, which leaves only r = 2.19. (a) 3y + y = sinx (i) 2nd order (ii) linear (iii) N/A
(b) y + sin y = 0 (i) 2nd order (ii) nonlinear (iii) N/A
(c) y(3) + (sin x)y(2) + y = x, y(0) = 1, y(0) = 0, y(0) = 2(i) 3rd order (ii) linear (iii) IVP
(d) y + exy = y4, y(0) = 0(i) 1st order (ii) nonlinear (iii) IVP
(e) y + y y (i) 2nd order (ii) linear (iii) N/A(f) y + exy + y2 = 0, y(0) = 1, y() = 0
(i) 2nd order (ii) nonlinear (iii) BVP
20. (a) y 3yy = x (i) 2nd order (ii) nonlinear (iii) N/A(b) y = sin x (i) 2nd order (ii) linear (iii) N/A
(c) y + 3y = 0, y(0) = 1, y(1) = 0(i) 2nd order (ii) linear (iii) BVP
(d) y = 0, y(1) = 1, y(1) = 2(i) 2nd order (ii) linear (iii) IVP
(e) y 4y + 4y = 0, y(0) = 1, y(0) = 1(i) 2nd order (ii) linear (iii) IVP
(f) x2y + y + (lnx)y = 0 (i) 2nd order (ii) linear (iii) N/A
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1.2. SEPARABLE DIFFERENTIAL EQUATIONS 5
1.2 Separable Differential Equations
1.
4xy dx + (x2 + 1) dy = 04x
x2 + 1dx = 1
ydy
4xx2 + 1
dx =
dy
y
Letting u = x2 + 1,
2
du
u=
dy
y
2 ln |u| + C1 = ln |y|2 ln(x2 + 1) + C1 = ln |y|
e2 ln(x2+1)+C1 = |y|
|y| = eC1(x2 + 1)2, since eC1 > 0y =
C
(x2 + 1)2, where C = eC1
2.
tan xdy + 2y dx = 0
dy
2y=
dx
tan x ln |y| = 2 ln |sin x| + C1
1|y| = |sinx|
2eC1
y = C csc2 x, c = eC1
3. (ex + 1) cos y dy + ex(sin y + 1) dx = 0, y(0) = 3. Then
cos ysin y + 1
dy = ex
ex + 1dx
Substituteu1 = sin y + 1 u2 = ex + 1
du1 = cos y dy du2 = ex dx
Then
ln |sin y + 1| = ln |ex + 1| + C sin y + 1 = A[
1(ex + 1)
]
Apply IC: (sin 3) + 1 =A
(e0 + 1) A = 2(sin 3) + 2
Therefore,
sin y + 1 =2 + 2 sin 3
ex + 1
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6 Section 1.2
4. We have 2x(y2 + 1) dx + (x4 + 1) dy = 0, y(1) = 1. Then
(x4 + 1) ds = 2x(s2 + 1) drdy
y2 + 1= 2x
x4 + 1dx =
u = x2
du = 2xdx
1u2 + 1
du
Therefore,
arctan(y) = arctan(u) + Carctan(y) = arctan(x2) + Carctan(1) = arctan(1) + C
+2 arctan(1) /4
= C C = 2
Finally,
arctan(y) = arctan(x2) + 2
y = tan(
2 arctan(x2)
)
5. xy dx + (x + 1) dy = 0
dy
y= x
x + 1dx. Substituting u = x + 1, x = u 1, du = dx, we
have
ln |y| =
(u 1)u
du = (
1u 1)
du = ln |x + 1| x + C
y = A(x + 1)ex
6.
(y2 + 1)1/2 dx = xy dydx
x=
yy2 + 1
dydx
x=
yy2 + 1
dy
Let u = y2 + 1 and du = 2y dy, so du2 = y dy. Then we havedx
x=
12
du
u
ln |x| + C = (1/2)
u
(1/2)
ln |x| + C =
y2 + 1
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Section 1.2 7
7. We have (x2 1)y + 2xy2 = 0 and y(2) = 1. Then
(x2 1)dydx
= 2xy2
dy
y2=
2xdxx2 1
dy
y2=
2xx2 1 dx
1y
= ln |x2 1| + C
y =1
ln |x2 1| + CIC:
11
= ln |2 1| + C C = 1
y =1
ln |x2 1| + 1
8. We have y cotx + y = 2 and y(0) = 1. Thendy
dxcotx = y + 2dy
y 2 =
tan xdx
ln |y 2| = ln |cosx| + C|y 2| = eC |cos x|
y = eC(cosx) + 2 = A(cosx) + 2
Since y(0) = 1, 1 = A + 2 A = 3. Thus
y = 3(cosx) + 2
9.
y = 10x+y dydx
= 10x10y
dy
10y= 10x dx
10y dy =
10x dx
Use au = eu ln a. Then
10y = 10x + C110y =
110x + C1
y = log10 |10x + C|
-
8 Section 1.2
10.
xdx
dt+ t = 1
xdx = (1 t) dtxdx =
(1 t) dt
12x2 = t 1
2t2 + C1
x2 = 2t t2 + 2C1x =
2t t2 + C, C = 2C1
11. Let u = y x
dudx
=dy
dx 1 du
dx+ 1 = cosu
du
dx= cosu 1
du
cosu 1 = dx
x =1
tan(u/2)
=1
tan(yx2 )
y = x + 2 tan1(1/x)
12. Let u = 2x + y 3 dudx
= 2 +dy
dx
u 2 = uu = u + 2
du
u + 2= dx
ln(u + 2) = xex = u + 2 = 2x + y 1y = ex 2x + 1
13. We have (x + 2y)y = 1, y(0) = 2. Let u = 2y + x. Thendu
dx= 2
dy
dx+ 1
Substituting into the ODE gives
u
(u 1
2
)= 1
This is separable:
u 1 = 2u
u = 2u
+ 1 =2 + u
u
u
2 + udu =
dx
-
Section 1.2 9
Note thatu
2 + u=
u + 2 22 + u
= 1 22 + u
Then (1 2
2 + u
)du =
dx u 2 ln |2 + u| = x + C
Substitute back u = x + 2y:
(x + 2y) 2 ln |2 + x + 2y| = x + CApply IC: 0 + 2(2) 2 ln |2 + 0 + 2(2)| = 0 + C
4 2 ln |2| = CThe solution is given in implicit form as
2y 2 ln |2 + x + 2y| + 4 + 2 ln 2 = 0
14. y =
4x + 2y 1. Let u = 4x + 2y 1 dudx = 4 + 2 dydx u = 2
u + 4
dx =du
2
u + 4x = 2 (ln (u + 2))u
= 2(ln(
4x + 2y 1 + 2))
4x + 2y 1
15. We have (y + 2) dx + y(x + 4) dy = 0, y(3) = 1. Then(x + 4)y dy = (y + 2) dx
y
y + 2dy =
dx
x + 4 (1 2
y + 2
)dy =
dx
x + 4y 2 ln |y + 2| = ln |x + 4| + C
ey2 ln |y+2| = e ln |x+4|+C1
|y + 2|2 ey = eC 1|x + 4|
ey
(y + 2)2= e
C
(x + 4)=
A
(x + 4)where A = eC
Apply IC: y(3) = 1 e1
(1 + 2)2 =A
(3 + 4) A = e1 e
y
(y + 2)2=
e1
(x + 4)
16. We have 8 cos2 y dx + csc2 xdy = 0, y(/12) = /4. Then
csc2 xdy = 8 cos2 y dxsec2 y dy = 8
sin2 xdx
tan y = 8(
12x 1
4sin 2x + C1
)tan y = 2 sin 2x 4x + C, where C = 8C1
-
10 Section 1.2
Using y(/12) = /4, we solve for C:
tan(
4
)= 2 sin
(6
)
3+ C
C =
3
Sotan y = 2 sin 2x 4x +
3
17. We havedy
dx=
y3 + 2yx2 + 3x
, y(1) = 1. Separating variables and integrating gives
1
y3 + 2ydy =
1
x2 + 3xdx
( y
2(y2 + 2)+
12y
)dy =
( 13(x + 3)
+13x
) 1
4ln |y2 + 2| + 1
2ln |y| = 1
3ln |x + 3| + 1
3ln |x| + C
Now apply IC:
14
ln(12 + 2) +12
ln(1) = 13
ln(1 + 3) +13
ln(1) + C
C = 13
ln 4 14
ln 3
The solution is then
14
ln(y2 + 2) +12
ln(y) = 13
ln |x + 3| + 13
ln |x| + 13
ln 4 14
ln 3
18. We have y = ex2, y(0) = 0. Then
dy
dx= ex
2
dy =
ex
2dx
y =
ex2dx
Apply IC: x0
y dy = x
0
et2dt
y(x) y(0) = x
0
et2dt
y(x) = x
0
et2dt
-
Section 1.2 11
19. We have y = xyex2, y(0) = 1. Then
dy
y=
xex2dx
ln |y| = 12ex
2+ C
C = ln |y| 12ex
2, when x = 0, y = 1
C = 12
Thus
ln |y| = 12x2 1
2|y| = e1/2e(1/2)x2
y = e1/2ex21
y is even since we have only an x2 term.
20. (a) We have x2y cos 2y = 1, limx+ y = 9/4. Then
x2dy
dx= cos 2y + 1
First we use a trigonometric identity as follows:
cos 2y + 1 = (cos2 y sin2 y) + (cos2 y + sin2 y)= 2 cos2 y
Thus we obtain
dy
2 cos2 y=
dx
x2
2
sec2 y dy =
dx
x2
2 tan y = 2x
+ C1
tan y = 1x
+ C2
C2 = tan y +1x
limx+C2 = tan
(94
)+ 0 = 1
Thus
tan y = 1 1x
or y = arctan(
1 1x
)
-
12 Section 1.2
(b) We have 3y2y + 16x = 2xy3. Then
3y2dy
dx= 2x(y3 8)
3y2
y3 8 dy =
2xdx
ln |y3 8| = x2 + C1|y3 8| = eC1ex2
y3 = eC1ex2 + 8y = [Cex
2+ 8]1/3 where C = eC1
Notice that no value for C bounds y since eC1 = 0.21.
dN
dt= rN
1N
dN = r dt1N
dN =
r dt
ln N = rt + C1
Note N is number of population, i.e. > 0.
elnN = ert+C1
N = eC1ert
N = C2ert
Note C2 is > 0, dealing with possible population. Thus N = C2ert is a solution to the ODE.
Case 1. For values r 0 we have exponential growth as t .Case 2. For values r < 0 we have exponential decay as t .
-
Section 1.2 13
22.
dN
dt= r
(1 N
K
)N
dN
rN(1 NK
) = dt(Let u = 1 N/K) du
dt= 1
K
dN
dt
K dudt
= ruK(1 u)du
dt= ru(u 1)
dt =du
ru(u 1) t = ln |u 1| ln |u|
r
=1r
ln
(1 1(
1 NK))
etr = 1 1(1 NK
)1 N
K=
11 etr
N = K K1 etr
limtN = K
23. We have (x + y) dx xdy = 0. Then(x + y)
x=
dy
dx 1 + y
x=
dy
dx
Letu =
y
x y = u = xu
Thenu + xu = 1 + u u = 1
x u = ln |x| + C
Substituting for u givesy = x(ln |x| + C)
24. We havedy
dx=
y2 + 2xyx2
=y2
x2+
2xyx2
Letu =
y
x y = u + xu
Then
u + xu = u2 + 2u u = u2 + ux
-
14 Section 1.2
This equation is separable: du
u2 + u=
1x
dx = ln |x| + C1
Use partial fractions to rewrite1
u2 + u.
