Instructors Solution Manual Advanced Engineering Mathematics 4th Edition By Merle C. Potter, Jack L.

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 Ann Arbor, Michigan Allendale, Michigan

Merle Potter Jack Lessing 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.

3. sin u  u  0 is nonlinear, first order. ( sin x  x 

x3  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).

9. u  x2  2 implies u( x) 

x3  2 x  C. 3

10. u  sin x  e x implies u( x)  cos x  e x  C. 11. u  x  cos2 x implies u( x) 

x2 x 1   sin 2 x  C. 2 2 4

12. u  2x implies u( x)  x2  C , which implies u( x)  13. u  x2 implies u  u( x) 

x3  Cx  D. 3

x3 x4  C , which implies u   Cx  D , which implies 3 12

x5 x2 C  Dx  E. 60 2

14. uiv  x  2 implies u 

x2  2 x  C1 2

 u 

x3  x2  C1 x  C2 6

 u 

x4 x3 x2   C1  C2 x  C3 24 3 2

x5 x4 x3 x2  u( x)    C1  C2  C3 x  C4 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 2 / 2  c1t  c2 . s(0)  s(0)  0. Thus,

c1  c2  0.  s(t)  9.81t 2 / 2 and s  8  9.81t g2 / 2.  t g  1.277 s 24. y( x)  Cx  C 2 implies y( x)  C and ( y)2  C 2 . x2 x x2 2 ( y )  xy   y  C  xC  xC  C  0. Also, y   , y    , ( y )  4 2 4 2 2 2 x x x    0 4 2 4 2

2

2

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

x2 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

u( / 2a)  A  20.  u( x)  20sin ax  10cos ax is a specific solution.

28. uiv  0.006. u ).006x  c1 . u  0.003x2  c1 x  c2 . x2 x3 x2 u  0.001x3  c1  c2 x  c3 . u  0.00025x 4  c1  c2  c3 x  c4 .

2 6 2 Now, u(0)  u(0)  0.  c3  c4  0.

10 3 10 2   c2  6 2    c1  0.03, c2  0.05 2 10 3 u(10)  0  0.001  10  c1  c2  10   2  2 5 Then, u(5)  umax  0.00025  54  0.03  0.05  52  0.625 m 2 Also, u(10)  0  0.00025  10 4  c1

29. The Maple input follows: > ode:=diff(u(x), x) = x^2 + 2: > dsolve(ode, u(x)); 30. > ode:=diff(u(x), x) = sin(x)+exp(x): > dsolve(ode, u(x)); 31. > ode:=diff(u(x), x) = x + (cos(x))^2: > dsolve(ode, u(x)); 32. > ode:=diff(u(x), x$2) = 2*x: > dsolve(ode, u(x)); 33. > ode:=diff(u(x), x$3) = x^2: > dsolve(ode, u(x)); 34. > ode:=diff(u(x), x$4) = x - 2: > dsolve(ode, u(x));

Section 1.3.1 1. u  10u. 

2.

u  10 and d(ln u)  d(10 x). So, ln u  10 x  ln C.  u( x)  Ce10 x u

u 1 1 1  10 implies d     d(10 x). So,   10 x  C and u( x)   2 u u 10x  C  u

u 1 1  1. So, d[ln(2u  3)]  d( x).  ln(2u  3)  x  C 2u  3 2 2 3 or ln(2u  3)  2 x  ln K.  2u  3  Ke 2 x or u( x)    Ke 2 x 2

3. u  2u  3. 

4. We have

u  sin x so d(ln u)  d(  cos x). u  ln x   cos x  ln C or u( x)  Ce  cos x

5. u tan u  sin x so d( ln cos u)  d( cos x).  ln cos u  cos x  ln C and cos u  Cecos x . Finally, u( x)  arccos(Ce cos x ) u 1 1 1 1   . Thus,    u  d(ln x) 2 1u x 2 1u 1 u 1 1 1  u or d   ln(1  u)  ln(1  u)  d(ln x) or d ln  d(ln x) 2 2  1  u  1/ 2 1 1 u 1u  Thus, ln  ln x  ln C.  ln    ln Cx 2 1u 1u 1 u Kx2  1 Hence,  (Cx)2  Kx2 .  u( x)  1u Kx2  1

6. xu  1  u2 so

u 1 1 1 1  x   1 1       .  d     d  ln  2 u x( x  2) 2  x x  2   u 2  x2 1 1 x 1 1 Cx 2 Thus,   ln  ln C  ln .  u( x)   Cx u 2 x2 2 2 x2 ln x2

7. x( x  2)u  u2 so

8. 5xdu  x2 udx  0 so Thus, ln u  C 

x2 .

10

5 x2 du  xdx  0.  5ln u   C. u 2 Finally, u( x)  Ke  x / 10 2

u . Set u  vx. Then u  v  xv. x 2 So v  xv  2  v or xv  2 or v   .  v  2 ln x  C x or v  2 ln Kx.  u( x)  2x ln Kx

9. xu  2x  u so u  2 

u u2  . Set u  vx so u  v  xv. x x2 v 1 1 Then v  xv  v  v2 or 2  . Then, d     d(ln Cx). v x  v 1 1 x Hence,   ln Cx so v   . Finally, u( x)   v ln Cx ln Cx

10. x2 u  xu  u2 so u 

3

2

u u 11. x  u  xu u  0 or 1       u  0. Set u  xv so u  v  xv x x 1 v3 Then 1  v3  v2 ( v  xv)  0 or v2 vx  1.  v 2 v  and  ln Cx x 3 3

3

2

Finally, v  ln(Cx)3 

1/ 3

so that u( x)  x[ln(Kx3 )]1/ 3

u u 12. 3u  (u  x)u  0 becomes 3    1  u  0. Let u  vx so u  v  xv. x x  v(1  v) 1 Then 3v  (v  1)(v  xv)  0. This leads to xv(1  v)  4v  v 2 or  2 4v  v x 1/ 4 3 / 4 1 1 3    . So ln v  ln( v  4)   ln(cx). or v    v v  4 x 4 4   v K u K   4 or  6 3 3 ( v  4) x (u  4 x) x

13. Let x  u  y or x  y  u. u  1  y and x(1  y)  y 2  x  y or  xy   y 2  y So

 1 y 1 1 1   or y       .  ln( y  1)  ln y   ln(cx) y y x x  y 1 y  y 1 K x x   or y   x  u.  u( x)  x  y x xK xK 2

14. Let x  2u  y so y  1  2u. Then ( x  2u  1)u  x  2u  4 becomes

 y  1 2   y  4 or  1   y   3.  y  2 ln( y  3)  3x  c.

2 y  3   Finally, x  2u  2 ln(3  x  2u)  3x  c ( y  1)

15. The general solution is ce 2 x 

1 1  u( x). Let x  0 so c   u(0)  2.

2 2

1 Then c  5 / 2.  u( x)  (5e 2 x  1) 2

u  cot x, we obtain ln(u  1)  ln( c sin x), so the general solution is u1 u( x)  1  c sin x. But, u(2)  0  1  c sin 2.  c  1/ sin 2.  u( x)  1  sin x / sin 2

16. From

c 17. We find the general solution: u( x)  x  . u(1)  10  1  c. x  c  9 and u( x)  x  9 / x

18. Set v  u  x in xu  (u  x)3  u to obtain x(v  1)  v3  v  x

  v v 1 v2 1  2 v  .  2  cx2 so u( x)  x  1   v v 1 x v 1 K  x2   1 u(1)  2  1  .  K  2 and u( x)  x 1  (2  x2 )1 / 2  K 1 or

19. The Maple input follows: > ode:=diff(u(x), x) = 10*u(x): > dsolve(ode, u(x)); 20. > ode:=diff(u(x), x) = 10*(u(x))^2: > dsolve(ode, u(x)); 21. > ode:=diff(u(x), x) = 2*u(x) + 3: > dsolve(ode, u(x)); 22. > ode:=diff(u(x), x) = u(x)*sin(x): > dsolve(ode, u(x)); 23. > ode:=diff(u(x), x) = cot(u(x))*sin(x): > dsolve(ode, u(x)); 24. > ode:=x^2*diff(u(x), x) + (u(x))^2 = 1: > dsolve(ode, u(x));

25. > ode:=x*(x+2)*diff(u(x), x) = (u(x))^2: > dsolve(ode, u(x)); 26. > ode:=5*x*diff(u(x), x) + x^2 = 0: > dsolve(ode, u(x)); 27. > ode:=x*diff(u(x), x) + 2*x = u(x): > dsolve(ode, u(x)); 28. > ode:=x^2*diff(u(x), x) = x*u(x) + (u(x))^2: > dsolve(ode, u(x)); 29. > ode:=x^3 + (u(x))^3 – x*(u(x))^2*diff(u(x), x) = 0: > dsolve(ode, u(x)); (Note: Maple also generates two complex-valued solutions; neither are appropriate here.)

