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Solution Manual for Advanced Engineering Mathematics 8th Edition ONeil 1305635159 9781305635159 Full download at link: https://testbankpack.com/?s=Advanced+Accounting+13th+Edition+Hoyle+Solutio ns+Manual&post_type=product INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY ADVANCED ENGINEERNG MATHEMATICS 8th EDITION

PETER V. O’NEIL

UNIVERSITY OF ALABAMA AT BIRMINGHAM

Contents 1 First-Order Differential Equations 1 1.1 Terminology and Separable Equations 1 1.2 The Linear First-Order Equation 12 1.3 Exact Equations 19 1.4 Homogeneous, Bernoulli and Riccati Equations 28 2 Second-Order Differential Equations 37 2.1 The Linear Second-Order Equation 37 2.2 The Constant Coefficient Homogeneous Equation 41 2.3 Particular Solutions of the Nonhomogeneous Equation 46 2.4 The Euler Differential Equation 53 2.5 Series Solutions 58 3 The Laplace Transform 69 3.1 Definition and Notation 69 3.2 Solution of Initial Value Problems 72 3.3 The Heaviside Function and Shifting Theorems 77 3.4 Convolution 86 3.5 Impulses and the Dirac Delta Function 92 3.6 Systems of Linear Differential Equations 93 iii

iv CONTENTS 4 Sturm-Liouville Problems and Eigenfunction Expansions 101 4.1 Eigenvalues and Eigenfunctions and Sturm-Liouville Problems 101 4.2 Eigenfunction Expansions 107 4.3 Fourier Series 114 5 The Heat Equation 137 5.1 Diffusion Problems on a Bounded Medium 137 5.2 The Heat Equation With a Forcing Term F (x, t) 147 5.3 The Heat Equation on the Real Line 150 5.4 The Heat Equation on a Half-Line 153 5.5 The Two-Dimensional Heat Equation 155 6 The Wave Equation 157 6.1 Wave Motion on a Bounded Interval 157 6.2 Wave Motion in an Unbounded Medium 167 6.3 d’Alembert’s Solution and Characteristics 173 6.4 The Wave Equation With a Forcing Term K(x, t) 190 6.5 The Wave Equation in Higher Dimensions 192 7 Laplace’s Equation 197 7.1 The Dirichlet Problem for a Rectangle 197 7.2 The Dirichlet Problem for a Disk 202 7.3 The Poisson Integral Formula 205 7.4 The Dirichlet Problem for Unbounded Regions 205 7.5 A Dirichlet Problem in 3 Dimensions 208 7.6 The Neumann Problem 211 7.7 Poisson’s Equation 217 8 Special Functions and Applications 221 8.1 Legendre Polynomials 221 8.2 Bessel Functions 235 8.3 Some Applications of Bessel Functions 251 9 Transform Methods of Solution 263 9.1 Laplace Transform Methods 263 9.2 Fourier Transform Methods 268 9.3 Fourier Sine and Cosine Transforms 271 10 Vectors and the Vector Space Rn 275 10.1 Vectors in the Plane and 3 Space 275 10.2 The Dot Product 277 10.3 The Cross Product 278 10.4 n Vectors and the Algebraic Structure of Rn 280 10.5 Orthogonal Sets and Orthogonalization 284 10.6 Orthogonal Complements and Projections 287 11 Matrices, Determinants and Linear Systems 291 11.1 Matrices and Matrix Algebra 291 11.2. Row Operations and Reduced Matrices 295 11.3 Solution of Homogeneous Linear Systems 299 11.4 Nonhomogeneous Systems 306 11.5 Matrix Inverses 313 11.6 Determinants 315 11.7 Cramer’s Rule 318 11.8 The Matrix Tree Theorem 320

17.2 Sine and Cosine Series 423 v 12 Eigenvalues, Diagonalization and Special Matrices 323 12.1 Eigenvalues and Eigenvectors 323 12.2 Diagonalization 327 12.3 Special Matrices and Their Eigenvalues and Eigenvectors 332 12.4 Quadratic Forms 336 13 Systems of Linear Differential Equations 339 13.1 Linear Systems 339 13.2 Solution of X′ = AX When A Is Constant 341 13.3 Exponential Matrix Solutions 348 13.4 Solution of X′ = AX + G for Constant A 350 13.5 Solution by Diagonalization 353 14 Nonlinear Systems and Qualitative Analysis 359 14.1 Nonlinear Systems and Phase Portraits 359 14.2 Critical Points and Stability 363 14.3 Almost Linear Systems 364 14.4 Linearization 369 15 Vector Differential Calculus 373 15.1 Vector Functions of One Variable 373 15.2 Velocity, Acceleration and Curvature 376 15.3 The Gradient Field 381 15.4 Divergence and Curl 385 15.5 Streamlines of a Vector Field 387 16 Vector Integral Calculus 391 16.1 Line Integrals 391 16.2 Green’s Theorem 393 16.3 Independence of Path and Potential Theory 398 16.4 Surface Integrals 405 16.5 Applications of Surface Integrals 408 16.6 Gauss’s Divergence Theorem 412 16.7 Stokes’s Theorem 414 17 Fourier Series 419 17.1 Fourier Series on [ L, L] 419 17.3 Integration and Differentiation of Fourier Series 428 17.4 Properties of Fourier Coefficients 430 17.5 Phase Angle Form 432 17.6 Complex Fourier Series 435 17.7 Filtering of Signals 438