1u(u + 1)
=A
u+
B
u + 1 1 = A(u + 1) + Bu = u(A + B) + A
A + B = 0A = 1
} B = 1
1
u(u + 1)du =
(1u 1
u + 1
)du = ln |u| ln |u + 1| = ln
uu + 1
Thus
ln uu + 1
= ln |x| + C uu + 1
= eln |x|+C1 = eC1 |x| u
u + 1= C x u = Cx(u + 1)
u(1 Cx) = Cx u = Cx
1 CxPlugging in u = y/x then gives
y =Cx2
1 Cx25. We have
2x2dy
dx= x2 + y2 2dy
dx= 1 +
y2
x2
Let u = yx y = u + xu. Substituting gives
2(u + xu) = 1 + u2 2xu = 1 + u2 2u = (u 1)2
This is separable: 2 du
(u 1)2 =
1x
dx
Letting w = u 1, dw = du,
2u 1 = ln |x| + C u 1 =
2ln |x| + C
u = 1 2ln |x| + C
Plugging in u = y/x gives
y = x 2xln |x| + C
-
Section 1.2 15
26.
xy y =
x2 + y2
xy =
x2 + y2 + y
y =
x2 + y2
x+
y
x
=
1 +( y
x
)2+( y
x
)(homogeneous)
Let y = vx, y = vx + v. Then
dv
dxx + v =
1 + v2 + v
xdv
dx=
1 + v2
11 + v2
dv =1x
dx
arcsinh v = ln |x| + C
Substituting back in, we have
y
x= sinh(ln |x| + C) y = x sinh(ln |x| + C)
27. We have (x + 2y) dx xdy = 0. Then
x + 2y xdydx
= 0
x + 2y = xdy
dxdy
dx= 1 + 2
(yx
)(homogeneous)
Let y = vx, y = vx + v. Then
dv
dxx + v = 1 + 2v
dv
dx= 1 + v
1v + 1
dv = dx
ln |v + 1| = x + Cv + 1 = Cex
y
x= Cex 1
y = Cxex x
28. We have
(y2 2xy) dx + x2 dy = 0 dydx
=2xy y2
x2= 2(y
x
)( y
x
)2
-
16 Section 1.2
Let u = yx , y = ux + u. Then
xu + u = 2u u2 ux = u u2 ln |u 1| + ln |u| = lnx + C1 u
u 1 = eln x+C1 = x C
y/x(y/x) 1 = xC
yxy x = Cx
2
yy x = Cx
y = xCy Cx2
y = Cx2
1 + Cx
29. We have 2x3y = y(2x2 y2). Thendy
dx=
y3(2x2 y2)2x3
dydx
=( y
x
)(1 1
2
(yx
)2)
Let u =y
x,
dy
dx= x
du
dx+ u. Then
xdu
dx+ u = u
(1 1
2u2)
dudx
x = 12u3
2u3 du =
1x
dx
u2 = ln x + C y2 = x2 ln x + Cx2 y = xln x + C
-
Section 1.2 17
30. We have (x2 + y2)y = 2xy. Then
dy
dx=
2xyx2 + y2
dvdx
x + v =2x2v
x2 + v2x2
dvdx
x + v =2v
1 + v2
dvdx
x =2v
1 + v2 v
dvdx
x =2v v v3
1 + v2
dvdx
x =v v31 + v2
1 + v2
v v3 dv =1x
dv
1v(1 v2) dv +
v
(1 v2) dv =1x
dx
ln |v| ln |v2 1| = lnx + Cv
v2 1 = Cx y(
yx
)2 1 = Cx2 y = Cx2
(yx
)2 Cx2
y = Cy2 Cx2 y = C(y2 x2)
31.
xy y = x tan(y/x)y =
y
x+ tan
( yx
)(Homogeneous)(
y = vx, y = v + xv, v =y
x
)v + xv = v + tan v
dv
tan v=
dx
xln |x| = ln | sin v| + C|x| = eC | sin v| = eC | sin y
x|
y
x= sin1 | x
eC|
y = x sin1 | xeC
|
32.
(2x + y) dx (4x + 2y) dy = 0dy
dx=
12
y =x
2+ C
-
18 Section 1.2
33. We have y2 + x2y = xyy. Then
y2 = y(xy x2) y = y2
xy x2
y = 4y x
dydx
=y
(y x) (y x) dy = y dx (y x) dy + (y) dx = 0
Let y = vx + v. Then
dv
dxx + v =
vx
vx y =v
v 1 dv
dxx =
v
v 1 v
dvdx
x =v v(v 1)
v 1 dv
dxx =
v v2 + vv 1
dvdx
x =2v v2v 1
v 12x v2 dv =
1x
dx 12
ln(v 2) 12
ln |v| = lnx + C
ln 1v 2
+ ln 1v = ln x + C
1v(v 2) = Cx
1(yx
) (yx 2
) = Cx1
1x2 (y)(y 2x)
= Cx
=1
y(y 2x) = C
34. (x y) + (y x)y = 0 y = 1( for y = x) y = x + C. If x = y, the equation is truetrivally.
-
Section 1.2 19
35.
(x + 4y)y = 2x + 3y
y =2x + 3yx + 4y
=2 + 3(y/x)1 + 4(y/x)(
y = vx, y = v + xv, v =y
x
)v + xv =
2 + 3v1 + 4v
dx
x=
(1 + 4v)dv2 + 2v 4v2(Partial fractions yields:)
ln |x| = 16[ln(1 + 2v)(2 2v)5]
ln |x| = 16[ln(1 + 2(y/x))(2 2(y/x))5]
36. We have
(x y) dx + (x + y) dy = 0 dydx
=y xx + y
=yx 11 + yx
Let u = yx , y = xu + u. Then
xu + u =u 11 + u
dudx
=u 11 + u
u
dudx
=1 u21 + u
1 + u1 + u2
du =
dx (1
1 + u2+
u
1 + u2
)du =
dx arctanu + 1
2ln |1 + u2| = x
arctan( y
x
)+
12
ln(
1 +(y
x
)2)= x
-
20 Section 1.2
37.
y dx = (2x + y) dydy
dx=
(y/x)2 + (y/x)
(y = vx, y = v + xv)
v + xv =v
2 + v
xv = v2 + vv + 2
2 + vv2 + v
dv =dx
x
ln( |v + 1|
v2
)= ln |x| + C
|v + 1|v2
= eC |x||(y/x) + 1|
(y/x)2= eC |x|
38.
y = 2(
y
x + y
)2= 2
((y/x)
1 + (y/x)
)2(y = vx, y = v + xv, v =
y
x
)v + xv =
2v2
(1 + v)2
xv =v2 v
(1 + v)2
(1 + v)2
v2 v dv =dx
x
ln(
(v 1)4v
)+ v = ln |x| + C
ln(
((y/x) 1)4(y/x)
)+ (y/x) = ln |x| + C
41. (a) M(1,y
x)
t= 1x= M(tx, ty)Homog. deg. n
= tnM(x, y)
=(
1x
)nM(x, y) M(x, y) = xnM(1, y
x)
(b) Repeat above exchanging M for N .
(c) M(x, y)dx + N(x, y)dy = 0 dydx
=M(x, y)N(x, y)
=M(1, yx)N(1, yx )
Define g1(t) = M(1, t), g2(t) = N(1, t), g(t) = g1(t)g2(t)
dydx
=M(1, yx )N(1, yx )
=g1( yx)g2( yx )
= g(y
x)
-
Section 1.2 21
42. In (23.)F (x, y) = (x + y)dx xdy = 0 F (tx, ty) = (tx + ty)t dx tx t dy = t2((x + y)dx xdy) = t2F (x, y) Homogeneous of degree 2. The others are similar.
43. Given that M(x, y) dx + N(x, y) dy = 0 is homogeneous, thendy
dx=
M(x, y)N(x, y)
=M(1, (y/x))N(1, (y/x))
and the substitution
x = r cos tdx = r sin t dt + cos t dry = r sin t
dy = r cos t dt + sin t drdy
dx=
M(1, tan t)N(1, tan t)
=MN
dy
dx=
r cos t dt + sin t drr sin t dt + cos t dr
MN
=r cos t dt + sin t drr sin t dt + cos t dr
(r cos t dt + sin t dr)N = M(r sin t dt + cos t dr)N sin t dr + M cos t dr = Nr cos t dt + Mr sin t dt
dr
dt= r
(N cos t + M sin tN sin t + M cos t
)Since the function M and N are functions of t, the right hand side is a function of r times afunction of t. Hence, it is seperable.
44. (a) From 43.
dr
dt= r
(r cos2 t + (cos t + sin t) sin t
cos t sin t + (cos t + sin t) cos t)
= r(
cos2 t + sin t cos t + sin2 tcos2 t
)
= r(
1 + sin t cos tcos2 t
)dr
r=(
1 + sin t cos tcos2 t
)dt
ln |r| = ln(
1cos2 t
)+ tan t
|r| = exp[ln
(1
cos2 t
)+ tan t + C
]
with r =
x2 + y2 and t = tan1(y
x
), we finally have
x2 + y2 = exp
[ln
(1
cos2(tan1(
yx
))
)+ tan(tan1
(yx
)) + C
]
(b)
-
22 Section 1.2
45. (a)dy
dx=
y xy + x
=(y/x) 1(y/x) + 1
which is homogeneous. As before, y = vx and
v + vx =v 1v + 1
vx =2
v + 1
(v + 1) dv =2 dx
xv2
2+ v + C = = 2 ln |x|
|x| = Ke(v2/4)(v/2) = K exp( y2
4x2 y
2x)
(b) i. (x y 1) dx + (y + x + 5) dy = 0 M(x, y) = x y 1 M(tx, ty) = tx ty 1 = tM(x, y). Hence, M is not homogeneous. Similarly forN , and so the equation is not homogeneous.