30. > ode:=3*u(x)+(x+u(x))*diff(u(x), x) = 0: > dsolve(ode, u(x)); 31. > ode:=x*diff(u(x), x) = (x-u(x))^2 + u(x): > dsolve(ode, u(x)); 32. > ode:=(x + 2*u(x) + 1)*diff(u(x), x) = x + 2*u(x) + 4: > dsolve(ode, u(x));

(Compare this answer to that of Problem 14.)

33. > ode:=diff(u(x), x) = 2*u(x) - 1: > dsolve({ode, u(0)=2}, u(x));

34. > ode:=diff(u(x), x)*tan(x) = u(x) + 1: > dsolve({ode, u(2)=0}, u(x)); 2 1 x 1 1 1 sin( x ) ( sin( 2 ) tan( 1 ) sin( 2 ) ) u( x )sin( x )   tan    2 2 2 x  2 sin( 2 ) tan( 1 )  tan     2 35. > ode:=x*diff(u(x), x) + u(x) = 2*x: > dsolve({ode, u(1)=10}, u(x));

36. > ode:=x*diff(u(x), x) = (u(x)-x)^3 + u(x): > dsolve({ode, u(1)=2}, u(x)); 42 x2 1   u( x )   x 2   2 ( 2x2 ) 2x  

Section 1.3.2 du  u  0. First order, linear and dx du dx homogeneous. The variables are separable:  . The solution is then u x u( x)  cx

1. Divide by dx and multiply by x2 to obtain x

2. The equation x

du  2u  0 is first order, linear, and homogeneous. It separates: dx du dx  2 .  u( x)  c / x2 u x

3. (2  x2 )du  2xudx  0. N  2  x2 , M  2 xu.

N M  2x  .  exact. x u

   M  2xu,   x2 u  h(u).   x2  h  N  x 2  2. x u 2  h  2 and h(u)  2u.   x u  2u and x2 u  2u  c.  u( x)  c(2  x2 )1

From

4. 3u2 du  x2 dx  0. N  3u2 , M  x2 .

N M 0 .  exact. x u

 1   M  x2 .   x3  h(u).  h  3u2 .  h  u2 . x 3 u 1 3 Thus,   x  u3  c and u( x)  (c  x3 / 3)1 / 3 3

5. sin 2 x du  2u cos 2 x dx  0. N  sin 2 x, M  2u cos 2 x.

N M  2 cos 2 x  x u

   2u cos 2 x.   u sin 2 x  h(u).  sin 2 x  h  sin 2 x. x u c  h  0 and h  0.   u sin 2 x  c. Finally, u( x)  sin 2 x

 exact.

6. e x du  ue x dx  0. N  e x , M  ue x .

N M  ex  .  exact . x u

   ue x ,   ue x  h(u).  e x  h  e x .  h  0 and   ue x  c. x u

Finally, u( x)  ce  x

7. f ( x, u)dx  du  0. M  f ( x, u), N  1.

f N M 0 .   0. x u x

 f ( x, u)  h( x). Suppose f ( x, u)  h( x). Then and    h( x) dx  g(u).

  h( x) x

  g   1.  g  u and    h( x) dx  u. u

Since   c we have u( x)   h( x) dx  K

8. Write



1 1 1 1  1 dx  du  0. M  , N .  . Thus g( x) h(u) g( x) h(u) x g( x) 1  1 1 dx  f (u).  f .  f (u)   du. g( x) u h(u) h(u)

Hence,   

dx du   C. g( x) h(u)



du dx C h(u) g( x)

p dx p dx M N  p0 e  0 ,  p0 e  0 . u x p dx p dx p dx   e  0 [ p0 u  g( x)].   u e  0 p0 dx   e  0 gdx  h(u) x p dx p dx p dx p dx  or   ue  0   e  0 gdx  h(u).  e  0  h  N  e  0 .  h(u)  0. u p dx p dx  p ( x ) dx  p ( x ) dx p ( x ) dx   ue  0  e  0 gdx  C.  u( x)  Ce  0 e  0 e 0 g( x)dx

9. M  e  0 [ p0 u  g( x)], N  e  0 . p dx

p dx



10.



du 2x K  dx. Thus, ln u   ln(1  x2 )C.  u( x)  2 u 1 x 1  x2

But, u(0)  1  K.

 u( x)  (1  x2 )1

u u u 11.  1   u   1. Let v  so u  v  xv.  (1  v)(v  xv)  v  1 x x x  (1  v)dv dx 1 K or  .  ln(1  2v  v2 )  ln Cx.  1  2v  v2  2 2 1  2v  v x 2 x 1 1 u(1)  0 so v(1)  0.  1  K.  1  2v  v2  2 or v2  2v  1  2  2 x x 1/ 2 1/ 2 1 u 1 1/ 2  v  1   2  2  and  1   2  2  .  u( x)   x  (2 x2  1) x  x x   

12. (u  u)e x  d(ue x )  0.  ue x  C. u(0)e 0  C  0.  u( x)  0 13.

( kM) ( kM) M ( kN ) N N M N k , k .  k if and only if  (k  0) u u x x u x u x  M  N M N However, ( f ( x) M)  f ( x) and ( fN )  f N  f . Since  u u x x u x M N N we have f  f  f N  f .  f ( x)N  0. u x x Since N  0, f ( x)  0 so f ( x)  Const

14.

 M  N [ M( x , u)  g( x)]  .

[N( x , u)  h(u)]  and the result is established. u u x x

15. The Maple input follows: > ode:=(1+x^2)*diff(u(x), x) + 2*x*u(x) = 0: > dsolve({ode, u(0)=1}, u(x));

16. These two commands lead to complex-valued solutions: > ode:=(x+u(x))*diff(u(x), x) + u(x) = x: > dsolve({ode, u(1)=0}, u(x));

u( x )I ( x I 2 x21 ), u( x )( x I 2 x21 ) I Modifying the dsolve command with the[exact] option yields the correct answer: > dsolve({ode, u(1)=0}, [exact], u(x));

u( x )x 2 x21 17. > ode:=(diff(u(x), x) + u(x))*exp(x) = 0: > dsolve({ode, u(0)=1}, u(x));

Section 1.3.3 ˆ ( x). So, we have 1. Let Fc ( x)  exp   pdx  c   cˆ exp  pdx  cF   1 K 1 K 1 Kˆ ˆ uˆ  F ( x ) g ( x ) dx   cF ( x ) g ( x ) dx   F ( x ) g ( x ) dx  c ˆ ( x)  ˆ ( x) F ( x)  Fc ( x)  Fc ( x) cF cF F ( x)

which is u(x) with K replaced by Kˆ .

3. For u  u  2, F( x)  e 

dx

 ex . 

d (ue x )  2e x . dx

Thus e x u  2e x  c. So u( x)  2  ce  x

4. For u  2u  2 x, F( x)  e 

d 2x ( e u)  2 xe 2 x and dx 1 1 ue 2 x   2 xe 2 x dx  c  xe 2 x  e 2 x  c.  u( x)  x   ce 2 x 2 2 2 dx

5. For u  xu  10, F( x)  e  x

 e2x . 

xdx

 e x / 2 u( x)  10 et / 2 dt  C. 2

2

0

6. For u  2u  e x , F( x)  e   e 2 x u( x)  e  x  c.

2

 ex / 2 .

2 d x2 / 2 ( e u)  10e x / 2 dx

x

 u( x)  10e  x / 2  et / 2 dt  Ce  x / 2 2

2

0

d 2 x ( e u)  e 2 x e x  e  x dx x  u( x)  e  ce 2 x 2 dx

 e 2 x .

2

7. For u  u  xe  x , F( x)  e   u( x) 

dx

d x x2 x e . ( e u)  x so e u( x)  c dx 2 x

1 2 x x e  ce  x 2

8. For u  u  cos x , F( x)  e 

 dx

d x ( e u)  e  x cos x so dx

 ex .

1  u( x)  (sin x  cos x)  ce x 2

e  x u( x)   e x cos x dx  c.

2  dx 2 d 2 9. For u  u  e x , F( x)  e  x  e 2 ln x  x 2 . ( x u)  x 2 e x so x dx

x

x2 u( x)   t 2 et dt  c.

0

x

 u( x)  x2  t 2 e t dt  cx 2 0

1

  2 dx 1 2 1 d 1/ x 2 1 10. For u  2 u  2 sin , F( x)  e x  e1 / x . (e u)  2 sin e1 / x x x x dx x x x 2 x 2 1 1  e1 / x u( x)   2 e1 / t sin dt  C.  u( x)  e 1 / x  2 e1 / t sin dt  Ce 1 / x 1 t 1 t t t

11. For u  2u  2e 2 x , F( x)  e 

d 2x ( e u)  2 so dx e 2 x u( x)  2x  C.  u( x)  2xe 2 x  Ce 2 x . u(0)  2  C.  u( x)  2e 2 x ( x  1)

12. For u  xu  e  x , F( x)  e  2

x

2 dx

xdx

 e2x .