vi CONTENTS 18 Fourier Transforms 441 18.1 The Fourier Transform 441 18.2 Fourier sine and Cosine Transforms 448 19 Complex Numbers and Functions 451 19.1 Geometry and Arithmetic of Complex Numbers 451 19.2 Complex Functions 455 19.3 The Exponential and Trigonometric Functions 461 19.4 The Complex Logarithm 467 19.5 Powers 468 20 Complex Integration 473 20.1 The Integral of a Complex Function 473 20.2 Cauchy’s Theorem 477 20.3 Consequences of Cauchy’s Theorem 479 21 Series Representations of Functions 485 21.1 Power Series 485 21.2 The Laurent Expansion 492 22 Singularities and the Residue Theorem 497 22.1 Classification of Singularities 497 22.2 The Residue Theorem 499 22.3 Evaluation of Real Integrals 505 23 Conformal Mappings 515 23.1 The Idea of a Conformal Mapping 515 23.2 Construction of Conformal Mappings 533

Chapter 1 First-Order Differential Equations

1.1 Terminology and Separable Equations

1. The differential equation is separable because it can be written 3y2 dy = 4x, dx or, in differential form, Integrate to obtain

This implicitly defines a general solution, which can be written explicitly as

= (2x2 + k)1/3 , with k an arbitrary constant.

2. Write the differential equation as dy x dx = y, which separates as 1 1

and y

Integrate to get

y3 = 2x2 + k.

3y2 dy = 4xdx.

c 1

if x /= 0

y dy = x dx 0.

ln |y| = ln |x| + k. Then ln |xy| =

, so

2 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

with c constant (c = ek). y = 0 is a singular solution, satisfying the original differential equation.

3. If cos(y) /= 0, the differential equation is

y sin(x + y)

= dx cos(y) sin(x) cos(y) + cos(x) sin(y)

= cos(y)

= sin(x) + cos(x) tan(y).

There is no way to separate the variables in this equation, so the differential equation is not separable.

4. Write the differential equation as

exey dy = 3x, dx which separates in differential form as

ey dy = 3xe x dx.

Integrate to get

ey = 3e x(x + 1) + c, with c constant. This implicitly defines a general solution.

5. The differential equation can be written

x dy = y2 y, dx or 1 y(y 1) dy = 1 dx, x

and is therefore separable. Separating the variables assumes that y /= 0 and y 1. We can further write 1 1 dy = 1 dx.

y 1 y x

Integrate to obtain

ln |y 1| ln |y| = ln |x| + k.

Using properties of the logarithm, this is xy

ln .y 1. = k.

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 3

y 1 = c, xy with c = ek constant. Solve this for y to obtain the general solution 1 y = . 1 cx y = 0 and y = 1 are singular solutions because these satisfy the differential equation, but were excluded in the algebra of separating the variables.

6. The differential equation is not separable.

7. The equation is separable because it can be written in differential form as sin(y) cos(y) dy = 1 dx. x

This assumes that x /= 0 and cos(y) /= 0. Integrate this equation to obtain ln | cos(y)| = ln |x| + k. This implicitly defines a general solution. From this we can also write sec(y) = cx with c constant. The algebra of separating the variables required that cos(y) 0. Now cos(y) = 0 if y = (2n+1)π/2, with n any integer. Now y = (2n+1)π/2 also satisfies the original differential equation, so these are singular solutions.

8. The differential equation itself requires that y = 0 and x = 1. Write the equation as x dy = y dx and separate the variables to get 1

2y2 + 1 x 1

y(2y2 + 1) dy = x(x + 1) dx. Use a partial fractions decomposition to write this as 1 2y dy = 1 1 dx.

Integrate to obtain

y 2y2 + 1 x x + 1

Then

y| 1 ln(1

2) =

x + 1| + c

ln |x|

4 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

with c constant. This implicitly defines a general solution. We can go a step further by writing this equation as

ln √ y ! = ln x + c and take the exponential of both sides to get √ y = k x , which also defines a general solution.

9. The differential equation is dy = ex y + sin(y), dx and this is not separable. It is not possible to separate all terms involving x on one side of the equation and all terms involving y on the other.