ii. x = X 2, y = Y 3, dx = dX , dy = dY and dydx = dYdX = Y XY +Xiii. |X | = K exp( Y
2
4X2 Y
2X) |x + 2| =
K exp( (y + 3)2
4(x + 2)2 y + 3
2(x + 2))
46. Lines are parallel, so there is no point of intersection.
47. Lines are parallel, so there is no point of intersection.
48.dy
dx=
2x + 3y 5x + 4y
which gives point of intersection to be (4,1). Thus, we let x = X 4 andy = Y + 1 and get to
dY
dX=
2X + 3YX + 4Y
=2 + 3(Y/X)1 + 4(Y/X)
Y = vX Y = vX + v
vX =1 v1 + 4v
ln |X | = 5 ln |1 v| 4Xln |x + 4| = 5 ln
1 y 1x + 4 4(x + 4)
-
Section 1.2 23
49. Point of intersection is (3,2). Thus we let x = X + 3 and y = Y 2 and getdY
dX=
Y
2X + YdX
dY=
2X + YY
= 2(x/y) + 1
X = uYdX
dY= uY + u
uY + u = 2u + 1du
u + 1=
dY
Yln |u + 1| = ln |Y | + C
|Y | = K|u + 1| = K|(X/Y ) + 1|y + 2 = K
(x 3y + 2
+ 1)
50. y = 2(
y + 2x + y 1
)2and as in 49.,
dY
dX= 2
(Y
X + Y
)2dX
dY=
12
(X
Y+ 1)2
X = uYdX
dY= uY + u
u + uX =12u2 + u +
12
uX =12(u2 + 1)
du
u2 + 1=
12
dX
X
tan1(u) =12
ln |X | + C
tan1(
x 3y + 2
)=
12
ln |x 3| + C
-
24 Section 1.3
1.3 Physical Problems with Separable Equations
1. r = kv2, ve = 60m/s, v0 = 0, m = 75kg, s0 = 1800m, se = 500m
mdv
dt= mg kv2
75v = 75(9.8) kv2= 735 kv2
75735 kv2 dv = dt
1.38 lnkv27.11
kv+27.11
k= t + C
kv 27.11kv + 27.11
= exp[
kt + C1.38
]From the given information lim
t v(t) = 60
k60 27.11k60 + 27.11
= 0 k = .20417v0 = 0 C = 0
Thus, we have.45v 27.11.45v + 27.11
= e(.326t)
and after some simplification
v(t) = 60(
1 + e.326t
1 e.326t)
2. r = kv2, r = .48N when v = 1 k = .48
.4dv
dt= (.4)(9.8) .48v2
dv
dt= 9.8 1.2v2
dv
9.8 1.2v2 = dt
If air resistance is neglected, thens(t) = 4.9t2 + 20t, v(t) = 9.8t + 20 = 0 t = 2.04, s(2.04) = 20.4ft
4. v(t) = gt + v0, v(3) = 3g + v0 = 0 v0 = 3g = 3s(32ft/s2) = 96ft/s
5. (a) v0 = 0, v(t) = gt = 32t, v(4) = 128ft/s
(b) Ave velocity=14
40
32t dt = 64ft/s
(c) v(t) = 32t, s(t)16t2 + s0, s(4) = 0 s(4) = 162 + s0 = 0 s0 = 256 cliff was 256 ft tall
6. a(t) = 32, v(t) = 32t + v0 = 32t, s(t) = 16t2 + s0,s(5) = 16(25) + s0 = 0 s0 = 400 cliff was 400 ft tall
-
Section 1.3 25
7. s0 = 350, v0 = 0, s(t) = 16t2 350 = 0 t =
3504
average velocity = 2km350/4
=8km350s
= .428km/s
8. T (t) = (T0 Ts)ekt + Ts, Ts = 2, T0 = 15, T (2) = 5 T (2) = 17e2k 2 = 5 k = 1
2ln(
717
). Then to find eating time (t = eat),
T (eat) = 17ek(eat) 2 = 0 eat = 2 ln(2/17)ln(7/17)
= 4.82hrs 5pm
9. Let t = 0 be 3pm. T (0) = 79, T (3) = 68, Ts = 60.T (t) = 19ekt + 60, T (3) = 19e3k + 60 = 68
19e3k = 8
k =13
ln .421 = .288T (t) = 19e.288t + 60
(Solve for t when T = 98.6) = 98.6t = 2.46
Thus the person died about 12:30pm
10. T (t) = 80ekt + 20, T (10) = 80e10k + 20 = 60 k = .069T (t) = 80e.069t + 20 = 25 t = 10 ln 16
ln 2= 40 min
13. N(t) = amount of N in tank after t seconds. N(0) = 16. dNdt = rate in - rate out = .1 N
20(.1)
200dN20 N = dt N(t) = 20 e
(t/200)+C ,
N(0) = 16 = 20 eC C = ln 4 1.39 N(t) = 20 e(t/200)+1.39. We want to solve N(t) =20(.99) = 19.8 for t. t = 200(ln(.2) 1.39) 599 seconds 10 min.
14. s(t)=amount of salt at time t. s(0) = 10.ds
dt= 0
(s(t)kg100L
)(5L/m) =
s(t)20
s(t) = e(t/20)+C C = ln 10s(t) = 10e(t/20)+ln10 s(60) = 4.97 5kg
15. (a)ds
dt= (4lb/gal)(2gal/m) (s(t)lb/50gal)(2gal/min) = 8 s(t)
25 s(t) = 16 + 9e(t/25) concentration= s(t)/25.
(b)ds
dt= 8 3s
50 s(t) = 400
3 325
3e(3t/50) conc= s(t)/25 t.
16. s(t) = amount of salt in pond, s0 = 0, v(t) =volume of pond, v0 = 10000, dvdt = 50 v(t) =10000 50t.ds
dt= (1250m3/d)
(5kg
1000m3
)(
s(t)kg(10000 5t)m3
)(1300m3/d)
= 6.25 1300s(t)10000 50t
Yes, after 200 days, there is no water in the pond which will result in no salt in the pond.
-
26 Section 1.3
17. C(t) = amount of CO2, C0 = (.0015)(200) = .3m3, dCdt =(
20m3
min
)(.0004)
(Cm3
200m3
)(20m3/min) =
.008 .1C This gives us that the concentration will be half of what it started (i.e. there willbe .15m3 of CO2) in 11.45 minutes.
18.dA
dt= kA,
dA
A= k dt A(t) = ekt+C = A0ekt.
A(30) = A0/2 = A0e30k
12
= e30k
k =ln(1/2)
30= .023
A(t) = .01A0 = A0e.023t
t =ln(.01).023 200 days
19. A(1) = .56 = ek k = .5798 t = 1.195 years20. p = amount of decayed tin, M=original amount of tin.
dp
dt= k(M p)(p) where M p is the amount of tin left
dP
p(M p) = k dt1M
ln(
p
M p)
= kt + C
p(t) = M M1 + eMkt+C
where C would be determined by the original amount of decayed tin in the organ.
22. Let (x, y) be the point of tangency and c be the x-intercept of the tangent line. Then 12 |c x||y| = a2 |c x| = 2a2|y| .
dy
dx=
y
|c x| =y
2a2
|y|
=y|y|2a2
dy
y2=
12a2
dx
y = 2a2
x
Because of symmetry, either graph will give the desired result.
23. Let (x, y) be the point of tangency and c be the x-intercept of the tangent line. Then |cx|+|y| = b.
dy
dx=
y
|c x| =y
b |y|b ln |y| |y| = x
This is an implicitly defined function.
-
1.4. EXACT EQUATIONS 27
33. Let A(t) be the amount of snow on the ground. Then A(t) = kt since it is falling at aconstant rate and there was none when it started. Let P (t) be the distance plowed. ThendPdt =
mA(t) =
mkt . Thus dP =
mktdt P = mk ln t + C, P (1) = 2 = C P (t) = mk ln t + 2.
P (2) = mk ln 2 +2 = 1 mk = 1.443. P (t) = 1.443 ln t + 2. Set P (t) = 0 to get t 4 whichsays it started snowing at about 8am.
1.4 Exact Equations
1. We have 2xy3 + (1 + 3x2y2) dydx = 0. Using M = 2xy3 and N = 1 + 3x2y2, we have
M
y= 6xy2,
N
x= 6xy2
so the equation is exact. We need to solve fx = M andfy = N :
f =
f
xdx =
2xy3 dx = x2y3 + g(y)
This f = x2y3 + g(y) must also satisfy fy = N .
1 + 3x2y2 =
y(x2y3 + g(y)) = 3x2y2 + g(y)
g(y) = 1 g(y) = y
Thus the solution is given by f = C:
x2y3 + y = C
2. We have (2xy + 1) + (x2 + 4y) dy = 0, y(0) = 1. Use M = 2xy + 1 and N = x2 + 4y. Then
M
y= 2x =
N
x
so the equation is exact. We set fx = M andfy = N and solve.
f =
f
ydy =
(x2 + 4y) dy = x2y + 2y2 + h(x)
This f must also satisfy fx = M :
2xy + 1 =
x(x2y + 2y2 + h(x)) = 2xy + h(x)
h(x) = 1 h(x) = x
The solution is given byx2y + 2y2 + x = C
Initial conditions yield C = 2, sox2y + 2y2 + x = 2
-
28 Section 1.4
3. We have (y sec2 x + secx tan x) + (tanx + 2y) dy = 0 with y(0) = 1. Setting M = (y sec2 x +secx tan x) and N = tan x + 2y leads to
M
y= sec2 x =
N
x
so the ODE is exact. Then,
F =
(y sec2 x + secx tan x) dx
F = y tan x + secx + g(y)tan x + 2y =
f
y= tanx + g(y)
2y = g(y) g(y) = y2
The solution is y tan x + secx + y2 = C. With initial conditions, 1(0) + 1 + 12 = C C = 2,so
y tan x + secx + y2 = 2
4. We have (2y sinx cos x + y2 sin x) + (sin2 x 2y cosx) dy = 0 with y(0) = 3. Let M =2y sinx cos x + y2 sin x and N = sin2 x 2y cosx. Then
M
y= 2 sinx cos x + 2y sinx
N
x= 2 sinx cos x 2y( sinx)
so the equation is exact. We set fx = M andfy = N and solve.
f =
f
ydy =
(sin2 x 2y cosx) dy = y sin2 x y2 cosx + h(x)
Then
2y sinx cos x + y2 sin x =
x(f) = 2y sin x cos x + y2 sin x + h(x)
h(x) = 0 h(x) = C1The solution is given by
y2 sin2 x y2 cosx = CInitial conditions give us 32 sin2 0 32 cos 0 = 0, so C = 9 and
y2 sin2 x y2 cosx = 9
5. We have 2xy + (x2 y2) dy = 0. Setting M = 2xy and N = x2 y2,M
y= 2x =
N
x
Then
f
x= 2xy f(x, y) = x2y + g(y)
f
y= x2 + g(y) = x2 y2 g(y) = y2
g(y) = 13y3
-
Section 1.4 29
Thus the solution satisfiesx2y 1
3y3 = C
6. We have (2 9xy2)x + (4y2 6x3)y dy = 0. Using M = (2 9xy2)x and N = (4y2 6x3)ygives
M
y= 18x2y =
N
x
Then
f
x= 2x 9x2y2 f(x, y) = x2 3x3y2 + g(y)
f
y= 6x3y + g(y) = 4y3 6x3y g(y) = y4
Thus the solution satisfiesx2 3x3y2 + y4 = C
With initial conditions y(1) = 1, 12 3(1)3(1)2 + 14 = C gives C = 1, so
x2 3x3y2 + y4 = 1
7. We have ey (2y + xe4)dy = 0 with y(1) = 3. Using M = ey and N = 2y xey wehave
M
y= ey = N
x
We need to solvef =
M dx =
ey dx = xey + g(y)
Substituting into fy = N ,
y(xey + g(y)) = xey + g(y) = 2y xey g(y) = 2y
g(y) = y2
The solution is given by xey + y2 = C. Now apply initial conditions: 1e3 (3)2 = C, soC = 9 e3. The solution is then
xey y2 = e3 9
8. We have (1 + y2 sin 2x) + (2y cos 2x) dy = 0. Then
y[1 + y2 sin 2x] = 2y sin 2x
x[2y cos 2x] = 4y sin 2x
so the equation is not exact.