2

 ex / 2 .

2 d x2 / 2 (e u)  e  x / 2 so dx

x

e x / 2 u( x)   e t / 2 dt  c.  u( x)  e  x / 2  e t / 2 dt  ce  x / 2 2

2

2

1

u(1)  0  ce

1 / 2

2

2

1

.  c  0.

13. For u  u  x , F( x)  e 

 u( x)  e

 x2 / 2

x

e 1

t2 / 2

dt

d x ( e u)  xe  x so e  x u   xe  x dx  C dx x  ( x  1)e  C.  u( x)  x  1  Ce x . u(0)  1  1  C.  C  2.  dx

 ex .

 u( x)  2e x  x  1

14. For u  2u  4, F( x)  e 2 x . u(0)  0  2  C.  C  2.

e 2 x u   4e 2 x dx  C. u( x)  2  Ce 2 x

u( x)  2e 2 x  2

15. u  u  9 is one of many equations.

16. The Maple input follows: > ode:=diff(u(x), x) = (exp(x)-u(x))/x: > dsolve({ode, u(1)=b}, u(x)); u( x )

e xe b x

In order for u(0) to exist, b = e – 1. > subs(b=exp(1)-1, %); u( x )

e x1 x

Here is an example of varying b slightly: > plot({(exp(x)-1)/x, (exp(x)-exp(1)+1.7)/x}, x=-2..2, > y=-3..3, axes=boxed, color=[red, blue], discont=true);

17. > ode:=diff(u(x), x) + u(x) = 2: > dsolve(ode, u(x)); 18. > ode:=diff(u(x), x) + 2*u(x) = 2*x: > dsolve(ode, u(x)); 19. > ode:=diff(u(x), x) + x*u(x) = 10: > dsolve(ode, u(x)); 20. > ode:=diff(u(x), x) – 2*u(x) = exp(x): > dsolve(ode, u(x));

21. > ode:=diff(u(x), x) + u(x) = x*exp(-x): > dsolve(ode, u(x)); 22. > ode:= diff(u(x), x) - u(x) = cos(x): > dsolve(ode, u(x)); 23. > ode:=x*diff(u(x), x) – 2*u(x) = x*exp(x): > dsolve(ode, u(x)); 24. > ode:=x^2*diff(u(x), x) - u(x) = 2*sin(1/x): > dsolve(ode, u(x)); 25. > ode:=diff(u(x), x) + 2*u(x) = 2*exp(-2*x): > dsolve({ode, u(0)=2}, u(x)); 26. > ode:=diff(u(x), x) + x*u(x) = exp(-x^2): > dsolve({ode, u(1)=0}, u(x)); 1 u( x )  2

 2 x 1  2 2 erf  2  2

27. > ode:=diff(u(x), x) - u(x) = x: > dsolve({ode, u(0)=1}, u(x)); 28. > ode:=diff(u(x), x) - 2*u(x) = 4: > dsolve({ode, u(0)=0}, u(x));



x2 

 2     2    e  erf  2 

Section 1.4 di di  10 5 i  12  10 4.   dt dt 12  104  105 i

1. Li   Ri  v. 104 i   10i  12 or

which gives  105 ln(12  104  105 i)  t  c. 12  04  105 i  e ( t c )10

5

If i(0)  0, 12  104  e 10 c .  c  1.17  10 4.  i(t)  1.2(1  e 10 t ) 5

5

At t  2  106 , i  1.2(1  e 0.2 )  0.2175 Amp di  2  10 4 i  200e 2t dt

2. Li   Ri  v. 103 i   20i  0.2e 2t or

F(t)  e 

2104 dt



 e 210 t .  i(t)  e 210 t  200e 2t e 210 t dt  C 4

4

4



4 4 200 20002t  e 210 t  e  C   0.01e 2t  Ce 210 t . i(0)  0  0.01  C  20002 

 C  0.01.



 i(t)  0.01 e 2t  e 210 t 4



5 dq dq 1 12  7 q or  2  105 q  sin 2t. F(t)  e 210 t dt 10 dt 50 5 5 12  q(t )  e 210 t   sin 2te 210 t dt  C  . This integral requires integration by parts  15  10 5 5  10  105 210 t twice:  sin 2te dt    cos 2t  sin 2t e 210 t .

2 2  

3. 12 sin 2t  50

5 5  12  105 1010  q(t )  sin 2t  cos 2t  Ce 20 t  12  1012 (cos 2t  105 sin 2t)  Ce 210 t .  50  2 2 

 i(t)  24  012 (sin 2t  105 cos 2t)  2  105 Ce 210 t 5

dq 6  5000q  . F(t)  e 5000t dt 100  q(t )  e 5000t  0.06e 5000t dt  C  12  106  Ce 5000t . q(0)  0.  C  12  10 6

4. 12  200q  q / 106 or





 q(t)  12  106 (1  e 5000t ). as t  , q  12  06. Let q  6  10 6 1 Then  1  e 5000t .  t  1.386  104 s 2

5. The mass of salt in solution remains constant at 0.2  0  2 kg of salt. Hence, dC 0.1 2  C(t )V (t ). V (t )  10  0.1t.   C  0 since C1  0 for fresh water. dt 20 dC C   0.005dt.  ln C  0.005t  ln K.   e 0.005t . At t  t f  100, C  0.1 C K 0.1   e 0.5 .  K  0.1649 and C(t)  0.1649e 0.005t for t  t f K

6.

dC 10 / 60 10 / 60  C C1 . To find C1 : CO2 entering  0.0016  18  300  0.04 dt 1500 1500 0.3456  0.3456 m 3 /min. Air entering = 10 m 3 /min.  C1  a 10 dC dC  1.111  104 ( a  C) or  1.111  10 4 dt.  ln(C  a)  1.111  10 4 t  ln K dt a C 4 or C(t)  Ke 1.11110 t  a. At t  0, C(0)  0.004.  K  0.03056 4

 C(t)  0.03456  0.03056e 1.11110 t 7. Exiting velocity v  2 gh . Flow out q  Av  9  106 2 gh . V   D  2h and V (t) 

 3

h3 . V  qt or

1 h D2 / 4.

3

dV  dh  q.  3h2   2 gh 9  10 6 dt 3 dt

2 5/ 2 h  9 2 g  10 6 t  K. h(0)  0.15 5 6  K  0.00349.  h(t)  (99.8  10 t  0.00873). Let h  0.  t  87.5 s h

3/ 2

dh  9 2 g  106 dt so that

h

8. v( y)  2 gy . The flow rate out q   v 0

V  9h. 

h 6 6 dy   2 gy dy  0.005657 h3 / 2 0 1000 1000

dV dh dh  9 . 9  0.005657 h3 / 2 or h3 / 2 dh  6.29  0 4 dt dt dt dt

2h1/ 2  6.29  104 t  K. h(0)  2.  K  1.414.  h(t)  (3.143 10 4 t  .707)2

For h  1, 1  (3.143  104 t  0.707)2 .

 t  932 s

dV .  Eq. is OK. dt dV dV a) 100  100  9.81  0.01V  0.01(V  9.81  10 4 ).   10 4 dt 4 dt V  9.81  10 4 4 4 4  ln(V  9.81  10 )  10 t  ln K or V (t)  9.81  10 (1  e 10 t ).

9. Fy  may . Sum forces in vertical direction:  D  Mg  M

4

Let V  50 : 50  9.81  104 (1  e 10 t ).

 t  5.098 s

dV dV  981  0.004V 2 or  4  105 dt or 2 5 dt V  2.453  10 1 1 dV V  495.3     4  105 dt.   Ce 0.0396t .   V  495.3  V  495.3 V  495.3  990.6  .0396 t 495.3(1  e ) V (0)  0.  C  1.  V (t)  . If V  50 t  5.116 s  .0396 t 1 e

b) 100

0  dr gR2 gR2 dv dv dr   2 or v   2 .  vdv   gR2 2 .  vdv   gR2  2 v0 R r dt dr r r r 2  v 1  0   gR2 r 1 R  gR2  0   .  v0  2  9.81  6 400 000  11210 m/s 2 R 

10.





11. With insulated sides (no heat transfer) the heat flux is constant at each x-location. T 1200 dT dT  10  kA  100 6 .   83.33 and T( x)  83.33 x  C x dx 10 dx T(2)  50  83.33  2  C. C  217 and T( x)  83.33x  217 12.

dT dT dT   A(T  T )  0.02  0.2(T  5).   0.004(T  5).  0.004dt dt dt T 5 Then ln(T  5)  0.004t  ln C. T(t)  5  Ce 0.004t . T(0)  80  5  C. C  75 T(t)  5  75e 0.004t . 8  5  75e 0.004t .