10. Substitute

sin(x y) = sin(x) cos(y) cos(x) sin(y), cos(x + y) = cos(x) cos(y) sin(x) sin(y), and cos(2x) = cos2(x) sin2(x) into the differential equation to get the separated differential form (cos(y) sin(y)) dy = (cos(x) sin(x)) dx.

Integrate to obtain the implicitly defined general solution

cos(y) + sin(y) = cos(x) + sin(x) + c.

11. If y /= 1 and x /= 0, we obtain the separated equation y2 y + 1 dy = 1 dx. x

To make the integration easier, write this as y 1 + 1 dy = 1 dx. 1 + y x

Integrate to obtain

+ 2

2 x + 1

+ 2y2 x + 1

1 y2

y + ln |1 + y| = ln |x| + c.

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS

5

This implicitly defines a general solution. The initial condition is y(3e2) = 2, so put y = 2 and x = 3e2 to obtain

Then c = 2 and the solution of the initial value problem is implicitly defined by

12. Integrate

that y /= 2, to obtain

This implicitly defines a general solution. To have y(2) = 8, let x = 2 and y = 8 to obtain

The solution of the initial value problem is implicitly defined by

We can take this a step further and write

By taking the exponential of both sides of this equation we obtain the explicit solution

13. With ln(yx) = x ln(y), we obtain the separated equation Integrate to obtain

For y(2) = e3 , we need

y) y dy = 3xdx.

2 2 + ln(3) = ln(3e2) + c.

so ln(3e2) = ln(3) + ln(e2) = ln(3) + 2, ln(3) = ln(3) + 2 + c.

Now

1 y2 y + ln |1 + y| = ln |x| 2 1 y + 2

dy = 3x2 dx, ln |2 + y| = x3 + c.

assuming

ln(10) =

8 +

ln |2 + y| = x3 + ln(10) 8.

ln 2 + y = x3 8

y = 10ex 3 8 2.

ln(

(ln(y))2 = 3x2 + c. (ln(e3))2 = 3(4) + c,

6 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

or 9 = 12 + c. Then c = 3 and the solution of the initial value problem is defined by (ln(y))2 = 3x2 3. Solve this to obtain the explicit solution y = e √ 3(x2 1) if |x| > 1.

14. Because ex y 2 = exe y 2 , the variables can be separated to obtain 2yey 2 dy = ex dx.

Integrate to get

To satisfy y(4) = 2 we need ey 2 = ex + c.

e4 = e4 + c

so c = 0 and the solution of the initial value problem is implicitly defined by ey 2 = ex , which reduces to the simpler equation x = y2

Because y(4) = 2, the explicit solution is y = √ x for x > 0.

15. Separate the variables to obtain

implicitly defines a general solution. For y

we need

Integrate to

1 y sin(3y) + 1 cos(3y) = x2 + c, 3 9

to 1 π sin(π) + 3 3 1 1 cos(π) = 9 4 4 + c. 9 9 = 9 + c, so c = 5/9 and the solution of the

by 1 y sin(3y) + 1 cos(3y) = x2 5 , 3 9 9 or 3y sin(3y) + cos(3y) = 9x2 1.

cos(3y) dy = 2xdx.

get

which

(2/3) = π/3,

This reduces

initial value problem

implicitly defined

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 7

16. Let T (t) be the temperature function. By Newton’s law of cooling, T ′(t) = k(T 60) for some constant k to be determined. This equation is separable and is easily solved to obtain:

To determine k, use the fact that T (10)

Now we know the temperature function completely:

We want to know T (20), so compute

Fahrenheit. To see how long it will take for the object to reach 65 degrees, solve for t in

T (t) = 60 + 30ekt .

= 88: T (10) = 60 + 30e10k = 88 Then e10k = 88 60 = 14 , 30 15 so 1 k = ln(14/15) 10

T (t) = 60 + 30ekt = 60 + 30 e10k t/10 14 t/10

T (20) = 60 + 30 14 2 15 ≈ 86.13

T (t) = 65 = 60 + 30 14 t/10 15 Then so 14 t/10 = 15 t 65 60 = 1 , 30 6 10 ln(14/15) = ln(1/6) = ln(6). The object reaches

degrees at time 10 ln(6) t = ln(14/15) ≈ 259 7 minutes. = 60 + 30 15

degrees

8 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

17. Suppose the thermometer was removed from the house at time t = 0, and let T (t) be the temperature function. Let A be the ambient temperature outside the house (assumed constant). By Newton’s law,

We are also given that T (0) = 70 and T (5) = 60. Further, fifteen minutes after being removed from the house, the thermometer reads 50 4, so T

We want to determine A, the constant outside temperature. From the differential equation for T ,

Integrate this, as we have done before, to get

Now use the other two conditions:

T ′(t) = k(t A)

(15) = 50.4.