-
30 Section 1.4
9. We have 3x2(1 ln y) +(
x3
y 2y
)dy = 0. Then
y[3x2 + 3x2 ln y] =
3x2
y
x
[x3
y 2y
]=
3x2
y
so the equation is exact.
f
x= 3x2 + 3x2 ln y f(x, y) = x3 + x3 ln y + (y)
fy
=x3
y+ (y) =
x3
y 2y (y) = y2 + C
Thenf(x, y) = x3 + x3 ln y y2 + C
10.(2 +
y
x2
)dx +
(y 1
x
)dy = 0 M
y=
1x2
,N
x=
1x2
the equation is exact. ThusFx =2 + yx2 F = 2x yx + g(y). Fy = 1x + g(y) = y 1x g(y) = y, g(y) = y2/2 + C F (x, y) = 2x yx + y
2
2 + C
11. We have(
x
sin y+ 2)
dx +(x2 + 1) cos y
cos 2y 1 dy = 0. Then
y
[x
sin y+ 2]
= x cot[y] csc[y]
x
[(x2 + 1) cos y
cos 2y 1]
=2x cos y
cos 2y 1We find
cos2 x =12
+12
cos(2y) 12
cos 2y = cos2 x 12
cos 2y = 2 cos2 y 1So
N
x=
2x cos y2(cos2 x 1) =
x cos ysin2 y
= x cot y csc y
so it is exact. Thenf
x=
x
sin y= 2
f(x, y) = 12
x2
sin y+ 2x + (y)
fy
= 12x2 cot y csc y + (y)
=x2 cos y + cos y
cos 2y 1=
x2 cos y2 sin2 y
+cos y
2 sin 2y
=12x2 cot y csc y 1
2cot y csc y
-
Section 1.4 31
So
(y) = 12
cot y csc y (y) = csc(y)2
+ C
andf(x, y) =
12x2 csc y + 2x +
12
csc(y) + C
12. Solve axa1y1a dx + ((1 a)xaya) dy = 0.(i) As a separable equation,
axa1y1a dx = (a 1)xaya dyaxa1
xadx =
(a 1)yay1a
dy
a
xdx =
(a 1)y
dy
a ln |x| + C = (a 1) ln(y)y(a1) = Cxa
(ii) As an exact equation,
y[axa1y1a] = a(1 a)xa1ya
y[(1 a)xaya] = a(1 a)xa1ya
f
x= aa1y1a f(x, y) = xay(1a) + (y)
fy
= (1 a)xaya + (y) = (1 a)xaya
(y) = Cf(x, y) = xay(1a) + C
13. Determine A R such that the equation is exact. Then solve the resulting equation.(a) (x2 + 3xy) dx + (Ax2 + 4y) dy = 0. Then
y[x2 + 3xy] = 3x
x[Ax2 + 4y] = 2Ax
so 2A = 3 and A = 32 . Then
f
x= x2 + 3xy f(x, y) = 1
3x3 +
32x2y + (y)
fy
=32x2 + (y) =
32x2 + 4y
(y) = 2y2 + CSo
f(x, y) =13x3 +
32x2y + 2y2 + C
-
32 Section 1.4
(b)(
Ay
x3+
y
x2
)dx +
(1x2
1x
)dy = 0. We have
y
[Ay
x3+
y
x2
]=
A
x3+
1x2
x
[1x2
1x
]= 2
x3+
1x2
so A = 2. Thenf
x=
y
x2 2y
x3 f(x, y) = y
x+
y
x2+ (y)
f
y= 1
x+
1x2
+ (y) =1x2
1x (y) = C
Thereforef(x, y) =
y
x2 y
x+ C
14. AssumeM
y=
N
x. We want to find F (x, y) such that
F
x= M,
F
y= N . As in the proof
of Thm. 1.4.1, proceed replacing the argument ofF
x= M(x, y) with one for
F
y= N(x, y) and continuing in the same manner.
15. Determine the most general function N(x, y) so that
(x3 + xy2) dx + N(x, y) dy = 0
is exact. For the equation to be exact,
y[x3 + xy2] = 2xy =
N
x
soN(x, y) =
2xy dx = x2y + (y)
16. Determine the most general function M(x, y) such that
M(x, y) dx + (2yex + y2e3x) dy = 0
is exact. We haveM
y=
x[2yex + y2e3x] = 2yex + 3y2e3x
SoM(x, y) =
2yex + 3y2e3x dy = y2ex + y3e3x + (x)
17. Show that(Ax + By) dx + (Cx + Dy) dy = 0
is exact if B = C. To prove this, assume that B = C. Then
y[Ax + By] = B
x[Cx + Dy] = C
-
1.5. LINEAR EQUATIONS 33
Then
y[Ax + By] =
x[Cx + Dy]
which by Theorem 2.8.1 implies that the equation is exact.
Now assume that the equation is exact. Then by Theorem 2.8.1 we know that
B =
y[Ax + By] =
x[Cx + Dy] = C
or that B = C.
18. Show that the homogeneous equation
(Ax2 + Bxy + Cy2) dx + (Dx2 + Exy + Fy2) dy = 0
is exact if and only if B = 2D and E = 2C. To prove this, assume that B = 2D and E = 2C.Then
y[Ax2 + Bxy + Cy2] = Bx + 2Cy = 2Dx + Ey =
x[Dx2 + Exy + Fy2]
Therefore, by Theorem 2.8.1, the equation is exact. Now assume that the equation is exact.Then by Theorem 2.8.1,
Bx + 2Cy =
y[Ax2 + Bxy + Cy2]
=
x[Dx2 + Exy + Fy2] = 2Dx + Ey
which implies that B = 2D and 2C = E.
1.5 Linear Equations
1. y + 1xy = x: Linear equation with a(x) =1x , b(x) = x. Let A(x) =
1x dx ln |x|. Then
eA(x) = |x| and eA(x) = 1|x| . The solution is given by
y(x) =1|x|[
x |x| dx]
+C
|x| =
1x(x3
3
)+
C
x if x < 0
1x
(x3
3
)+
C
xif x 0
Then
y(x) =x2
3+
C
x
for all x.
2. y 2x1+x2 y = x2. Linear with a(x) = 2x1+x2 , b(x) = x2. Then
A(x) = 2x
1 + x2dx = ln |1 + x2| = ln(1 + x2)
because 1 + x2 is always positive. Substituting u = 1 + x2, du = 2xdx leads to
eA(x) = e ln(1+x2) =
11 + x2
; eA(x) = eln(1+x2) = 1 + x2
-
34 Section 1.5
Then b(x)eA(x) dx =
x2(
11 + x2
)dx =
(1 1
1 + x2
)dx = x arctanx
The solution is therefore
y = (1 + x2)(x arctanx) + (1 + x2) + C
3. (2x + 1)y = 4x + 2y leads to
y 2y2x + 1
=4x
2x + 1
Using P = 22x+1 and Q =4x
2x+1 gives 22x + 1
dx = ln |2x + 1|
e
p(x) = e ln |2x+1| =1
|2x + 1|e
p(x) = eln |2x+1| = |2x + 1|4x
2x + 1|2x + 1| dx = 2x2 sgn(2x + 1)
where
sgn(2x + 1) =
{1 if 2x + 1 < 01 if 2x + 1 0
The solution is y = |2x + 1|(2x2 sgn(2x + 1) + C) ory = (2x + 1)(2x2 + C1)
4. We have y + y tan x = secx with y() = 1. Then
N(x) = secx ddx
[y secx] = sec2 x
so y secx = tanx + C.
y = sinx + C cosx 1 = sin + C cos 1 = C C = 1
y = sinx cosx
5. We have dy =(
2x +xy
x2 1)
dx. Then
dy
dx x
x2 1y = 2x Q(x) = x
x2 1So
N(x) = e
(x/(x21))dx
Let u = x2 1 = du/2x = dx. Then
N(x) = e(1/2)
(1/u) du = eln |(1/
x21)| =1
x2 1
-
Section 1.5 35
Sod
dx
[y
x2 1
]=
2xx2 1
yx2 1 =
2x
x2 1Let u = x2 1. Then du/2x = dx and
2xx2 1 dx =
1u
du =
u1/2 du = 2
x2 1 + Cso
yx2 1 = 2
x2 1 + C
andy = 2(x2 1) + C
x2 1
6. y y = 4ex and y(0) = 4. Let P (x) = 1, Q(x) = 4ex. Thendx = x e
p(x) = ex; e
p(x) dx = ex
4ex ex dx =
4 dx = 4x
The solution isy = ex(4x + C)
With initial conditions, 4 = e0(4(0) + C) C = 4, thusy = 4ex(x + 1)
7. y + xy = 2x, P (x) = x, Q(x) = 2x, (x) = ex2/2
2xex
2/2 dx
= 2ex2/2 + C.
Thus the general solution is y = ex2/2(2ex
2/2 + C)
8. P (x) =1x
, Q(x) = ex, (x) = eln x = x
xex dx = xex ex.
Thus the general solution is y = e ln x (xex ex + C) = 1x
(xex ex + C)
9. P (x) = 2, Q(x) = 3x, (x) = e2x
3xe2x dx
=3xe2x
2+
34e2x + C. Thus the general solution is
y = e2x(3xe2x
2+
34e2x + C
)
10. We havedy
dx+
1x
y =cosx
xwith y
(2
)=
4
and x > 0. This time Q(x) = 1x so N(x) = x,which implies that
d
dx[xy] = cos(x) xy = sin(x) + C
So with y(2 ) =4 , we have that
2 4
= sin(
2
)+ C 2 = 1 + C
so C = 1 and
y(x) =sin(x)
x+ x1
-
36 Section 1.5
11.
P (x) = 3x2Q(x) = x2
(x) = ex3
Q(x)(x) dx = e
x3
3+ C
y = ex3
(e
x3
3+ C
)
12.
P (x) = 2xQ(x) = x3
(x) = ex2
Q(x)(x) dx = (x
2 + 1)ex2
2+ C
y =
13.
P (x) = 1Q(x) = cosx(x) = ex
Q(x)(x) dx =ex
2(cos x + sinx) + C
y = ex(
ex
2(cosx + sin x) + C
)
14.