13.

 t  805 s

de   e.  ln e   t  ln C.  e(t )  Ce  t . e(0)  e 0  C.  e(t )  e0 e  t dt e e(20)  0  e0 e 20 .   0.0347. 0.05e0  e0 e .0347 t .  t  86.3 min 2

14. The Maple input follows: a) > ode:= diff(y(x), x)=x*y(x)^2; > dfieldplot(ode, y(x), x=-3..1, y=-1..4, arrows=MEDIUM); b) > dsolve({ode, y(-2)=3}, y(x)); y( x )

2 14 x2 3

c) > DEplot(ode, y(x), x=-3..1, [[y(-2)=3]], y=-1..4, arrows=MEDIUM); d.) > ode2:= diff(y(x), x)=cos(x)/y(x); > dfieldplot(ode2, y(x), x=-3..1, y=-2..2, arrows=MEDIUM); > dsolve({ode2, y(-Pi/2)=1}, y(x)); y( x ) 2 sin( x )3

> DEplot(ode2, y(x), x=-3..1, [[y(-Pi/2)=1]], y=-2..2, arrows=MEDIUM);

15. > ode:= 10^(-4)*diff(i(t), t) + 10*i(t)=12: > dfieldplot(ode, i(t), t=0..1, i=0..2, arrows=LARGE);

16. > ode:= 10^(-3)*diff(i(t), t) + 20*i(t)=0.2*exp(2*t): > dfieldplot(ode, i(t), t=0..2, i=0..0.2, arrows=LARGE); 17. > ode:= 50*diff(q(t), t) + 10^7*q(t)=12*sin(2*t): > dfieldplot(ode, q(t), t=0..6, q=-0.000005..0.000005, arrows=LARGE); 18. > ode:= 200*diff(q(t), t) + 10^6*q(t)=12: > dfieldplot(ode, q(t), t=0..6, q=0..0.0001, arrows=LARGE); 19. > ode:= diff(C(t), t) + 0.1/20*C(t)=0: > dfieldplot(ode, C(t), t=100..120, C=0.009..0.011, arrows=LARGE); 20. > ode:= diff(C(t), t) + 10/1500*C(t)=.3456/1500: > dfieldplot(ode, C(t), t=0..2, C=0.0399..0.0401, arrows=LARGE);

Section 1.5.1 1. A discontinuity at 0, not a jump because lim(ln x)   .  x0

2. A discontinuity at 0, not a jump because lim(ln x)   .  x0

3. A jump discontinuity.

1 4. Not a jump discontinuity because lim  2    . x0  x 

5. No discontinuity at x = 0.

sin x  sin x  6. Not a jump discontinuity because lim   1  lim   . x0  x  x0  x  x 0 x 0

7. Not a discontinuity because lim  x sin  / x   0 . x0

8. Not sectionally continuous because lim  ln x    . x0

9. This function is continuous in each interval [1, M].

10. Not sectionally continuous because the singularity at x = 0 is not a jump discontinuity.

11. This is sectionally continuous.

12. Not sectionally continuous because the singularity at x = 0 is not a jump discontinuity.

13. This is sectionally continuous because the discontinuities at 0 and 1 are jump discontinuities.

14. Suppose p(x) is continuous and g(x) is sectionally continuous in a < x < b and a < x0 <b.

Then, there is one and only one solution to the initial-value problem: du  p( x)u  g( x), u( x0 )  u0 x  a, x0  b dx 15. In standard form, the equation is

du 2  u  0 and p(x) is not continuous at x = 0. dx x

16. > f6:= x -> piecewise(x < 0, sin(x)/x, x=0, 0, x > 0, sin(x)/x); sin( x ) sin( x )  f6 := x piecewise x0, , x0, 0, 0x,  x x   > plot(f6(x), x=-3..3); 17. > f7:= x -> piecewise(x < 0, x*sin(Pi/x), x=0, 0, x > 0, x*sin(Pi/x)): 18. > f10:= x -> piecewise(x < 0, 1/x, x=0, 0, x > 0, 1/x): 19. > f11:= x -> piecewise(x < 0, -1, x=0, 0, x > 0, 1):

20. > f12:= x -> piecewise(x < 0, 0, x>=0, abs(sin(x))): 21. > f13:= x -> piecewise(x >= 0 and x <=1, 1):

Section 1.5.2 3. L[c1u  c2 v]  L[c1u]  L[c2 v]  c1 L[u]  c2 L[v] 4. L[u  v]  L[u]  L[v]  g1  g2 5. L[c1u  c2 v]  c1 L[u]  c2 L[v]  c1 g  c2 g  (c1  c2 )g  g

Section 1.5.3 1. p0 ( x)   .  W ( x)  Ke 

  dx

2. p0 ( x) 

 Ke  x

1  dx / x .  W ( x)  Ke   Ke  ln x  Ke ln1/ x  K / x x

1 dx / x 3. p0 ( x)   .  W ( x)  Ke   Ke ln x  Kx x

4. p0 ( x)  0.  W( x)  K 5. u1 ( x)  e  x , p0  2 .  W  Ke 2 x  e  x u2   e  x u2 . Thus, u2   u2  Ke  x d  ( e x u2 )  K and e x u2  Kx  C. Thus, u2 ( x)  Kxe  x  Ce  x dx Therefore, a second solution is u2  xe  x 8.

u1  u2 u1  u2

u1  u2  u1u1  u1u2  u2 u1  u2 u2  u1u1  u1u2  u2 u1  u2 u2 u1  u2  2(u2 u1  u1u2 )  2

u1 u1

u2 u2

u1  u2 u1  u1u1  u1u2  u1u1  u2 u1  u1u2  u2 u1  u1  u2 u1

9.

u1 u1

u2 u2

10.

 u1   u2  u1   u2   u1u1   u1u2   u2 u1   u2 u2   u1u1   u2 u1  u1   u2  u1   u2  u1u2   u2 u2   (u1u2  u2 u1 )   (u2 u1  u1u2 )  (   )

    0

11.

u1 u1

u1 u1

u2  (   )W u2

ku1  ku1u1  ku1u1  0 ku1

Section 1.6  u( x)  c1e 3x  c2 e 2 x

1. m2  m  6  (m  3)(m  2)  0.  m  3,  2.

 u( x)  c1e 3x  c2 e 3x

2. m2  9  (m  3)(m  3)  0.  m  3,  3.

 u( x)  c1e 3ix  c2 e 3ix

3. m2  9  ( m  3i)(m  3i)  0.  m  3i ,  3i.

4. 4m2  1  (2m  i)(2m  i)  0.  m  i / 2,  i / 2.

5. m2  4m  4  (m  2)2  0.  m  2, 2.

6. m2  4m  4  (m  2)2  0.  m  2,  2.

 u( x)  c1e ix / 2  c2 e ix / 2

 u( x)  c1e 2 x  c2 xe 2 x  u( x)  c1e 2 x  c2 xe 2 x

7. m2  4m  4  0.  m  2  2 2.

 u( x)  c1 e(22 2 ) x  c2 e(22 2 ) x

8. m2  4m  4  0.  m  2  2 2.

 u( x)  c1 e (22 2 ) x  c2 e (22 2 ) x

9. m2  4  0.  m  2.

 u( x)  c1e 2 x  c2 e 2 x

10. m2  4m  8  0.  m  2  2i.

11. m2  2m  10  0.  m  1  3i.

 u( x)  c1e(22i ) x  c2 e(22i ) x  u( x)  c1e (13i ) x  c2 e (13i ) x

3 12. 2m  6m  5  0.  m    i.

2 2

3 3 (  i ) x (  i ) x 2 2  u( x)  c1 e  c2 e

1 5 5 x x  x 13. From No.1, u( x)  e 2  c1 e 2  c2 e 2     

14. From No.2, a  0 so u( x)  c1 e 3 x  c2 e 3 x 15. From No.3, a  0 so u( x)  c1 cos 3x  c2 sin 3x 16. From No.4, a  0 so u( x)  c1 cos x / 2  c2 sin x / 2



17. From No.7, u( x)  e 2 x c1 e 2 2x  c2 e 2 2x





18. From No.8, u( x)  e 2 x c1 e 2 2x  c2 e 2 2x



19. From No.9, u( x)  c1 e 2 x  c2 e 2 x 20. From No.10, u( x)  e 2 x  c1 cos 2x  c2 sin 2x  21. From No.11, u( x)  e  x  c1 cos 3x  c2 sin 3x 

22. m2  5m  3  0, m 

5 13  .

