Advanced Engineering Mathematics, 4th Edition (2020) – Potter – Solutions Manual PDF

ALL 11 CHAPTERS COVERED




SOLUTIONS MANUAL

, PREFACE
This manual provides solutions to the problems in our book, ADVANCED
ENGINEERING MATHEMATICS. In many cases, the solutions are not as
detailed as the examples in the book; they are intended to provide the primary
steps in each solution so the instructor is able to quickly review how a problem
is solved. The discussion of a subtle point, should one exist in a particular
problem, is left as a task for the instructor. In general, some knowledge of a
problem may be needed to fully understand all of the steps presented. This
manual is not intended to be a self-paced workbook for the student; the
instructor is critically needed to provide explanations and discussions of
significant points in many of the problems.

The degree of difficulty and length of solution for each problem varies
considerably. Some are relatively easy and others quite difficult. This allows for
flexibility in assignments or in practice sessions. Typically, the easier problems
are the first problems for a particular section.

The problems have been carefully solved with the hope that errors have not
been introduced. Even though extreme care is taken and problems and equations
are reviewed, errors creep in. We would appreciate knowing about any errors
that you may find. They can be eliminated in future printings and/or included on
an appropriate web page.

We’d also like to thank Professor Matthew Boelkins of Grand Valley State
University for his contributions to the solutions to the problems in Chapters 4, 5,
and 6.


East Lansing, Michigan Merle Potter
Ann Arbor, Michigan Jack Lessing
Allendale, Michigan Edward Aboufadel

,1. Ordinary Differential Equations

Section 1.2

1. We can simplify the equation as u  2u  u  cos x Hence, the equation is linear,
homogeneous, and 1st order.

2. uu  1  x is nonlinear, 1st order.

x3
3. sin u  u  0 is nonlinear, first order. ( sin x  x   )
3!

4. u  2u  u  cos x is linear, nonhomogeneous, 2nd order.

5. u  x2 is linear, nonhomogeneous, 2nd order.

6. u  u  0 is linear, homogeneous, 2nd order.

7. u  u2  0 is nonlinear, 2nd order.

8. (u2 )  u  0 is linear, 1st order, homogeneous (divide by u).

x3
9. u  x2  2 implies u( x)   2 x  C.
3

10. u  sin x  e x implies u( x)  cos x  e x  C.

x2 x 1
11. u  x  cos2 x implies u( x)    sin 2 x  C.
2 2 4

x3
12. u  2x implies u( x)  x2  C , which implies u( x)   Cx  D.
3

x3 x4
13. u  x2 implies u   C , which implies u   Cx  D , which implies
3 12
x5 x2
u( x)  C  Dx  E.
60 2

x2 x3
14. uiv  x  2 implies u   2 x  C1  u   x2  C1 x  C2
2 6

, x4 x3 x2 x5 x4 x3 x2
 u    C1  C2 x  C3  u( x)    C1  C2  C3 x  C4
24 3 2 120 12 6 2

16. Let u  Ax  B( x3  1). Then u  A  3Bx2 and u  6Bx. Hence,
(2x3  1)6Bx  6x2 ( A  3Bx2 )  6x( Ax  Bx3  B)
= 12Bx4  6Bx  6 Ax2  18Bx 4  6 Ax 2  6Bx 4  6Bx  0

17. A( x  C1 )( x  C2 )  B( x  C3 )  C  A( x2  2)(C1  C2 )x  C1C2  Bx  C3 B  C
 Ax2  Dx  E
where D  2 A(C1  C2 )  B and E  AC1C2  BC3  C

20. u2  x2  10 implies 2uu  2x  0.

21. u  e 3 x  12e 2 x  u  3e 3 x  24e 2 x  u  9e 3 x  48e 2 x
So, u  5u  6u  9e 3 x  48e 2 x  5(3e 3 x  24e 2 x )  6( e 3 x  12e 2 x )  0

22. s  20. s  20t  c1 . s  10t 2  c1t  c2 . s(0)  100.  c1  100.
s(0)  0.  c2  0.  s(t)  10t 2  100t. s  20t  100  0.  t  5 at smax .
 smax  10  52  100  5  250 m

23. s  9.81. s  9.81t  c1 . s  9.82t  c1t  c2 . s(0)  s(0)  0. Thus,
c1  c2  0.  s(t)  9.81t and s  8  9.81t g.  t g  1.277 s

24. y( x)  Cx  C 2 implies y( x)  C and ( y)2  C 2 .
x2 x x2
( y )  xy   y  C  xC  xC  C  0. Also, y   , y    , ( y ) 
2 2 2 2

4 2 4
2 2 2
x x x
   0
4 2 4

25. From No. 24 we choose C so that y(2)  C  2  C 2  1.  C  1. Thus,
x2
y( x)  x  1 and y   are two such solutions.
4

26. We have y1  1 so ( y)2  xy  y  1  x  (x  1)  0. For y2  2( x  2),
y2  2, ( y2 )2  4. Hence, ( y)2  xy  y  4  2x  (2x  4)  0.

27. If u( y)  A sin ax  B cos ax, u  aA cos ax  aB sin ax and
u  a2 sin ax  a2 B cos ax.  u  a2 u  0. Now, u(0)  B  10 and