1 T A dT = kdt.

T (t) = A + cekt . Now, so c = 70 A and T (0) = 70 = A + c, T (t) = A + (70 A)ekt .

T (5) = A + (70 A)e5k = 15.5 and T (15) = A + (70 A)e15k = 50.4. From the

T (5), solve for e5k to get e5k = 60 A . 70 A Then e15k = e5k 3 = 60 A 3 . 70 A Substitute this into the equation T (15) to get Then (70 A) 70 A = 50 4 A. (60 A)3 = (50 4 A)(70 A)2 The cubic terms cancel and this reduces to the quadratic equation 10.4A2 1156A + 30960 = 0, with roots 45 and (approximately) 66 15385. Clearly the outside temperature must be less than 50, and must therefore equal 45 degree.

equation for

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 9

18. The amount A(t) of radioactive material at time t is modeled by A′(t) = kA, A(0) = e3 together with the given half-life of the material,

text)

19. The problem is like Problem 18, and we find that the amount of Uranium235 at time t is with t in years. Then

20. At time t there will be A(t) = 12ekt grams, and A(4) = 12e4k = 9 1. Solve this for k to get

The half-life of this element is the time t∗ it will take for there to be 6 grams, so

A(ln(2)) = 1 e3 2

1 t/ ln(2) A(t) = 2 e3 . Then A(3) = e3 1 3/ ln(2) = 1 tonne

Solve this (as in the

obtain

U(t) = 10 1 t/(4 5(109)) U(109) = 10 1 1/4 5 ≈ 8 57 kg.

k = 1 ln 9 1 .

A(t∗) = 6 = 12eln(9 1/12)t∗/4 t∗ = 4 ln(1/2) 10 02 minutes ln(9.1/12) 21.

I(x) = ∞ e t2 (x/t)2 dt. 0 The

I(3). Compute I′(x) = 2x ∫ ∞ 1 e t 2 (x/t)2 dt. 2 2 2 0 ,

Solve this to get

Let

integral we want is

10 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

Let u = x/t, so t = x/u and

Then I(x) satisfies the separable

with general solution of the form

in which we used a standard integral

arises often in statistics. Then

Finally, put x = 3 for the particular integral of interest:

22. Begin with the logistic equation

in which a and b are positive constants. Then

and the variables are separated. To make the integration easier, write this equation as

Using

of the

we can write this equation as

x dt = u2 du. ′ ∫ 0 u2 2 2 x = 2I(x).

I(x) = ce 2x Now observe

I(0) = ∫ ∞ e t 2 dt = √ π = c,

that

√ π I(x) = e 2x . 2

Then

differential equation I′ =

that

I(3) = ∫ ∞ e t 2 (9/t)2 dt = √ π e 6

P ′(t) = aP (t) bP (t)2 ,

dP = P (a bP ) dt so 1 P (a bP ) dP = dt

1 1 + b 1 dP = dt. a P Integrate to obtain 1 a a bP b a ln(P ) a ln(a bP ) = t + c.

P (t)

a bP (t) > 0.

ln P = at +

a bP ∞ x2 u2 0 (x/u) du 0

> and

properties

logarithm,

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 11

in which k = ac is still constant. Then

in which K = ek is a positive constant. Now suppose the initial population (say at time zero) is p0. Then

We now have

It is a straightforward algebraic manipulation to solve this for P (t):

This is the solution of the logistic equation with P (0) = p0 Because a bp0 > 0 by assumption, then

This means that this population function is bounded above. Further, by multiplying the numerator and denominator of P (t) by e( at, we have

23. With a and b as given, and p0 = 3, 929, 214 (the population in 1790), the logistic population function for the United States is 123, 141 5668

+ 0

If we attempt an exponential model Q(t) = Aekt , then take A = Q(0) = 3, 929, 214, the population in 1790. To find k, use the fact that

(10) = 5308483 = 3929214e10k and we can solve for k to get

P a bP = eat+k = ekeat = Keat ,

p0 = K. a bp0

P (0) = p0 and

P a bP = p0 a bp0 eat

P (t) = ap0 eat a bp0 + bp0

bp0eat < a bp0 + bpeat , so P(t) < ap0 bp0eat eat = a . b

lim P (t) = lim ap0 t→∞ t→∞ (a bp0)e at + bp0 = lim ap0 = a t→∞ bp0 b

P (t) =

t e0 03134t

0 03071576577

0006242342283e0 03134

k = 1 ln 5308483

0 03008667012

12 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

The exponential model, using these two data points (1790 and 1800 populations), is

Table 1.1 uses Q(t) and P (t) to predict later populations from these two initial figures. The logistic model remains quite accurate until about 1960, at which time it loses accuracy quickly. The exponential model becomes quite inaccurate by 1870, after which the error becomes so large that it is not worth computing further. Exponential models do not work well over time with complex populations, such as fish in the ocean or countries throughout the world.