2 2

23. m2  5m  6  0, a  1, b  6.

13 13   x  x  u( x)  e 5 x / 2  c1 e 2  c2 e 2     

5  u( x)  c1 e x / 2 sinh  x  c2  2 

24. m2  9  0, a  0, b  9.

 u( x)  c1 sinh  3x  c2 

25. m2  9  0, a  0, b  9.

 u( x)  c1 cos  3x  c2 

26. m2 

1 1  0, a  0, b  .

4 4

 u( x)  c1 cos  x / 2  c2 

27. m2  4m  4  0, a  4, b  4.



 u( x)  c1 e 2 x sinh 2 2 x  c2



28. m2  4m  4  0, a  4, b  4.

29. m2  4  0, a  0, b  4.



 u( x)  c1e 2 x sinh 2 2 x  c2



 u( x)  c1 sinh  2x  c2 

30. m2  4m  8  0, a  4, b  8.

 u( x)  c1e 2 x cos  2x  c2 

31. m2  2m  5  0, a  2, b  5.

 u( x)  c1e  x cos  2x  c2 

32. m2  5m  3  0, a  5, b  3.

 u( x)  c1e 2.5 x cos

 13x / 2  c  2

33. m2  9  0, a  0, b  9.

 u( x)  c1 cos 3x  c2 sin 3x. u(0)  0  c1 1 u(0)  1  3c2 .  c2  1/ 3 and u( x)  sin 3x 3

34. m2  5m  6  0, a  5, b  6. u(0)  0  2c1  3c2 .

 c1  6, c2  4

35. m2  4m  4  0, a  4, b  4. u(0)  2  c1  2c2 .

 u( x)  c1e 2 x  c2 e 3x . u(0)  2  c1  c2

and

u( x)  6e 2 x  4e 3x

 u( x)  (c1 x  c2 )e 2 x . u(0)  0  c2

 c1  2, c2  0

and

u( x)  2 xe 2 x

36. m2  4, a  0, b  4.  u( x)  c1e 2 x  c2 e 2 x . u(0)  2  c1  c2 5 3 u(0)  1  2c1  2c2 .  c1  5 / 4, c2  3 / 4 and u( x)  e 2 x  e 2 x 4 4 37. See No.33: u( x)  c1 cos(3x  c2 ). u(0)  c1 cos c2 . u(0)  1  3c1 sin c2 1  c2   / 2, c1  1/ 3 and u( x)   cos(3x   / 2) 3 x 38. See No.34: u( x)  c1 e 5 x / 2 sinh(  c2 ). u(0)  2  c1 sinh c2 .

2 5 1 u(0)  0   sinh c2  c1 cosh c2 .  10  c1 cosh c2 and tanh c2  .

2 5 x  c1  9.8, c2  0.2027 and u( x)  9.8 sinh( 2  0.2027)

39. See No.36: u( x) 

5 2 x 3 2 x e  e 4 4

40. See No.33: u( x) 

1 sin 3x 3

41. See No.38: u( x)  e 5x / 2 ( A sinh x / 2  B cosh x / 2). u(0)  2  B, 5 u(0)  0   B  2 A.  A  10 and u( x)  2e 5 x / 2 (5sinh x / 2  cosh x / 2) 2 42. See No.36: u( x) 

5 2 x 3 2 x e  e 4 4

44. m3  1  ( m  1)( m2  m  1).  m  1, Hence, u( x)  c1 e  e x

x / 2

1  i 3 2

,

1  i 3 2

(c2 cos 3x / 2  c3 sin 3x / 2)

45. m3  2m2  m  2  0.  m  1,  1, 2.  u( x)  c1e x  c2 e  x  c3 e 2 x 46. m4  m2  0.  m  0, 0, 1,  1.  u( x)  c1  c2 x  c3 e x  c4 e  x 47. m4  1  0.  m  1,  1, i , -i.  u( x)  c1e x  c2 e  x  c3 cos x  c4 sin x 48. m4  m3  0.  m  0, 0, 0, 1.  u( x)  c1  c2 x  c3 x2  c4 e x 49. > ode:= diff(u(x), x$2) - 4*u(x) = 0: > dsolve(ode, u(x)); 50. > ode:= diff(u(x), x$2) - 4*diff(u(x), x) + 8*u(x) = 0: > dsolve(ode, u(x)); 51. > ode:= diff(u(x), x$2) - 2*diff(u(x), x) + 5*u(x) = 0: > dsolve(ode, u(x)); 52. > ode:= diff(u(x), x$2) + 5*diff(u(x), x) + 3*u(x) = 0: > dsolve(ode, u(x)); 53. > ode:= diff(u(x), x$2) + 9*u(x) = 0: > dsolve({ode, u(0)=0, D(u)(0)=1}, u(x));

54. > ode:= diff(u(x), x$2) + 5*diff(u(x), x) + 6*u(x) = 0: > dsolve({ode, u(0)=2, D(u)(0)=0}, u(x)); 55. > ode:= diff(u(x), x$2) + 4*diff(u(x), x) + 4*u(x) = 0: > dsolve({ode, u(0)=0, D(u)(0)=2}, u(x)); 56. > ode:= diff(u(x), x$2) - 4*u(x) = 0: > dsolve({ode, u(0)=2, D(u)(0)=1}, u(x)); 57. See Problem 53.

58. > ode:= diff(u(x), x$3) - u(x) = 0: > dsolve(ode, u(x)); 59. > ode:= diff(u(x), x$3) - 2*diff(u(x), x$2) - diff(u(x), x) + 2*u(x) = 0: > dsolve(ode, u(x)); 60. > ode:= diff(u(x), x$4) - diff(u(x), x$2) = 0: > dsolve(ode, u(x)); 61. > ode:= diff(u(x), x$4) - u(x) = 0: > dsolve(ode, u(x)); 62. > ode:= diff(u(x), x$4) - diff(u(x), x$3) = 0: > dsolve(ode, u(x));

Section 1.7.1 1.  Ft  matan .  W sin  

WL d2 s g dt 2

Small angles: sin    . L  s. W  

WL d2  g dt 2

L tan s W

g g d2  g    0. m2   . m   i 2 L L L dt   (t)  A cos( g / L t)  B sin( g / L t) d2 s d2 s .

10  10s  0. m2  1.  m1  i , m2  i 2 2 dt dt  s(t)  A cos t  B sin t. s(0)  10  A, s(0)  0  B. s(t)  10cos t

2.  F  ma.  Ks  m

3. 4 y  100 y  0. 4m2  100.  m  5i.  y(t )  A cos 5t  B sin 5t f 

5  0.7928 hertz 2

4.  F  ma.  Ky  my. 0.03y  0.5y  0. m2  16.67.  m  4.08i  y(t)  A cos 4.08t  B sin 4.08t.

4.08t  2 .

 tcycle  1.54 s

5. 4 y  100 y  0. m2  25. m  5i y(t)  A cos 5t  B sin 5t 1 1 a) y(0)   A. y(0)  0  B.  y(t)  cos 5t 2 2 b) y(0)  0  A. y(0)  10  5B.  y(t)  2sin 5t

(b) 2 /5

t (a)

6. 2 y  50 y  0. m2  25. m  5i.  y(t )  A cos 5t  B sin 5t y(0)  2  A, y(0)  10  5B.  y(t)  2(cos 5t  sin 5t)

 y(t)  A cos 5t  B sin 5t 7. 2 y  50 y  0. y(0)  0  A, y(0)  10  5B.  y(t)  2sin 5t M=2

M = 10 2 /5

10 y  50 y  0. m  2.236i  y(t)  B sin 2.236t y(0)  10  2.236B.  y(t)  4.472sin 2.236t

t

Section 1.7.2 1. 4 y  32 y  64 y  0. m2  8m  16  0. m1  4, m2  4.  y(t)  ( A  Bte 4t )

y(0)  1  A. y(0)  0  4 A  B.  B  4  y(t)  e 4t (1  4t) 1 y( 2 )  0.406,

y(1)  0.09, y(2)  0.003

y 1 t 0.5

1

2. 4 y  40 y  64 y  0. m2  10m  16  0. m1  8, m2  2.  y(t)  Ae 8t  Be 2t 25 25 y(0)  0  A  B, y(0)  50  8 A  2 B.  A   B  .  y(t )  ( e 8t  e 2t ) 3 3 50 y  17.7 y  50 y  0. m1  1.735  5.215i , m2  1.735  5.215i 9.8  y(t)  ( A cos 5.215t  B sin 5.215t)e 1.735t

3. a)

y(t)  (5.215sin 5.215t  5.215cos 5.215t)e 1.735t 1.735e 1.735t ( A cos 5.215t  B sin 5.215t)

y(0)  2  A, y(0)  0  5.215B  1.735 A .  B  0.6654  y(t)  (2 cos 5.215t  0.6654 sin 5.215t)e 1.735t

b)