P (x) = 1Q(x) = ex
(x) = exQ(x)(x) dx =
e2x
2+ C
y = ex(
e2x
2+ C)
-
Section 1.5 37
15. y + 2xy = 1 with y(2) = 1. Then we have
A(x) =
2xdx = x2
eA(x) = ex2, eA(x) = e4x
2b(x)eA(x) =
ex
2dx x
2
y(t)eA(t) = x
2
b(t)eAt
16. We have (x + y2) dy = y dx gives
x + y2 = ydx
dyor
dx
dy 1
yx = y
Thus a(y) = 1y , b(y) = y, A(y) = 1y dy = ln y. Then
eA(y) = e ln y =1y, eA(y) = y
b(y)eA(y) dy =
y0(
1y
)dy =
1 dy = y
Then x = y[y] + Cy andx = y2 + Cy
17. We have (2ey x)y = 1 or (2ey x) dydx = 1. So 2eyx = dxdy and dxdy +x = 2ey. With Q(y) = 1we have N(y) = ey.
d
dy[xey] = 2e2y and xey = e2y + C
so the solution isx = ey + Cey
18. We have (sin 2y + x cot y)y = 1. Rewriting, we have
(sin 2y + x cot y) =dx
dy
dx
dy x cot y = sin 2y
With Q(x) = cot y we have N(y) = e ln(sin x) csc y, sod
dy[x csc y] =
sin 2ysin y
ddy
[x csc y] =2 sin y cos y
sin y
x csc y =
2 cos y dy
x csc y = 2 sin y + C x(y) = 2 sin2 y + C sin y
19. (a) y = 0, y = 0, y + P (x)y = 0 + 0 = 0
-
38 Section 1.5
(b) y = y1 is a solution. y = ky1, y = ky1.Then ky1 + P (x)ky1 = k(y
1 + P (x)y1) = k(0) = 0
(c) y = y1 + y2, y = y1 + y2
Then y1 + y2 + P (x)(y1 + y2) = (y
1 + P (x)y1) + (y
2 + P (x)y2)
= 0 + 0(Since y1 and y2 are solutions) = 0
20. (a) y = y1 + y2, y = y1 + y2
Then y1 + y2 + P (x)(y1 + y2) = (y
1 + P (x)y1) + (y
2 + P (x)y2)
= 0 + r(x) (Since y1 and y2 are solutions) = r(x)(b) y = y1 + y2, y = y1 + y
2
Then y1 + y2 + P (x)(y1 + y2) = (y
1 + P (x)y1) + (y
2 + P (x)y2)
= r(x) + q(x) (Since y1 and y2 are solutions)(c) Solution to y + 2y = ex is y = e2x(ex + C) and y + 2y = cosx is
y = e2x(
e2x
5(2 cosx + sin x) + C
).
Thus y = e2x(
ex +e2x
5(2 cosx + sin x) + C
)is a solution to the original equation.
21. (a)
y = P (x)ydy
y= P (x) dx
ln |y| =
P (x) dx
yc = e P (x) dx
(b) y = A(x)yc, y = A(x)yc + A(x)yc.Then Q(x) = y + P (x)y = A(x)yc + A(x)yc + P (x)A(x)yc =A(x)yc + A(x)(yc + P (x)yc) = A
(x)yc(c) A(x)e
P (x)dx = Q(x) A(x) = Q(x)e P (x) dx A(x) =
Q(x)e
P (x) dx dx
(d) yp = ycA(x), y = yc + yp = yc + ycA(x) = yc(1 + A(x)) =
e P (x) dx( Q(x)e P (x) dx dx)
22. Let y be the amount of pollutant at time t. Then
y = 12000(2) 10000(
y
500000 + 2000t
)= 24000 10y
500 + 2t
Then y +10y
500 + 2t= 24000 is linear with solution
y(t) =1
(t + 250)5(4000(t + 250)6 + C
). C = 4000(250)6.
y(t) = 4000(t + 250) 4000(250)6
(t + 250)5, y(10) = 218072
Concentration = 218072520000 = .419g/gal
-
Section 1.5 39
23. (a)dN
dt= rN + rA dN
dt rN = rA N(t) = a + Cert. N(0) = 0
A = C N(t) = Aert A(b) r = ln 2. So, with A = 3, N(t) = 3ert 3 = 3 2t 3.
N(24) = 50, 331, 645N(36) = 206, 158, 430, 205
27.dx
dt+ a(t)x = f(t), P (t) = a(t), (t) = e
a(t) dt
x(t) = e a(t) dt( e a(t) dtf(t) dt + C).
limtx(t) = limt
(e a(t) dt( e a(t) dtf(t) dt + C)) = 0 since a(t) > 0, e a(t) dt 0
29. (a) See section 1.5.1. (x) = exp(
1x2y x(1 (2xy 1)) dx
)=
1x2
which gives an exact
DE and F (x, y) = 2x yx
+y2
2+ C
(b)1N
(M
y N
x
)=
1x
(1 (1)) = 2x
. So (x) =1x2
which gives an exact De and
F (x, y) =x3
3 ln |x| 1
xy+ C
31. M and N change roles, but the argument is exactly the same as (1.31)
32.M
y= 2y,
N
x= 0 1
N
(M
y N
x
)=
1y(2y) = 2 which a function only in x so (x) = e2x
33.M
y=
1 + x2,N
x=
x21 + x2
+
1 + x2 1N
(M
y N
x
)=
x1 + x2
which a function
only in x so (x) =1
1 + x2
34.M
y= x + 2y,
N
x= y 1
M
(N
x M
y
)=
1y
which a function only in y so (y) =1y
35.M
y= 3y2 + 1,
N
x= y2 2x + 1 Which does not yield anything of a function in x or y
36.M
y= 1,
N
x= 1 1
N
(M
y N
x
)=
2x
which a function only in x so (x) =1x2
37.M
y= 1,
N
x= ex 1
N
(M
y N
x
)= 1 which a function only in x so (x) = ex
38.M
y= 2y + 1, N
x= 2y 1 1
N
(M
y N
x
)=
2x
which a function only in x so (x) =1x2
40.M
y= 2x cos y, N
x= 2x cos y 1
N
(M
y N
x
)=
4x(x2 + 1)2
which a function only in x
so (x) =1
(x2 + 1)2
41. n = 2, v = y1 v v = x which is linear and gives solutionv = x + 1 + Cex y = 1
x + 1 + Cex
-
40 Section 1.5
43. y = y4 cosx + y tan x, sody
dx+ (y tan(x))y = y4 cosx
Let v = y3. Thendv
dx= 3y4 dy
dx= 7
dv
dx
[1
3y4]
=dy
dx
So
dv
dx
(1
3y4)
+ ( tan(x))y = y4 cosxdv
dx+ (3 tan(x))v = 3 cosx
N = e3
tan x dx = eln(1/(cos3 x)) = sec3 x
d
dx[v sec3 x] = 3 cos2 x
d
dx[v sec3 x] = 3
2+
32
cos 2x
v sec3 x =34
sin 2x 32x + C
y3 sec3 x =34
sin 2x 32x + C
44. n = 4, v = y3 v 9v = 3 which is linear and givesv = Ce9x +
13 y3 = 3
3Ce9x + 1 y = 3
3
3Ce9x + 1
45. We have xy2y = x2 + y3. Then
y y3
xy2=
x2
xy2 y 1
xy3 = xy2
This is a Bernoulli equation with r = 2, so 1 r = 1 (2) = 3. Thus we need to solveu + 3
( 1x)u = 3(x). Take a(x) = 3x , b(x) = 3x, and A(x) = 3x dx = 3 ln |x|. TheneA(x) = e3 ln |x| =
1|x|3 , e
A(x) = e3 ln |x| = |x|3b(x)eA(x) dx =
3x(
1|x|3)
dx
x 0
3x(
1x3
)dx =
3(
1x2
)dx = 3
x
x < 0
3x( 1
x3
)dx =
3
x2dx =
3x
=
3|x|
u = |x|3( 3|x|
)+ |x|3 C = 3x2 + C |x|3
y3 = 3x2 + C |x|3 y = (3x2 + C|x|3)1/3
46. y = 0 is a trivial solution, so if y = 0 y
y= 4 2x dy
y= (4 2x) dx
ln |y| = 4x x2 + C y = e4xx2+C
-
Section 1.5 41
47. We have (x + 1)(y + y2) = y. Thendy
dx+ y2 =
yx + 1
dydx
+y
x + 1= y2
Let v = y1, sodv
dx= y2 dy
dx dv
dx(y2) = dy
dx
and thatdv
dx(y2) +
(1
x + 1
)y = y2
which implies that
dv
dx(
1x + 1
)y1 = 1 dv
dx(
1x + 1
)v = 1
Q(x) = 1x+1 , N(x) = e
(1/(x+1)) dx = 11+x , so
d
dx
[v 1
1 + x
]=
11 + x
v1 + x
= ln 11 + x
v = (ln |1 + x|)(1 + x) 1
y= (ln |1 + x|)(1 + x)
y = 1(x + 1) ln |x + 1|
48. We have xy 2x2y = 4y. Thendy
dx 2xy = 4
xy
dy
dx+( 4
x
)y = 2xy1/2 (Bernoulli equation)
Let v = y1(1/2) = y1/2, so dvdx =12y
1/2 dydx . Then
dv
dx[2y1/2] +
( 4
x
)y = 2xy1/2
dv
dx+( 2
x
)v = x
(N(x) =
1x2
)d
dx
[y
x2
]=
1x
y
x2= ln |x| + C
y = x2 ln |x| + Cx2
-
42 Section 1.5
49. We have xy + 2y + x5y3ex = 0. Then
xdy
dx+ 2y + x5y3ex = 0
dy
dx+
2x
y = (x5ex)y3 (Bernoulli equation)
v = y2 dvdx
= 2y3 dydx
dvdx
(1
2y3)
=dy
dx
dv
dx
(1
2y3)
+2x
y = (x5ex)y3
dv
dx 4
xy = (2x5ex)
Now use N(x) = 1x4 , soddx
[v 1x4
]= [2xex]. Then
v 1x4
= 2ex(x 1) + C v = 2x4ex(x 1) + Cx4
y = 1x2
2ex(x 1) + C
50. We have xy dy = (y2 + x) dx. Then
dy
dx(xy) = y2 + x dy
dx= x1y + y1
dy
dx+( 1
x
)y = y1 (Bernoulli equation)
Let v = y1+1 = y2, so dvdx(
12y
1) = dydx . Thendv
dx
(12y1)
+( 1
x
)y = y1 dv
dx+( 2
x
)v = 2
Using N(x) = 1x2 we haveddx
[vx2
]= 2x2 , so
v
x2= 2
x+ C
y2 = 2x + Cx2
51. Use the Bernoulli method to solve the logistic equation
dN
dt= rN
(1 N
K
)We have
dN
dt= rN r
KN2 dN
dt+ (r)N = r
KN2
Let v = N1, so dvdt = N2 dNdt and dvdt (N2) = dNdt . Thendv
dt(N2) + (r)N = r
KN2
dv
dt+ rv =
r
K
-
1.6. CHAPTER 1: ADDITIONAL PROBLEMS 43
Using N = ert gives ddt [vert] = rK e
rt, so
vert =1K
ert + C v = 1k
+ Cert
1N
=1K
+ cert N(t) = 11K + Ce
rt
1.6 Chapter 1: Additional Problems
1. False - It is linear, but no initial condition is given
2. True - y2 makes it non-linear
3. False - y(x) = f(x) is an explicit solution
4. The point is to get linear DE into an exact DE by multiplying by an integrating factor. Soeven if this were true, there would be no reason to do so.