1.2 The Linear First-Order Equation

1. With p(x) = 3/x, and integrating factor is

for x > 0. Multiply the differential equation by x 3 to get

year population P (t) percent error Q(t) percent error 1790 3,929,213 3,929,214 0 3,929,214 0 1800 5,308,483 5,336,313 0.52 5,308,483 0 1810 7,239,881 7,228,171 -0.16 7,179,158 -0.94 1820 9,638,453 9,757,448 1.23 7,179,158 0.53 1830 12,886,020 13,110,174 1.90 13,000,754 1.75 1840 17,169,453 17,507,365 2.57 17,685,992 3.61 1850 23,191,876 23.193,639 0.008 23,894,292 3.03 1860 31,443,321 30,414,301 -3.27 32,281,888 2.67 1870 38,558,371 39,374,437 2.12 43,613,774 13.11 1880 50,189,209 50,180,383 -0.018 58,923,484 17.40 1890 62,979766 62,772,907 -0.33 79,073,491 26.40 1900 76,212,168 76,873,907 0.87 107,551,857 41.12 1910 92,228,496 91,976,297 -0.27 145,303,703 57.55 1920 106,021,537 107,398,941 1.30 196,312,254 83.16 1930 123,202,624 122,401,360 -0.65 1940 132,164,569 136,329,577 3.15 1950 151,325,798 148,679,224 -1.75 1960 179,323,175 150,231,097 -11.2 1970 203,302,031 167,943,428 -17.39 1980 226,547,042 174,940,040 -22.78

Table 1.1: Census data for Problem 23

Q(t) = 3929214e0 03008667012t .

e , ( 3/x) dx = e 3 ln(x) = x 3

x 3y ′ 3x 4 = 2x 1

1.2. THE LINEAR FIRST-ORDER EQUATION 13

) + c, with c an arbitrary constant. For

we have a general solution

by replacing ln(x) with ln x to derive the solution

the last integration, we can allow

2. e dx = ex is an integrating factor. Multiply the differential equation by ex to get

3. e 2 dx = e2x is an integrating factor. Multiply the differential equation by e

or d

x 3

dx x Integrate

x 3y = 2 ln(x

x >

y = 2x3 ln(x) + cx3

x < 0

for x 0. y = 2x3 ln |x| + cx3

(

y) = 2 .

to get

Then y ′ ex + yex = 1 (e2x 1). 2 (exy)′ = 1 (e2x 1) 2 and an

us exy = 1 e2x 1 x + c. 4 2 Then y = 1 ex 1 xe x + ce x is a general solution,

c an arbitrary constant.

integration gives

with

2x: or Integrate to get y ′ e2x + 2ye2x = xe2x , (e2xy)′ = xe2x . e2xy = 1 xe2x 1 e2x + c. 2 4 giving us the general solution y = 1 x 1 + ce 2x .

14 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

4. For an integrating factor, compute

e , sec(x) dx = eln | sec(x)+tan(x)| = sec(x) + tan(x).

Multiply the differential equation by this integrating factor:

y ′(sec(x) + tan(x)) + sec(x)(sec(x) + tan(x))y

= y ′(sec(x) + tan(x)) + (sec(x) tan(x) + sec2(x))y

= ((sec(x) + tan(x))y)′

= cos(x)(sec(x) + tan(x))

= 1 + sin(x)

We therefore have

Integrate to get

Then

((sec(x) + tan(x))y)′ = 1 + sin(x)

y(sec(x) + tan(x)) = x cos(x) + c.

y = x cos(x) + c sec(x) + tan(x)

This is a general solution. If we wish, we can also observe that to obtain

1 = sec(x) + tan(x) cos(x) 1 + sin(x)

y = (x cos(x) + c) cos(x)

1 + sin(x)

x cos(x) cos2(x) + c cos(x) = . 1 + sin(x)

5. First determine the integrating factor

e , 2 dx = e 2x

Multiply the differential equation by e 2x to get (e 2xy)′ = 8x2e 2x .

This yields the general solution y

2e 2x dx = 4x2e 2x + 4xe 2x + 2e 2x + c.

Integrate to get e 2xy

∫

= 4x2 + 4x + 2 + ce2x .