50 y  40 y  50 y  0. m1  1.56, m2  6.29.  y(t)  Ae 1.56t  Be 6.29t 9.8 y(0)  2  A  B  1.56 t  0.66e 6.29t   A  2.66,  0.66 and y(t)  2.66e y(0)  0  1.56 A  6.29 B

4. 4 y  20 y  64 y  0.  m  2.5  3.12i and y(t)  e 2.5t ( A cos 3.12t  B sin 3.12t)

y(0)  1  A, y(0)  0  3.12B  2.5 A.  B  0.801  y(t)  e 2.5t (cos 3.12t  0.801sin 3.12t)

y 1 0.72 1.73

t

5. 2 y  32 y  100 y  0.  m  8  3.742 and y(t)  e 8t ( Ae 3.742t  Be 3.742t ) y(0)  0  A  B, y(0)  10  8( A  B)  3.742 A  3.742B.  A  B  1.336  y(t)  1.336e 8t (e 3.742t  e 3.742t )

6. y(t)  e (C / 2 M )t ( Aet  Be t ). Recall  sinh(   )  (sinh  cosh   cosh  sinh  )    (cosh   sinh  )e  (sinh   cosh  )e  . Let   t and 2 2  (cosh   sinh  )e  A   2  . If A and B are known find and Let  c1    (sinh   cosh  )e  B  2            and c2. Then we can write  C 2  4 MK  y(t )  c1 e (C / 2 M )t sinh(t  c2 )  c1 e (C / 2 M )t sinh  t  c2    2M  

7. y(t)  1.336(e 4.258t  e 11.742t ).  y(t)  5.689e 4.258t  15.69e 11.742t Set y  0 and solve for t. This can be done by trial and error. Try t = 0.1: y  1.13.

Try t = 0.12: y  0.42. Try t = 0.13: y  0.14. Try t = 0.14: y  0.10.  t  0.136 s It would be nice to have an expression to find t at ymax. See No.8.

8. From No.6, if y(0) = 0, then c2 = 0.  y(t)  c1e (C / 2 M )t sinh t. Then set C ( C / 2 M ) t m y(tm )  0  c1 e (C / 2 M )tm  cosh tm  c1 e sinh tm where tm is at ymax.

2M  tanh tm 

2M 1 2M 2M C 2  4KM .  tm  tanh 1  tanh 1 C  C C C 2  4KM

9. Substitute t = 0.136 s back into the expression for y(t): ymax  1.336(e 4.2580.136  e 11.7420.136 )  0.478 m

10. For curve 1, y(0)  0.  c2  0. From the solution to No.8:

y(0)  v0  c1.  c1  v0 / . Using tmax from No.8 we have: ymax 

v0 (C / 2 M ) tanh1 (2 M / C ) e sinh(tanh 1 (2M / C ), 



C 2  4KM 2M

11. Using the results of Example 1.7.1 we have:  1  1 2 t1   tan 1 (C / 2 M), t2   tan 1 (C / 2 M)       2 2  2 M 2  2  30  t2  t1     0.9734 s  4KM  C 2 4  2000  30  3002 12. y(t)  2e t sin t.  C  2 M , 

C  CC

D

4KM  4 M 2  1.  K  2 M. D  2M

 0.707.  C  70.7% of CC

2

4(2 M ) M  4 M 2

 2



D  4 ymax occurs at y  0,  tan tm  1.  tm   / 4 s  2

2  M

t /4

  5y  30 y  100 y  0.  m  3  3.32i.  y(t)  c1e m1t  c2 e m2t . y(0)  1  c1  c2

y(0)  0  m1c1  m2 c2 .  c1  0.5  0.452i , c2  0.5  0.452i  y(t)  (0.5  0.452i)e (33.32 i )t  (0.5  0.452i)e (33.32 i )t

Now, let y(t)  e 3t ( A cos 3.32t  B sin 3.32t). y(0)  1  A. y(0)  0  3 A  3.32 B

 B  0.904 and y(t)  e 3t (cos 3.32t  0.904 sin 3.32t)

Finally, y(t)  Ae 3t cos(3.32t   ). y(0)  1  A cos  . y(0)  0  3 A cos   3.32 A sin   A  1.35,   0.737 rad and y(t)  1.35e 3t cos(3.32t  0.737)

14. > ode:= 3*diff(y(t), t$2) + 2*diff(y(t), t) + 170/3*y(t)=0: > dsolve({ode, y(0)=-10, D(y)(0)=0.24}, y(t)); 

t



t

   232   3   13 t  13 t   3 y( t ) e sin cos  10 e  325  3   3 

> sol:=rhs(%): > plot(sol, t=0..4); 15. > s19:=-10*exp(-1/3*t)*sin(13*t/3+1.5); Since the sine function varies between –1 and 1, it is sufficient to solve 10e t / 3  0.002 .

The spring will appear to be at rest after 25.5 sec.

Section 1.7.3 1. a) Li   Ri   i / C  0. 103 i   Ri   i / 2  10 5  0. m  If R2 < 200 then oscillatory. Rcrit = 14.14 ohms

 R  R2  200 0.002

b) For a parallel circuit: 2  105 v  v / R  v / 103  0. m 

2.

in in 2



4 L / C  R2  2L

0.02  2  e R / L where   0.01  R2104 4

 2  e 400 R 10 . ln 2 

R2  R2  8  102 . R2  0.08.  Rcrit  3.54 4  105

2 R

4  10 4 / 10 6  R2 400  R2  2  10 4 2  10 4

 R  2.19 ohms 400  R 2 20  400  800  10 5  10 5 j (here j  1) 3. 104 i   20i   6 i  0.  m  10 2  10 4 2

2

or 40 R2  192.

 i(t)  e 10 t ( A cos105 t  B sin105 t). i(0)  10  A, i (0)  0  10 5 A  10 5 B.  B  10 5

 i(t)  10(cos105 t  sin105 t)e 10 t 5

d2 .  T(t )  c   k  I . The equation that describes the oscillations of a circular shaft is dt 2 then I   c   k  T(t ) . Thus, I  M , c  C, k K

4. T  I

Section 1.8 1. Let up ( x)  Ax  B. Then 0  2( Ax  B)  2x.  A  1, B  0.

 up ( x)  x

1 2. Let up ( x)  Ax  B. Then 0  A  2( Ax  B)  2 x.  A  1, B   .

2

3. Let up ( x)  Ae  x . Then Ae  x  Ae  x  e  x .  2 A  1.

 up ( x) 

 up ( x)  x 

1 2

1 x e 2

4. Let up ( x)  Axe  x since ex is a solution to the homogeneous eqn. Then

up  Ae x  Axe x , up  2 Ae x  Axe x .  2 Ae x  Axe x  Axe x  Axe x  e x 1  2 A  1 and up ( x)  xe x 2 5 5. Let up ( x)  A sin x. Then  A sin x  10 A sin x  5sin x.  9 A  5 and up ( x)  sin x 9

6. Let up ( x)  Ax sin 3x. up  A sin 3x  3 Ax cos 3x. up  6 A cos 3x  9 Ax sin 3x 1 1  6 A cos 3x  9 Ax sin 3x  9 Ax sin 3x  cos 3x. A  and up ( x)  x sin 3x 6 6 7. The characteristic eqn of the homogeneous eqn is m2  4m  4. Hence, m = 2 is a double root and e 2 x and xe 2 x are solutions. For the particular solution we assume up  Ax2 e 2 x .