5. False - Solutions can be defined on a restricted portion of the real line
6. True
7. True -dy
f(y)= dx
8. y = C4x3ex4 4x3(y 1) = 4x3(1 + Cex4 1) = 4x3(Cex4) = y
9. y = Cex 2x 2 x2 + y = x2 + Cex x2 2x 2= Cex 2x 2 = y
10. y2 =1
C 2x y3 = (C 2x)3/2, y =
1
C 2x,y = (C 2x)3/2 = y3. Valid when C = 2x
11. y =1
(x + C)2, (y 3)2 =
( 1(x + C)
)2=
1(x + C)2
= y
12. We have 2x2yy + y2 = 2. Then
2x2ydy
dx= 2 y2
2y dy2 y2 =
dx
x2
2y dyy2 2 =
dx
x2
ln |y2 2| = 1x
+ C1
|y2 2| = eC1e1/xy2 2 = Ce1/x, where C = C1
y =
Ce1/x + 2
-
44 Chapter 1 Review
13. We have y = 3 3
y2 with y(2) = 0. Rewrite the equation as y = 3y2/3. Separate variablesand integrate:
y2/3 dy =
3 dx
3y1/3 = 3x + C
It is easiest to apply the initial conditions now: 3(0)1/3 = 3(2) + C leads to C = 6, so3y1/3 = 3x 6y1/3 = x 2
y = (x 2)3
14. We have y xy2 = 2xy. Thendy
dx= x(y2 + 2y)
dy
y2 + 2y= xdx (by partial fractions) (1/2
y+
1/2y 2
)dy =
xdx
12
ln |y| + 12
ln |y 2| = 12x2 + C1
lny 2y
= x2 + 2C11 2y = e2C1ex22y
= 1 e2C1ex2
y =2
1 e2C1x2y =
21 + Cx2
where C = e2C1
15. We have ex(1 + dxdt
)= 1. Then
dx
dtex = 1 ex
ex
1 ex dx = dt
Letting v = 1 ex, du = ex,du
u=
dt
ln |u| = t + C1|u| = eC1et
1 ex = eC1etex = eC1et 1x = ln |eC1et 1|
x = ln |Cet 1| where C = eC1
-
Chapter 1 Review 45
16. We have xy + y = y2 with y(1) = .5. Then
xy = y2 y
dy
y2 y =
dx
x
(
1y 1
1y
)dy =
dx
x
ln |y 1| ln |y| = ln |x| + C y 1y
= Ax
Apply initial conditions:
12 1
12
= A(1) A = 1 1 1y
= x
Then we have1 + x =
1y 1
1 + x= y
17. We have xy = y xey/x. Rewriting we have y = yx ey/x. Let u = yx , dydx = xdudx + u. Then
xdu
dx+ u = u eu
xdu
dx= eu
eu du =
1
xdx
eu = lnx + C eye1/x = lnx + Cey =
lnx + Ce1/x
y = ln( lnx + C
e1/x
)
18. y yx
= (1 +y
x) ln(1 +
y
x) which is Homogeneous. Let v =
y
x
v + vx v = (1 + v) ln(1 + v)dv
(1 + v) ln(1 + v)=
dx
x
[u = ln(1 + v)]du
u=
dx
x
|u| = |x|eC
| ln(1 + yx
)| = |x|eC
ln(1 +y
x) = xeC
-
46 Chapter 1 Review
19. We have xy y cos (ln ( yx)) = 0. Thenyx = y cos
(ln( y
x
))y =
y
xcos(ln(y
x
))(homogeneous)
dv
dxx + v = v cos(ln(v))
dv
dxx = v(cos(ln(v))) v
dv
dxx = v(cos(ln(v)) 1)
1v(cos(ln(v)) 1) dv =
1x
dx
cot(
12
ln(v))
= ln x + C
cot(
12
ln(y
x
))= ln x + C
20. We have (y +
xy) dx = xdy. We use M(x, y) = y +
xy and N(x, y) = x, so
M(tx, ty) = ty +
t2xy = t(y +
xy) + tM(x, y)N(tx, ty) = tx = t(x) = tN(x, y)
We need to solve
dy
dx=
y
x+
xy
x dv
dxx + v = v +
v
dvdx
x =
v
1v
dv =1x
dx
2v = ln |x| + C
Then
y = 2
x ln x + 2C
x
y = 4x(ln y + C)2
21. We have x dydx =
x2 y2 + y. Then
dy
dx=
1 (y
x
)2+(y
x
)(homogeneous)
dvdx
x + v =
1 v2 + v1
1 v2 dv =1x
dx
arcsin(v) = lnx + C arcsin
( yx
)= lnx + C
-
Chapter 1 Review 47
22.
dy
dx=
4y 2xx + y
=4(y/x) 21 + (y/x)
Homogeneous (y = vx)
v + vx =4v 21 + v
1 + vv2 + 3v 2 dv =
dx
x
ln(1 v)2(2 v)3
= ln |x| + C[(1 (y/x))2(2 (y/x))3
]= eCx
23. We have (y/x) + (y3 + lnx) dy = 0. We find
y
(yx
)=
1x
x[y3 + lnx] =
1x
so the equation is exact. Then,
f
x=
4x
f(x, y) = y ln x + (y) f
y= lnx + (y) = y3 + lnx
(y) =14y4 + C
Sof(x, y) = y ln x +
14y4 + C
24. We have(
3x2 + y2
y2
)+(2x3 5y
y3
)dy
dx= 0. Then
y
[3x2 + y2
y2
]=
6x2y3
x
[2x3 5yy3
]=
6x2y3
-
48 Chapter 1 Review
so the equation is exact. Solving, we get
f
x=
3x2
y2+ 1
f(x, y) = x3
y2+ x + (y)
fy
=2x3y3
+ (y) =2x3y3
5y2
(y) = 5y2
(y) =5y
+ C
f(x, y) =x3
y2+ x +
5y
+ C
25. We have 2x(1 +
x2 y) dx x2 y dy = 0. Then
y
[2x + 2x
x2 y
]=
xx2 y
x
[
x2 y]
= 2x2
x2 y =xx2 y
so the equation is exact. Solving,
f
x= 2x + 2x
x2 y
f(x, y) = x2 +
2x
x2 y dx
f(x, y) = x2 + 23(x2 y)3/2 + (y)
fy
=
x2 y + (y)
=
x2 y
(y) = 0 leads to (y) = C, so
f(x, y) = x2 +23(x2 y)3/2 + C
26. y +1x
y =1x2
y2 which is Bernoulli n = 2, v = y3, dvdx
= 3y2dy
dx. Then we have
dv
dx+
3x
v =3x2
which is linear with solution v = y3 =32x
+C
x3
y = 3
32x
+C
x3
27.M
y= 2y,
N
x= y 3x2 1
N
(M
y N
x
)=
3y + 3x2
x(y + x2) which is only a function of x
which gives (x) = x3 and F (x, y) =y22x2
y + C
-
Chapter 1 Review 49
28.
dy
dx=
y
x y2
(Homog) (y = vx)v + vx = v v2x2
dv
v2 = xdx1v
=x2
2+ C
x
y=
x2
2+ C
29. Not linear, exact, homogeneous, Bernoulli, seperable
30. Not linear, exact, homogeneous, Bernoulli, seperable
31. y +y
x= x3 Linear y = x
4
5+
C
x
32. Linear y = ex[ex
2(cosx + sin x) + C
]
33. Linear y = e2x2[58e2x
2(2x2 1) + C
]
34. Bernoulli n = 4, v = y3, v 3x
v = 3 (linear) v = y3 = 3x2
+ Cx3
35. Bernoulli n = 3, v = y2, v + 2v = 2x (linear)v = y2 = e2x
[e2x2
(2x 1) + C]
36. Bernoulli n = 3, v = y2, v xv = 4x (linear)v = y2 = ex
2/2(4ex
2/2 + C)
37. T (t) = (T0 Ts)ekt + Ts = 95ekt + 5, k = .055 soup will be edible in about 34 minutes.38. T (t) = 50ekt + 70, k = .105 coffee is drinkable in about 5 minutes.39. Let y(t) = amount of salt at time t, y(0) = 6
dy
dt= 2 6 + 2t
15 + 2t
=24 2t15 + 2t
y =392
ln |2t + 15| t + Cy(0) = 6 C = 46.8
The tank will fill in 15 minutesy(15) = 12.43lbs of salt
-
50 Chapter 1 Review
40. y(t) lbs of sugar at time t, y(0) = 25
dy
dt= p 25 + pt
100 t=
100p 25 2pt100 t
dy =100p 25 2pt
100 t dty = (25 100p) ln |100 t|
200p ln |t 100| 2pt + CC = 25 25 ln 100
We want y(95) = 2.5 p = .074lbs/min
41. y(t) amount of nitric acid, y(0) = 1
dy
dx= 1.2 8
(100 + 1.2t200 2t
)=
560 12t200 2t
y(t) = 880 ln |t 100|+ 6t + 1 880 ln100
The solution reaches 10% when y(t) = 20 .2t which is when t = 138 minutes. Since the tankempties in 100 minutes, this will never happen.
42. If (x, y) is the point of tangency, then (x2 , 0) is the point of intersection anddy
dx=
yx2
=2yx
.
Thendy
y=
2 dxx
y = x2 is such a curve.
-
Chapter 2
Geometrical and NumericalMethods for First-Order Equations
2.1 Direction Fieldsthe Geometry of Differential Equations
1. Looking at the point (2, 1), y = 92 , which matches graph b.
2. Looking at the point (2, 1), y = 18, which matches graph c.
3. Looking at the point (2, 1), y = 29 , which matches graph a.
4. Looking at the point (2, 1), y = 118 , which matches graph d.
5. Graph B
6. Graph C
7. Graph D
8. Graph A
9. y = y4 #9 #10
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
10. y = cos y
51
-
52 Section 2.1
11. y = sin y
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
12. y = ey
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
13. y = x4
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
Section 2.1 53
14. y = cos x
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
15. y = sin x
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
16. y = ex
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
54 Section 2.1
17. y = xy
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
18. y = x + y
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
19. y = (x2 + 1)(y + 1)
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
Section 2.1 55
20. y = ex2
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
21. y =x 1y 1
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
22. y =x2 1y2 + 1
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
56 Section 2.2
23. y = y(y2 2)
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
24. xy(x2 + 2)
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
25. y =x3(y2 + 1)
y2 + x2
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
2.2 Existence and Uniqueness for First-Order Equations
1. (a) y = y2 x2, y(0) = 0, f = y2 x2, fy
= 2y. The theorem guarantees that a solution
exists and is unique on some interval.
-
Section 2.2 57
(b) y = y2 x2, y(0) = 0, f = 1y2
x2, fy
= y3. The theorem does not guarantee thata solution exists or is unique on some interval.
(c) y = y +1
1 x , y(1) = 0, f = y +1
1 x ,f
y= 1. The theorem does not guarantee that
a solution exists or is unique on some interval.
2. (a) f(x, y) = 3x(y + 2)2/3,f
y=
2x(y + 2)1/3
, which is discontinuous when y = 2, so asolution exists and is unique everywhere except possibly along y = 2.
(b) f(x, y) = (x y)1/5, fy
=1
5(x y)4/5 , which is discontinuous when y = x, so solutionsexist and are unique everywhere except possibly along y = x. Alternatively, f(x, y) =(x y)
5,
f
y= 1
5, so solutions exist and are unique everywhere.