1.2. THE LINEAR FIRST-ORDER EQUATION 15

6. e 3 dx = e3x is an integrating factor. Multiply the differential equation by

, ,

8. e ( 1) dx = e x is an integrating factor. Multiply the differential equation

Integrate this equation: (e3xy)′ = 5e5x 6e3x . e3xy = e5x 2e3x + c.

we have a general solution y = e2x 2 + ce 3x . We need y(0) = 2 = 1 2 + c, so c = 3. The unique solution of the initial value problem is y = e2x + 3e 3x 2

3x to get

Now

e , (1/(x 2)) dx = eln(x 2) = x 2. Multiply the differential equation by x 2 to get ((x 2)y)′ = 3x(x 2) Integrate to get (x 2)y = x3 3x2 + c. This gives us the general solution y = 1 (x3 3x2 + c). x 2 Now we

y(3) = 27 27 + c = 4, so c = 4 and

initial value problem is y = 1 (x3 3x2 + 4) x 2

7. x 2 is an integrating factor for the differential equation because

need

the solution

the

Integrate to get (ye x)′ = 2e3x ye x = 2 e3x + c, 3

by e x

get:

16 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS and

solution y = 2 e4x + cex 3 We need 2 y(0) = 3 + c = 3, so c =

problem has the solution y = 2 e4x 11 ex 3 3

we have the general

11/3 and the initial value

integrating factor e , (2/(x+1)) dx = e2 ln(x+1) = eln((x+1)2) = (x + 1)2 . Multiply the differential equation by (x + 1)2 to obtain (x + 1)2y ′ = 3(x + 1)2 Integrate to obtain Then Now (x + 1)2y = (x + 1)3 + c. c y = x + 1 + (x + 1)2 y(0) = 1 + c = 5 so c = 4 and the initial value problem has the solution 4 y = x + 1 + (x + 1)2

An integrating factor is e , (5/9x) dx = e(5/9) ln(x) = eln(x5/9) = x5/9 Multiply the differential equation by x5/9 to get (yx5/9) ′ = 3x32/9 + x14/9 Integrate to get yx5/9 = 27 x41/9 + 9 x23/9 + c. 41 23 Then y = 27 x4 + 9 x2 + cx 5/9 Finally, we need 41 23 27 9 y( 1) = 41 + 23 c = 4 Then c = 2782/943, so the initial value problem has the solution y = 23 x4 + 9 x2 2782 x 5/9 .

9. First derive the

10.

1.2. THE LINEAR FIRST-ORDER EQUATION 17

11. Let (x, y) be a point on the curve. The tangent line at (x, y) must pass through (0, 2x2), and so has slope

This is the linear differential equation

in which c can be any number.

12. Let A(t) be the number of pounds of salt in the tank at time t ≥ 0. Then dA

= rate salt is added rate salt is removed

We must solve this subject to the initial condition A(0) = 25. The differential equation is

which is linear with integrating factor

y ′ = y 2x2 x

y ′ 1 y = 2x. An

e (1/x) dx = e ln(x) = eln(1/x) = 1 , x so

1/x to

1 y ′ 1 y = 2. This is Integrate to get x x2 1 y ′ = 2 Then 1 x y = 2x + c. y = 2x2 + cx,

integrating factor is

multiply the differential equation by

get

= 6 2

(t)

50 + t

A′ + 2 50 + t

A = 6, e , 2/(50+t) dt = e2 ln(50+t) = (50 + t)2 Multiply the differential equation by (50 + t)2 to get (50 + t)2A′ + 2(50 + t)A = 6(50 + t)2

18 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

This is

A ′ = 6(50 + t)2 . Integrate this equation to get

which we will write as

+ t)

We need c so that

= 187, 500. The number of pounds of salt in the tank at time t is

13. Let A1(t) and A2(t) be the number of pounds of salt in tanks 1 and 2, respectively, at time t. Then

Solve the linear initial value problem for

to get

Substitute this into the differential equation for A

Solve this linear problem to obtain

2 2 2

(50

(50 + t)2A = 2(50 + t)3 + c,

c A(t) = 2(50 + t) + (50 + t)2 c A(0) = 100 + 2500 = 25,

187, 500 A(t) = 2(50 + t) (50 + t)2

A′ (t) = 5 5A1(t) ; A (0) = 20 and 1 2 100 1 A′ (t) = 5A1(t) 5A2(t) ; A (0) = 90

A1(t)

A1(t) = 50 30e t/20

A′ + 1 A = 5 3 e t/20; A (0) = 90

) to get

A2(t) = 75 + 90e t/20 75e t/30 . Tank 2

when A2 ′ (t) = 0, and this occurs

2.5e t/30 4.5e t/20 = 0. This

et/60 = 9/5, or t = 60 ln(9/5).

5450

A2(t)min = A2(60 ln(9/5)) = 81 2

has its minimum

when

occurs when

Then

pounds.

1.3. EXACT EQUATIONS 19

1.3 Exact Equations

In these problems it is assumed that the differential equation has the form M(x, y) + N(x, y)y ′ = 0, or, in differential form, M(x, y) dx + N(x, y) dy = 0.