Then up  2 Axe 2 x  2 Ax2 e 2 x , up  2 Ae 2 x  8 Axe 2 x  4 Ax2 e 2 x  2 Ae 2 x  8 Axe 2 x  4 Ax2 e 2 x  8 Axe 2 x  4 Ax2 e 2 x  e 2 x .  A 

1 1 and up ( x)  x 2 e 2 x 2 2

8. Since sin 3x is a solution to the homogeneous eqn, we assume

up ( x)  Ax2  Bx  C  Dx cos 3x Then substitute into the given eqn: 2 A  6D sin 3x  9Dx cos 3x  9 Ax2  9Bx  9C  9Dx cos 3x  x2  sin 3x 1 1  9 A  1, 9 B  0, 9C  2 A  0,  6 D  1.  up ( x)  (9 x 2  2)  x cos 3x 81 6

9. m2  1  0. m  i.  uh ( x)  A cos x  B sin x. Let up ( x)  Ce 2 x . 4Ce 2 x  Ce 2 x  e 2 x 1 1 C  and u( x)  A cos x  B sin x  e 2 x 5 5 10. m2  4m  4  0. m1  m2  2.  uh ( x)  c1 e 2 x  c2 xe 2 x . Let up ( x)  Ax2  Bx  C 2 A  8 Ax  4B  4 Ax2  4Bx  4C  x2  x  4.  4 A  1, 8 A  4B  1, 2 A  4B  4C  4

A 

1 1 9 1 1 9 , B   , C  .  u( x)  c1 e 2 x  c2 xe 2 x  x2  x  4 4 8 4 4 8

11. m2  9. m  3i.  uh ( x)  c1 cos 3x  c2 sin 3x. Let up ( x)  Ax2  Bx  C  D sin 2x 2 A  4D sin 2x  9 Ax2  9Bx  9C  9D sin 2x.  9 A  1, 9B  0, 2 A  9C  0, 5D  1 1 2 1 1 1  A  ,  0, C   , D  .  u( x)  c1 cos 3x  c2 sin 3x  (9 x 2  2)  sin 2 x 9 81 5 81 5

12. m2  4. m  2i.  uh ( x)  c1 cos 2x  c2 sin 2x. Let up ( x)  Ax cos 2x 1 4 A sin 2 x  4 Ax cos 2 x  4 Ax cos 2 x  sin 2 x. 4 A  1 and A   4 1  u( x)  c1 cos 2 x  c2 sin 2 x  x sin 2 x 4 13. m2  16  0. m  4.  uh ( x)  c1 e 4 x  c2 e 4 x . Let up ( x)  Axe 4 x 8 Ae 4 x  16 Axe 4 x  16 Axe 4 x  e 4 x . Let up ( x)  Axe 4 x . 8 Ae 4 x  16 Axe 4 x  16 Axe 4 x  e 4 x 1 1  8 A  1 and A  .  u( x)  c1 e 4 x  c2 e 4 x  xe 4 x 8 8

14. m2  5m  6  0. m1  2, m2  3.  uh ( x)  c1 e 2 x  c2 xe 3 x Let up ( x)  A sin 2 x  B cos 2 x. There results: 4 A sin 2 x  4 B cos 2 x  10 A cos 2 x  10 B sin 2 x  6 A sin 2 x  6 B cos 2 x  3 sin 2 x cos 2 x :  4 B  10 A  6 B  0   sin 2 x :  4 A  10 B  6 A  3   u( x)  c1 e 2 x  c2 e 3 x 

A 

3 15 , B 52 52

3 (sin 2 x  5 cos 2 x) 52

15. m1  m2  2.  uh ( x)  c1 e 2 x  c2 xe 2 x . Let up ( x)  Ax2  Bx  C 2 A  8 Ax  4 B  4 Ax 2  4 Bx  4C  x 2 .  4 A  1, 8 A  4 B  0, 2 A  4 B  4C  0 1 1 3 1 , B   , C  .  u( x)  c1 e 2 x  c2 xe 2 x  (2 x 2  4 x  3) 4 2 8 8 u(0)  0  c1  3 / 8  3 1 1 1  c1   , c2  .  u( x)  (2 x  3)e 2 x  (2 x 2  4 x  3)  u(0)  1/ 2  2c1  c2  1/ 2  8 4 8 8

A 

16. m  2i.  uh ( x)  c1 cos 2x  c2 sin 2x. Let up ( x)  A sin x.  A sin x  4 A sin x  2sin x 2 2 2 A  and u(x)=c1 cos 2 x  c2 sin 2 x  sin x. u(0)  1  c1 . u(0)  0  2c2  3 3 3 1 1 2  c2   and u( x)  cos 2 x  sin 2 x  sin x 3 3 3 17. m  2  i.  uh ( x)  e 2 x (c1 cos x  c2 sin x). Let up ( x)  Ax2  Bx  C 2 A  8 Ax  4B  5 Ax2  5Bx  5C  x2  5.

5 A  1, 8 A  5B  0, 2 A  4 B  5C  5

 A  0.2, B  0.32, C  1.176 and u( x)  e

2 x

(c1 cos x  c2 sin x)  0.2 x2  0.32 x  1.176

u(0)  0  c1  1.176, u(0)  0  2c1  c2  0.32.  c1  1.176, c2  2.032  u( x)  e 2 x (1.176 cos x  2.032 sin x)  0.2x2  0.32x  1.176

18. m  2i.  uh ( x)  c1 cos 2x  c2 sin 2x. Let up ( x)  Ax cos 2x 4 A sin 2x  4 Ax cos 2x  4 Ax cos 2x  2sin 2x 1 1  A  1/ 2. u( x)  c1c cos 2 x  c2 sin 2 x  x cos 2 x. u(0)  0  c1 . u(0)  0  2c2  2 2 1 1 1  c2  and u( x)  sin 2 x  x cos 2 x 4 4 2

19. m  3  i.  uh ( x)  e 3 x (c1 cos x  c2 sin x). Let up ( x)  A cos 2x  B sin 2x) 4 A cos 2x  4B sin 2x  12 A sin 2x  12B cos 2x  10 A cos 2x  10B sin 2x  cos 2x cos 2 x :  4 A  12 B  10 A  1 1 1 , B  A sin 2 x :  4 B  12 A  10 B  0  30 15 u( x)  e 3 x (c1 cos x  c2 sin x) 

 c1  

1 71 , c2   .

30 30

1 1 4 (cos 2 x  2 sin 2 x). u(0)  0  c1  . u(0)  0  3c1  c2  30 30 30

 u( x)  

1 3 x [e (cos x  7 sin x)  cos 2 x  2 sin 2 x] 30

20. m  4.  uh ( x)  c1 e 4 x  c2 e 4 x . Let up ( x)  Axe 4 x . 8 Ae 4 x  16 Axe 4 x  16 Axe 4 x  2e 4 x 1 1 1  8 A  2 and A  .  u( x)  c1 e 4 x  c2 e 4 x  e 4 x . u(0)  0  c1  c2 . u(0)  0  4c1  4c2  4 4 4

22. If k is a root, c1 e kx is a solution to u  au  bu  0. Then Ae kx is also a solution and

up ( x)  Ae kx results in 0 = g(x).  Ae kx is no good.

23. If k = 0, then b = 0. Then n(n  1) An xn2  (n  1)(n  2) An1 xn2  n1

 2 A2

a(nAn x   2 A2 x  a1 )  c0  c1   cn x . There is no x on the left hand side.

Therefore, no solution of the proposed form exists. n

n

25. > ode:= diff(u(x), x$2) + 4*diff(u(x), x) + 4*u(x) = x^2: > dsolve({ode, u(0)=0, D(u)(0)=1/2}, u(x)); 26. > ode:= diff(u(x), x$2) + 4*u(x) = 2*sin(x): > dsolve({ode, u(0)=1, D(u)(0)=0}, u(x)); 27. > ode:= diff(u(x), x$2) + 4*diff(u(x), x) - 5*u(x) = x^2 + 5: > dsolve({ode, u(0)=0, D(u)(0)=0}, u(x)); 28. > ode:= diff(u(x), x$2) + 4*u(x) = 2*sin(2*x): > dsolve({ode, u(0)=0, D(u)(0)=0}, u(x)); 29. > ode:= diff(u(x), x$2) + 7*diff(u(x), x) + 10*u(x) =cos(2*x): > dsolve({ode, u(0)=0, D(u)(0)=0}, u(x)); 30. > ode:= diff(u(x), x$2) - 16*u(x) = 2*exp(4*x): > dsolve({ode, u(0)=0, D(u)(0)=0}, u(x));

Section 1.9.1 and 1.9.2 1. Mm2  Cm  0. m1  0. m2  C / M.  yh (t )  c1  c2 e Ct / M . Because the y-term is missing, let yp (t)  At.  CA   Mg.  A   Mg / C.  y(t)  c1  c2 e Ct / M  Mgt / C d2 y Fy  M 2 .  Mg  Cy  My dt  My  Cy  Mg  0

Drag = Cdy/dt

W = Mg

2. y(t)  c1  c2 e 0.2t  49.1t. y(0)  0  c1  c2 . y(0)  100  0.2c2  49.1.  c1  746, c2  746  y(t)  746(1  e 0.2t )  49.1t. To find ymax : y(t)  0  0.2  746e 0.2t  49.1.  tm  5.557 s

 y(tm )  746(1  e 0.25.557 )  49.1  5.557  228 m

3. The time to rise to the maximum height is 5.557 s (see No.2). For a falling body, the drag force acts opposite the weight. Thus, My  Cy  Mg with y positive downward.

 yh (t)  c1  c2 eCt / M and yp (t)  Mgt / C.  y(t )  c1  c2 e Ct / M  Mgt / C . Measure y(t) from the maximum height position: y(0)  0  c1  c2 . y(0)  0  c2C / M  Mg / C.  c2  M 2 g / C 2  245 and c1  245.  y(t )  245(e 0.2t  1)  49.1t. From No.2, the

distance to fall is 288 m. Thus, set 228  245(e 0.2t  1)  49.1t. Trial and error: t = 8.8 s.