(c) f(x, y) = x2y1,f
y= x2y2, which are both discontinuous when y = 0, so solutions
exist and are unique everywhere except possibly along y = 0.
(d) f(x, y) = (x + y)2,f
y=
2(x + y)3
, which are both discontinuous when y = x, sosolutions exist and are unique everywhere except possibly along y = x.
3. (a)f
y=
43xy(1/3) which is not continuous at y = 0. Hence a solution exists and is unique
everywhere except possibly along y = 0.
Actual solution: y =x6
27. This solution crosses y = 0 when x = 0.
(b)f
y= 2x(2/3) which, along with f(x, y), are continuous everywhere. Hence solution exists
and is unique.Actual solution: y = 0.
(c)f
y= 2
3xy(5/3) which is not continuous at y = 0. Hence a solution exists and is unique
everywhere except possibly along y = 0.
Actual solution: y =(
56
)(3/5)x(6/5) This solution crosses y = 0 when x = 0.
(d)f
y=
23xy(1/3) which is not continuous at y = 0. Hence a solution exists and is unique
everywhere except possibly along y = 0.
Actual solution: y =x6
216which crosses y = 0 when x = 0.
4. f(x, y) = 2
y is continuous for y 0 and fy
=1y
is continuous for y > 0. Hence, by the
theorem, a solution exists and is unique for y(1) = 3.Actual solution is given by: y = (x +
3 1)2.
5. f(x, y) = 5(y 2)3/5 and y(0) = 2. yy
=3
(y 2)2/5 is discontinuous when y = 2, so solutionsexist and are unique everywhere except possibly along y = 2. Solving for the initial value,
y = 5(y 2)3/5
(y 2)3/5 dy =
5 dx
52(y 2)2/5 = 5x + C
-
58 Section 2.2
Applying initial conditions,
52(2 2)2/5 = 5(0) + C C = 0
(y 2)2/5 = 2x y 2 = (2x)5/2 y = 2 (2x)5/2
Solutions passing through (0, 2) are NOT unique.
6.f
y= x(1/3)(2y) which is continuous everywhere (as is f), therefore, unique solution will
exist everywhere.
7.f
y= (y 1)(2/3) which is continuous everywhere except y = 1, so solutions exist and are
unique everywhere except possibly along y = 1.Actual solution: 1 2
2x(3/2)
8.f
y= 2xy3 which is continuous everywhere except y = 0, so solutions exist and are unique
everywhere except possibly along y = 0.
Actual solution:(3
2+
x2
2
)1/3
9. f(x, y) andf
yare not continuous at y = 1, so solutions exist and are unique everywhere
except possibly along y = 1.Actual solution: y = 1 + 3
3 3 cosx which IS unique.
10.f
yDNE at y = 0; if y > 0, then fy = 1 and if y < 0, then
fy = 1. Hence,
f
y= lim
h0f(x, y + h) f(x, y)
hDNE. Therefore, the solutions will exist and be unique except
possibly at y = 0.If y 0, y = eCex which will never be 0. If y < 0, y = eCex which will never be 0. Thus,the only solution to this DE is y = 0 which IS unique.
-
2.3. FIRST-ORDER AUTONOMOUS EQUATIONSGEOMETRICAL INSIGHT 59
2.3 First-Order Autonomous EquationsGeometrical Insight
1. y = 2y + 3, 2y + 3 = 0
Root y = 32Multiplicity 1
End behavior: +y
(i) y y
-3 -2 -1 1 2 3y
-2
2
4
6
8
y
(ii) By the phase line diagram, y = 32 is an unstable equilibrium point.(iii) y > 32 , y as x
y < 32 , y as x (iv) xy-plane
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
60 Section 2.3
2. y = y2 + 4y + 4, y2 + 4y + 4 = 0, (y + 2)2 = 0
Root y = 2Multiplicity 2
End behavior: y2
(i) y y
-5 -4 -3 -2 -1 1y
2
4
6
8
y
(ii) By the phase line diagram, y = 2 is a half-stable equilibrium point.(iii) y > 2, y as x
y < 2, y 2 as x (iv) xy-plane
-3 -2 -1 1 2 3
-6
-4
-2
2
-
Section 2.3 61
3. x = x2 x 6 = (x 3)(x + 2)(i) y y
-2 -1 1 2 3y
-6
-4
-2
2
y
(ii) x = 2 is a stable equilibriumx = 3 is an unstable equilibrium
(iii) If x0 > 3, then x(t) as t .If 2 < x0 < 3, then x(t) 3 as t .If x0 < 2, then x(t) 3 as t .
(iv) xy-plane
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
4
-
62 Section 2.3
4. x = x(x + 2)(x 3)
Equilibrium Multiplicityx = 0 1
x = 2 1x = 3 1
Highest power and coefficient: x(+x)(+x) = +x3
(i) y y
-2 -1 1 2 3y
-7.5
-5
-2.5
2.5
5
7.5
10y
(ii) x = 2, 3 unstable; x = 0 stable(iii) For x0 (,2), x as t
For x0 (2, 3), x 0 as t For x0 (3,), x as t
(iv) xy-plane
-3 -2 -1 1 2 3
-4
-2
2
4
-
Section 2.3 63
5. x = (x 2)3(x2 9) = (x 2)3(x + 3)(x 3)
Equilibrium 2 3 3Multiplicity 3 1 1
Highest power coefficient: (x)3(x2) = +x5
(i) y y
-4 -2 2 4y
-4
-3
-2
-1
1
2
3
4y
(ii) x = 3, 3 unstable, x = 2 stable(iii) For x0 (,3), x as t
For x0 (3, 3), x 2 as t For x0 (3,), x as t
(iv) xy-plane
-3 -2 -1 1 2 3
-4
-2
2
4
-
64 Section 2.3
6. y = sin y, 2 < y < 2sin y = 0, 2 < y < 2
Roots y = y = 0 y = Multiplicity 1 1 1
End behavior: trigonometric sine wave
(i) y y
-6 -4 -2 2 4 6y
-1
-0.5
0.5
1
y
(ii) By the phase line diagram, y = is a stable equilibrium point.y = 0 is an unstable equilibrium point.y = is a stable equilibrium point.
(iii) For y > , y as x For 0 < y < , y as x For < y < 0, y as x For y < , y as x
(iv) xy-plane
-6 -4 -2 2 4 6
-6
-4
-2
2
4
6
-
Section 2.3 65
7. y = ey2/2 e2, ey2/2 e2 = 0. When
y22
= 2y2 = 4
y2 = 4
(i) y y
-3 -2 -1 1 2 3y
0.2
0.4
0.6
0.8
y
(ii) By the phase line diagram, y = 2 is stabley = 2 is unstable
(iii) For y > 2, y For 2 < y < 2, y For y < 2, y
(iv) xy-plane
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
66 Section 2.3
8. y = y2, y2 = 0
Root y = 0Multiplicity 2
End behavior: y2
(i) y y
-3 -2 -1 1 2 3y
-4
-3
-2
-1
y
(ii) By the phase line diagram, y = 0 is a half-stable equilibrium point.
(iii) For y > 0, y 0 as x .For y < 0, y as x .
(iv) xy-plane
-3 -2 -1 1 2 3
-4
-2
2
4
-
Section 2.3 67
9. x = x2(2 x)
Equilibrium Multiplicityx = 0 2x = 2 1
Highest power and coefficient: x2(x) = x3
(i) y y
-3 -2 -1 1 2 3y
-1
1
2
3
y
(ii) x = 0 is a half-stable point; x = 2 is a stable point.
(iii) For x (, 0), x 0 as t .For x (0,), x 2 as t
(iv) xy-plane
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
68 Section 2.3
10. x = (x 1)2(x 2)3(1 + x)
Equilibrium 1 2 1Multiplicity 2 3 1
Highest power and coefficient: (x)2(x)3(+x) = +x6
(i) y y
-3 -2 -1 1 2 3y
-1
-0.75
-0.5
-0.25
0.25
0.5
0.75
1y
(ii) x = 1 is stable; x = 1 is half-stable; x = 2 is unstable.(iii) If x0 > 2, then x as t .
If 1 < x0 < 2, then x 1 as t .If 1 < x0 < 1, then x 1 as t .If x0 < 1, then x 1 as t .
(iv) xy-plane
-3 -2 -1 1 2 3
-1
-0.5
0.5
1
1.5
2
2.5
3
-
Section 2.3 69
11. y = cos y + 1, 2 < y < 2. cos y + 1 = 0, cos y = 1.
Roots y = y = Multiplicity 1 1
End behavior: cosine wave
(i) y y
-6 -4 -2 2 4 6y
0.5
1
1.5
2
y
(ii) By the phase line diagram, y = is a half-stable equilibrium point. y = is a half-stableequilibrium point.
(iii) For y > , y as x .For < y < , y as x .For y < , y as x .
(iv) xy-plane
-6 -4 -2 2 4 6
-6
-4
-2
2
4
6
-
70 Section 2.3
12. x = x3(2 + x)(5 + x)7(3 x)(x 2)4
Equilibrium 0 2 5 3 2Multiplicity 3 1 7 1 4
Highest power and coefficient: x3(+x)(+x)7(x)(x)4 = x16
(i) y y
-6 -4 -2 2 4y
-1107-7.5106
-5106-2.5106
2.51065106
7.5106
y
(ii) x = 5, 0 are unstable points; x = 2, 3 are stable points; x = 2 is a half-stable point.(iii) If x0 < 5, then x as t .
If 5 < x0 < 2, then x 2 as t .If 2 < x0 < 0, then x 2 as t .If 0 < x0 < 2, then x 2 as t .If 2 < x0 < 3, then x 3 as t .If x0 > 3, then x 3 as t .
(iv) xy-plane
-10 -5 5 10
-10
-5
5
10
-
Section 2.3 71
13. v = g kmv. Equilibrium satisfies g (k/m)v = 0, so v =gm
k
(i) v v
kmv
g
v
(ii) v =gm
kis a stable equilibrium point.
(iii) For v [0,), v gmk as t (iv) Figures will vary. With g = 32, m = .25, k = 2 as in example 2 in 1.3 then
1 2 3 4 5 6
1
2
3
4
5
6
14. v = g kmv2. Equilibrium satisfies g (k/m)v2 = 0, so v2 = gm/k yields v =
gm/k. Notethat this is a free-fall problem where v > 0 in the downward direction. We thus ignore v < 0.
(i) Graphs will vary
(ii)
gm
kis a stable equilibrium point
(iii) For v [0,), v +gm/k as t (iv) Graphs will vary
15. x = x2(2 x)(x + 3)2, x2(2 x)(x + 3)2 = 0
Roots x = 3 x = 0 x = 2Multiplicity 2 2 1
End behavior: x2(x)(x)2 = x5
-
72 Section 2.3
-4 -3 -2 -1 1 2 3x
-20
20
40
x
(i) x x(ii) By the phase line diagram, x = 2 is a stable equilibrium point; x = 0 is a half-stable
equilibrium point; x = 3 is a half-stable equilibrium point.(iii) For x > 2, 2 as t .
For 0 < x < 2, 2 as t .For 3 < x < 0, 0 as t .For x < 3, 3 as t .