1. With M(x, y) = 2y2 + yexy and N(x, y) = 4xy + xexy + 2y Then

∂N = 4y + exy + xyexy = ∂M ∂x ∂y

for all (x, y), so the differential equation is exact on the entire plane. A potential function ϕ(x, y) must satisfy

∂ϕ = M(x, y) = 2y2 + yexy ∂x and ∂ϕ = N(x, y) = 4xy + xexy + 2y. ∂y

Choose one to integrate. If we begin with ∂ϕ/∂x = M, then integrate with respect to x to get

ϕ(x, y) = 2xy2 + exy + α(y),

with α(y) the “constant” of integration with respect to x. Then we must have

This requires that α ′(y) = 2y, so we can choose α(y) = y2 to obtain the potential function ϕ(x,

The general solution is defined implicitly by the equation 2xy2 + exy + y2 = c, , with c an arbitrary constant.

2. ∂M/∂y = 4x = ∂N/∂x for all (x, y), so this equation is exact on the entire plane. For a potential function, we can begin by integrating

∂y

∂ϕ = 4xy + xexy + α ′(y) = 4xy + xexy + 2y.

y) = 2

2 + exy + y

∂ϕ

∂y

∂ϕ

∂x

= 2x2 + 3y2

to get Then ϕ(x, y) = 2x2y + y3 + c(x).

= 4xy + 2x = 4xy + c ′(x)

20 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

Then c ′(x) = 2x so we can choose c(x) = x2 to obtain the potential function

ϕ(x, y) = 2x2y + y3 + x2 . The general solution is defined implicitly by 2x2y + y3 + x2 = k, with k an arbitrary constant.

3. ∂M/∂y = 4x + 2x2 and ∂N/∂x = 4x, so this equation is not exact (on any rectangle).

4. ∂M ∂N

∂y = 2 sin(x + y) + 2x cos(x + y) = ∂x , for all (x, y), so this equation is exact on the entire plane. Integrate ∂ϕ/∂x = M or ∂ϕ/∂y = N to obtain the potential function

ϕ(x, y) = 2x cos(x + y).

The general solution is defined implicitly by 2x cos(x + y) = k with k an arbitrary constant.

5. ∂M/∂y = 1 = ∂N/∂x, for x = 0, so this equation is exact on the plane except at points (0, y). Integrate ∂ϕ/∂x = M or ∂ϕ/∂y = N to find the potential function

ϕ(x, y) = ln |x| + xy + y3 for x /= 0. The general solution is defined by an equation

ln |x| + xy + y3 = k.

6. For the equation to be exact, we need ∂M = αxyα 1 = ∂N

This will hold if α = 2. With this choice of α, the (exact) equation is

∂y ∂x = 2xyα 1

2y 3y

ϕ(x, y) = x3 x2 + 2y2

3 x2 x + 2y2 = k,

+ xy

′ = 0 Routine integrations produce a potential function

The general solution is defined by the equation for

1.3. EXACT EQUATIONS 21

7. For this equation to be exact, we need

This will be true if α = 3. By integrating, we find a potential function

and

general solution is defined implicitly by

8. We have

for all (x, y), so this equation is exact over the entire plane. By integrating

respect to

A general solution is defined implicitly by

xy2) = k.

For the solution satisfying y(1) = 2, put x = 1 and y = 2 into this implicitly defined solution to get

= k.

The solution of the initial value problem is defined implicitly by

9. Because ∂M/∂y = 12y2 = ∂N/∂x, this equation is exact for all (x, y). Straightforward integrations yield the potential function

(x, y) = 3xy4 x.

∂M = 6xy2 3 = ∂N ∂y ∂x = 3 2αxy2

ϕ(x, y) = x2y3 3xy 3y2

x2y3 3xy 3y2 = k.

∂M ∂y = 2 2y sec2(xy2) 2xy3 sec2(xy2) tan(xy2) = ∂N , ∂x

∂ϕ/∂x = 2y y2 sec2(xy2) with

ϕ(x, y) = 2xy tan(xy2) + c(y).

∂ϕ = 2x 2xy sec2(xy2) ∂y = 2x 2xy sec2(xy2) + c ′(y).

c ′(y) = 0

c(y) =

ϕ(x, y) = 2xy tan(xy2)

we find that

Then

and we can choose

0 to obtain the potential function

tan(

tan(4)

2xy tan(xy

2) = 4 tan(4)

22 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

A general solution is defined implicitly by 3xy4 x = k.

To satisfy the condition y(1) = 2, we must choose k so that 48 1 = k, so k = 47 and the solution of the initial value problem is specified by the equation 3xy4 x = 47.

In this case we can actually write this solution explicitly with y in terms of x

10. First,

so the equation is exact for all (x, y) with x = 0. For a potential function, we can begin with

and integrate with respect to y to get

Then we need

This requires that c ′(x) = 1 and we can choose c(x) = x Then ϕ(x, y) = xey/x + x.

The general solution of the differential equation is implicitly defined by xey/x + x = k.

To have y(1) = 5, we must choose k so that e 5 + 1 = k.

The solution of the initial value problem is given by xey/x + x = 1 + e 5 .

This can be solve for y to obtain the explicit solution for x + 1 > 0.

y = x ln x + 1

∂M

= y

= 1 ey/x 1 ey/x y ey/x

ey/x = ∂N ,

∂ϕ

∂y

= ey/x

(x, y) = xey/x + c(x)

∂ϕ = 1 + ey/x y ey/x = ey/x y ey/x + c ′(x)

1.3. EXACT EQUATIONS 23

11. First, ∂M

∂N

∂y = 2x sin(2y x) 2 cos(2y x) = ∂x , so the differential equation is exact for all (x, y). For a potential function, integrate

∂ϕ

∂y

= 2x cos(2y x)

with respect to y to get

ϕ(x, y) = x sin(2y x) + c(x)

Then we must have

∂ϕ

∂x = x cos(2y x) sin(2y x)

= x cos(2y x) sin(2y x) + c ′(x)

c ′(x) = 0 and we can take c(x) to be any constant. Choosing c(x) = 0 yields

Then

ϕ(x, y) = x sin(2y x) The general solution is defined implicitly by x sin(2y x) = k.

To satisfy y(π/12) = π/8, we need

π 12 sin(π/6) = k,

so choose k = π/24 to obtain the solution defined by

x sin(2y x) = 24 which of course is the same as

We can also write

π x sin(2y x

y = 1 x + arcsin π 12.12. for x /= 0. 2 24x ∂M = ey = ∂N ∂y ∂x so the differential

is exact. Integrate ∂ϕ = ey ∂x

) =

equation

24 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

with respect to x to get

ϕ(x, y) = xey + c(y). Then

∂ϕ = xey + c ′(y) = xey 1, ∂y so c ′(y) = 1 and we can let c(y) = y. This gives us the potential function ϕ(x, y) = xey y. The general solution is given by xey y = k.

For y(5) = 0 we need 5 0 = k so k = 5 and the solution of the initial value problem is given by xey y = 5

13. ϕ + c is also a potential function if ϕ is because and

∂y The function defined implicitly by ϕ(x, y) = k is the same as that defined by ϕ(x, y) + c = k if k is arbitrary.

14. (a) ∂M = 1 and ∂y ∂N ∂x = 1

so this equation is not exact over any rectangle in the plane.

(b) Multiply the differential equation by x 2 to obtain yx 2 x 1y ′ = 0. This is exact because ∂M ∗ = x 2 = ∂N ∗ ∂y ∂x

= ∂x

∂ϕ ∂(ϕ + c)

∂x ∂ϕ ∂(ϕ + c)

∂y

1.3. EXACT EQUATIONS 25

This new equation has potential function ϕ(x, y) = yx 1 and so has general solution defined implicitly by

This also defines a general solution of the original differential equation.

This is exact on any region of the plane not containing y = 0, because

The new equation has potential function ϕ(x, y) = xy 1 , so its general solution is defined implicitly by

It is easy to check that this also defines a solution of the original differential equation.

(d) Multiply the differential equation by xy

to obtain

This is exact (on any region not containing y = 0) because

y x = k.

y 1 xy 2y ′ = 0.

∂M ∗∗ = y 2 = ∂N ∗∗

1 =

xy 2 x2y 3y ′ = 0

∂M ∗∗∗ = 2xy 3 = ∂N ∗∗∗ . Integrate ∂ϕ/∂x = xy 2 with respect to x to obtain ϕ(x, y) = 1 x2y 2 + c(y). 2 Then ∂ϕ = x 2y 3 + c ′(y) = x2y 3 , ∂y

so c ′ = 0 and we can choose c(y) = 0. Then ϕ(x, y) = 1 x2y 2 2

and we can define a general solution of this differential equation as x2y 2 = k.

Here we absorbed the factor of 1/2 into the arbitrary constant c Thisagain defines a solution of the original differential equation.

Advanced engineering mathematics 8th edition oneil solutions manual 1

PETER V. O’NEIL

UNIVERSITY OF ALABAMA AT BIRMINGHAM

Chapter 1

First-Order Differential Equations

1.1 Terminology and Separable Equations

2 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 3

5

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 7

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 11

1.2 The Linear First-Order Equation

1.2. THE LINEAR FIRST-ORDER EQUATION 13

1.3. EXACT EQUATIONS 19

1.3 Exact Equations

20 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

1.3. EXACT EQUATIONS 21

22 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

1.3. EXACT EQUATIONS 23

1.3. EXACT EQUATIONS 25

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PETER V. O’NEIL