(iv) tx-plane
-3 -2 -1 1 2 3
-4
-3
-2
-1
1
2
-
Section 2.3 73
16. x = (2 x)3(x2 + 4), (2 x)3(x2 + 4) = 0
Roots 2 2iMultiplicity 3 ignore
End behavior: (x)3(x2) = x5
(i) x x
0.5 1 1.5 2 2.5 3x
-3
-2
-1
1
2
3
4
x
(ii) By the phase line diagram, x = 2 is a stable equilibrium point.
(iii) For x > 2, 2 as t .For x < 2, 2 as t .
(iv) tx-plane
-3 -2 -1 1 2 3
0.5
1
1.5
2
2.5
3
-
74 Section 2.3
17. y = (2 y)3(y2 + 4)2
Root 2Multiplicity 3
Highest power and coefficient: (y)3(y2)2 = y7
(i) x x
0.5 1 1.5 2 2.5 3x
-10
10
20
x
(ii) By the phase line diagram, y = 2 is a stable equilibrium point.
(iii) If y0 < 2, then y 2 as x .If y0 > 2, then y 2 as x .
(iv) tx-plane
-3 -2 -1 1 2 3
0.5
1
1.5
2
2.5
3
-
Section 2.3 75
18. y = y2(4 y)(9 y2)
Roots 0 4 3 3Multiplicity 2 1 1 1
Highest power and coefficient: y2(y)(y2) = y5
(i) x x
-1 1 2 3 4 5x
-40
-30
-20
-10
10
20
x
(ii) y = 3, 4 are stable; y = 0 is half-stable; y = 3 is unstable.(iii) If y0 < 3, y 3 as x .
If 3 < y0 < 0, y 3 as x .If 0 < x < 3, y 0 as x .If 3 < x < 4, y 4 as x .If x > 4, y 4 as x .
(iv) tx-plane
-4 -2 2 4
-4
-2
2
4
-
76 Section 2.3
19. x = x5(1 x)(1 x3), x5(1 x)(1 x3) = 0
Roots 0 1 two normalMultiplicity 5 2 ignore
End behavior: x5(x)(x3) = x9
(i) x x
-1 -0.5 0.5 1 1.5 2x
-0.1
-0.05
0.05
0.1x
(ii) By the phase line diagram, = 1 is a half-stable equilibrium point; = 0 is an unstableequilibrium point.
(iii) For x > 1, as t .For 0 < x < 1, 1 as t .For x < 0, as t .
(iv) tx-plane
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
Section 2.3 77
20. x = x(x 3)(1 + x3)(1 x2)2, x(x 3)(1 + x3)(1 x2)2 = 0
Roots 0 3 1 two non-real 1Multiplicity 1 1 1 ignore 2
End behavior: x(x)(x3)(x2)2 = x9
(i) x x
-4 -2 2 4x
-2
-1.5
-1
-0.5
0.5
1
1.5
2x
(ii) By the phase line diagram, x = 3 is an unstable equilibrium point; x = 1 is a half-stableequilibrium point; x = 0 is a stable equilibrium point; x = 1 is an unstable equilibriumpoint.
(iii) If x > 3, as t .If 1 < x < 3, 1 as t .If 0 < x < 1, 0 as t .If 1 < x < 0, 0 as t .If x < 1, as t .
(iv) tx-plane
-3 -2 -1 1 2 3
-2
-1
1
2
3
-
78 Section 2.3
21. x = x2(1 2x)3(x2 1), x2(1 2x)3(x2 1) = 0
Roots 0 12 1Multiplicity 2 3 1
End behavior: x2(x)3(x2) = x7
(i) x x
-4 -2 2 4x
-0.02
0.02
0.04
0.06
0.08x
(ii) By the phase line diagram, x = 1 is a stable equilibrium point; x = 12 is an unstableequilibrium point; x = 1 is a stable equilibrium point.
(iii) If x > 1, 1 as t .If 12 < x < 1, 1 as t .If 1 < x < 12 , 1 as t .If x < 1, 1 as t .
(iv) tx-plane
-3 -2 -1 1 2 3
-1.5
-1
-0.5
0.5
1
1.5
-
Section 2.3 79
22. x = x3(x2 + 5)(x 4)2(x + 5)
Roots 0 4 5Multiplicity 3 2 1
End behavior: x3x2x2x = x8
(i) x x
-6 -4 -2 2 4x
-75000
-50000
-25000
25000
50000
x
(ii) By the phase line diagram, x = 1 is a stable equilibrium point; x = 12 is an unstableequilibrium point; x = 1 is a stable equilibrium point.
(iii) If x > 1, 1 as t .If 12 < x < 1, 1 as t .If 1 < x < 12 , 1 as t .If x < 1, 1 as t .
(iv) tx-plane
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-
80 Section 2.3
23. y = 1, f (y) = 2y, f (1) = 2 Unstablef (1) = 2 Stable24. y = 1, f (y) = 2y, f (1) = 2 Unstablef (1) = 2 Stable25. y = 0, , f (y) = cos y, f (0) = 1 Unstablef () = 1 Stable26. y = 1, f (y) = 3y2, f (1) = 3 Unstable27. y = 0, f (y) = 3y2, f (0) = 0 Inconclusive. However, from the phase line diagram, y = 0
is stable.
28. If we Taylor expand the function about y and keep the lowest order non-zero term, we seethat we have y = f (3)(y)y3 as the approximate solution near the equilibrium point. Phaseline analysis then shows the equilibrium point is stable.
29. (a) y = ry y3
sqrt rsqrt ry
y
-1.5 -1 -0.5 0.5 1 1.5y
-0.4
-0.2
0.2
0.4y
sqrt rsqrt ry
y
Figure 2.1: Phase line for y = ry y3 : r < 0, r = 0, r > 0
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
Figure 2.2: Pitchfork bifurcation for y = ry y3
-
Section 2.3 81
(b) y = ry + y3
sqrt rsqrt ry
y
-1.5 -1 -0.5 0.5 1 1.5y
-0.4
-0.2
0.2
0.4y
sqrt rsqrt ry
y
Figure 2.3: Phaseline for y = ry + y3 : r < 0, r = , r > 0
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
Figure 2.4: Pitchfork bifurcation for y = ry + y3
-
82 Section 2.4
2.4 Population Modeling: An Application of AutonomousEquations
1. (a) x = 0 is half-stable, x = 1 is stable. For logistic equation, x = 0 is unstable and x = k isstable. Yes, the are different.
(b) For small x, logistic model growth is larger.
2. x = 0 - Stable, x = 1 - Unstable, x = 6 - Stable
3. x = 0 - Stable, x = 2 - Unstable, x = 10 - Stable
4. (a) x = 0 - Half-stable, x = 1 - Unstable, x = 4 - Stable
(b) For small x, Allee effect model has larger growth rate.
5. (a) Exponential growth - Unlimited growth rate. No limitations placed on organisms.
(b) Logistic model - Growth rate dependent on factors such as population amount or foodavailability.
(c) Allee effect - Growth rate dependent on factors such as population amount or food avail-ability as well as a sufficient population to sustain itself.
6. (a) x = 0 - Unstable, x = a - Stable, x = 5 - Unstable
(b) For x0 > 5, bacteria grows uninhibited.
(c) The parameter a could represent the strength of the immune system or ability of thebody to fight off the given bacteria. A healthy person would likely have an a-value that iscloser to 0 than to 5 because the lower value of a represents a lower value of the bacteria(that is stable).
7. (a) For a > 0,x = 0 - Unstablex = 5 a - Stablex = 5 +
a - Unstable
x = 10 - Stable
(b) For a = 0,x = 0 - Unstablex = 5 - Half-stablex = 10 - Stable
(c) For a < 0,x = 0 - Unstablex = 10 - Stable
(d) Saddle-node
(e) Bacteria grows unchecked to a level of 10. No, the bacteria population eventually reachesand levels off at 10, which above the fatal level.
(f) The parameter a could again represent the strength of the immune system or ability ofthe body to fight off the given bacteria.
8. x = x(1 x)(x 6)(x 10)9. x = x(2x 1)(x 1)2(x 2)(x 8) or x = x(2x 1)2(x 1)(x 2)(x 8)
10. (a) x = x2(2 x)2(x 4)
-
2.5. NUMERICAL APPROXIMATION WITH THE EULER METHOD 83
(b) x = 0 - Half-stable, x = 2 - Half-stable, x = 4 - Unstable
11. (a) x = rx(x a)(x 1) r, a > 0(b) i. 0 < a < 1, x = 0 - Unstable, x = a - Stable, x = 1 - Unstable
ii. a = 1, x = 0 - Unstable, x = 1 - Half-stableiii. a > 1, x = 0 - Unstable, x = 1 - Stable, x = a - Unstable
(c) When a < 1, the bacteria goes to the stable level a if it starts with a level less than 1and grows without bound otherwise. When a = 1, the bacteria goes to the half-stablelevel a = 1 if it starts with a level less than 1 and grows without bound otherwise. Whena > 1, the bacteria goes to the stable level of 1 if it starts with a level less than a andgrows without bound otherwise.
2.5 Numerical Approximation with the Euler Method
1. dy/dx = x3, y(1) = 1; explicit solution: y = 14 (x4 + 3)
Euler Explicitxi yi y(xi)1.0 1 11.1 1.1 1.11601.2 1.2331 1.26841.3 1.4059 1.46401.4 1.6256 1.7104
2. dy/dx = x4y, y(1) = 1; explicit solution: y = e(x51)/5
Euler Explicitxi yi y(xi)1.0 1 11.1 1.1 1.12991.2 1.2611 1.34671.3 1.5225 1.72051.4 1.9574 2.4004
3.dy
dx= y2 cosx, h = 0.1, y(0) = 1; explicit solution y = 1
1 + sinx.
x0 = 0x1 = 0 + (0.1)(1) = 0.1x2 = 0 + (0.1)(2) = 0.2x3 = 0 + (0.1)(3) = 0.3x4 = 0 + (0.1)(4) = 0.4y0 = 1y1 = 1 + (0.1)((1)2 cos(0)) = .9y2 = .9 + (0.1)((.9)2 cos(0.1)) = .8194y3 = .8194 + (0.1)((.8194)2 cos(0.2)) = .7536y4 = .7536 + (0.1)((.7536)2 cos(0.3)) = .6993
-
84 Section 2.5
Explicit:
y(0) =1
1 + sin 0= 1
y(0.1) =1
1 + sin(0.1)= .909228
y(0.2) =1
1 + sin(0.2)= .834258
y(0.3) =1
1 + sin(0.3)= .7718907
y(0.4) =1
1 + sin(0.4)= .719725
Euler Explicitxi yi y(xi)0.0 1 10.1 .9 .9092280.2 .8194 .8342580.3 .7536 .77189070.4 .6993 .719725
4. y =sin xy3
, y() = 2, h = .1.
x0 = x1 = + .1x2 = + .2x3 = + .3x4 = + .4y0 = 2
y1 = 2 + (.1)(
sin()23
)= 2
y2 = 2 + (.1)(
sin( + .1)23
)= 1.99875
y3 = 1.995 + (.1)(
sin( + .2)(1.995)3
)= 1.99626
y4 = 1.985 + (.1)(