This compares to 5.557 s found in No.2 for the body to rise the same distance.

4. The solution is (see Nos. 2 & 3) y(t)  c1  c2 e 0.049t  200t where C / M  0.5 /(100 / 9.81) and Mg / C  100 / 0.5. y(0)  0  c1  c2 . y(0)  0  0.049c2  200.  c2  4080, c1  4080

 y(t)  4080(e 0.049t  1)  200t so that v  y  200e 0.049t  200. As t  , v  200 2 Let v  0.99  200  198  200e 0.049t  200.   e 0.049t . t  94 s 200

5. My  Ky  F0 cos t. Mm2  K  0. m   K / M i.  yh (t)  c1 cos 0 t  c2 sin 0 t where

0  K / M . Let yp (t)  A cos t.

 AM 2 c cos t  KA sin t  F0 cos t

 A(K  M 2 )  F0 and y(t )  c1 cos 0 t  c2 sin 0 t 

F0 cos t K  M 2

6. 2 y  32 y  0.1sin 4t. 2m2  32  0.  m  4i.  yh (t)  c1 cos 4t  c2 sin 4t.

Let y (t)  At cos 4t. 16 A sin 4t  32 At cos 4t  32 At cos 4t  0.1sin 4t. 16 A  0.1 p t 1 A  1/ 160.  y(t)  c1 cos 4t  c2 sin 4t  cos 4t. y(0)  0  c1 . y(0)  0  4c2  160 160 y  y(t)  1/ 160(sin 4t  4t cos 4t) 0.5 t Ignore the sine term (it’s at most 1) and let cos 4t  1 .

Then 0.5  4t / 640 and t  80 s 7. m2  9  0.  m  3i and yh (t)  c1 cos 3t  c2 sin 3t. Let yp (t)  A cos 2t. Then 8 8 4 A  9 A  8.  A  .  y(t )  c1 cos 3t  c2 sin 3t  cos 2t.

5 5 8 8 y(0)  0  c1  , y(0)  0  3c2 .  y(t)   (cos 3t  cos 2t) 5 5 8. m2  9  0.  m  3i and yh (t)  c1 cos 3t  c2 sin 3t. Let yp (t)  At sin 3t. Then 4 6 A cos 3t  9 At sin 3t  9 At sin 3t  8 cos 3t.  A  4 / 3. y(t)  c1 cos 3t  c2 sin 3t  t sin 3t 3

y(0)  0  c1 , y(0)  0  3c2 .

 y(t) 

4 t sin 3t 3

9. m2  16  0.  m  4i and yh (t)  c1 cos 4t  c2 sin 4t. Let yp (t)  At sin 4t. Then 1 1 8 A cos 3t  16 At cos 4t  16 At cos 4t  2 sin 4t.  A  . y(t)  c1 cos 4t  c2 sin 4t  t cos 4t 4 4

1 y(0)  2  c1 , y(0)  0  4c2  .

4

 y(t)  2 cos 4t 

1 t sin 4t  cos 4t 16 4

10. m2  16  0.  m  4i and yh (t)  c1 cos 4t  c2 sin 4t. Let yp (t)  A sin t. Then 2 2  A  10 A  2.  A  .  y(t )  c1 cos 4t  c2 sin 4t  sin t 15 15 2 37 2 y(0)  0  c1 , y(0)  10  4c2  .  y(t)  sin 4t  sin t 15 15 15 11. m2  25  0.  m  5i and yh (t)  c1 cos 5t  c2 sin 5t. Let yp (t)  At  B. Then 1 2 1 2 25 At  25B  t  2.  A  ,  and y(t)  c1 cos 5t  c2 sin 5t  t  25 25 25 25 2 1 1 y(0)  1  c1  , y(0)  4  5c2  .  y(t)  1.08 cos 5t  0 / 808 sin 5t  (t  2) 25 25 25 12. m2  1  0.  m  i and yh (t)  c1 cos t  c2 sin t. Let yp (t)  Ae t . Then Ae t  Ae t  2e t .  A  1.  y(t)  c1 cos t  c2 sin t  e t

y(0)  0  c1  1, y(0)  2  c2  1.

 y(t)  3sin t  cos t  e t

13. Use Li   Ri   i / C  v. Then 0.5i   i / 0.02  100sin 2t. m2  100  0.  m  10i  ih (t)  c1 cos10t  c2 sin10t. Let ip (t)  At cos10t.  20 A sin10t  100 At cos10t 100 At cos10t  100sin10t. 20 A  100.  i(t)  c1 cos10t  c2 sin10t  5t cos10t i(0)  0  c1 , i (0)  0  10c2  5.  i(t)  0.5sin10t  5t cos10t 14. Li   Ri   i / C  v. i   10000i  12000sin 2t. m2  10000  0.  m  100i  ih (t)  c1 cos100t  c2 sin100t. Let ip (t)  At cos100t.  200 A sin100t  10 000 At cos100t

10000 At cos100t  12000sin100t.  A  60.  i(t)  c1 cos10t  c2 sin10t  60t cos10t

i(0)  0  c1 , i (0)  0  100c2  60.

 i(t)  0.6sin100t  60t cos100t

15. Li   Ri   i / C  v. i   10000i  2400sin 2t. m2  10000  0.  m  100i  ih (t)  c1 cos100t  c2 sin100t. Let ip (t)  A cos10t.  10 A  1000 A  240.  A  0.242

i(t)  c1 cos100t  c2 sin100t  0.242cos10t. i(0)  0  c1  0.242, i (0)  0  10c2  i(t)  0.242(cos10t  cos100t)

16.

20 20 2 y  98 y  2 cos7t. m  98  0.  m  6.933i and yh (t)  c1 cos 0 t  c2 sin 0 t 9.81 9.81

Let yp (t)  A cos7t. Then  99.898 A  98 A  2.  A  1.054

y(t)  c1 cos6.933t  c2 sin6.933t  1.054cos7t. y(0)  0  c1  1.054, y(0)  0  0.6933c2

7  6.933  0.0335.   7 2 Therefore, using Eq. 1.9.19, y(t)  2.1sin 0.0335t sin7t. The beat frequency is 0.0335/2 which is 0.00533 hertz. The maximum amplitude occurs when sin0.0335t and sin7t are 1.  ymax  2.10 m

 y(t)  1.054(cos 6.933t  cos7t). Using Eq.1.9.18,  =

17. 0.1i   103 i  120cos101t. m  100i. ih (t)  c1 cos100t  c2 sin100t. Let ip (t )  A cos101t 0.1  1012 A  103 A  120.  A  5.79 and i( y)  c1 cos100t  c2 sin100t  5.79cos101t i(0)  0  c1  5.79, i (0)  0  100c2 .  i(t)  5.79(cos100t  cos101t)  11.94sin0.5t sin100t 101  100 1 0.5   .  frequency   0.0796 hertz and max amplitude  11.94 2 2 2

18. > ode:= diff(y(t), t$2) + 9*y(t) = 8*cos(2*t): > dsolve({ode, y(0)=0, D(y)(0)=0}, y(t)); 19. > ode:= diff(y(t), t$2) + 9*y(t) = 8*cos(3*t): > dsolve({ode, y(0)=0, D(y)(0)=0}, y(t)); 20. > ode:= diff(y(t), t$2) + 16*y(t) = -2*sin(4*t): > dsolve({ode, y(0)=2, D(y)(0)=0}, y(t)); 21. > ode:= diff(y(t), t$2) + 16*y(t) = 2*sin(t): > dsolve({ode, y(0)=0, D(y)(0)=10}, y(t)); 22. > ode:= diff(y(t), t$2) + 25*y(t) = t-2: > dsolve({ode, y(0)=1, D(y)(0)=4}, y(t)); 23. > ode:= diff(y(t), t$2) + y(t) = 2*exp(-t): > dsolve({ode, y(0)=0, D(y)(0)=2}, y(t));

Section 1.9.3 1. My  Ky  F0 sin t. Assume yp (t)  A sin t. Then we have  MA 2  KA  F0 A 

F0 K . Let 02  .

2 K  M M

Then A 

F0 M(02   2 )

2. Let yp (t)  At sin 0 t. We know by experience that Bt cos 0 t will not remain since B = 0 with a cos 0 t input function. Then 2 A0 cos 0 t  A02 t sin 0 t  02 At sin 0 t   2 A0 

F0 F0 and a  .

M 20 M

 y(t ) 

F0 cos 0 t M

F0 t sin 0 t 20 M

3. Let yp (t)  A sin 2t  B cos 2t. Then  4 AS  4BC  2 AC  2BS  4 AS  4BC  2S 4 A  2 B  4 A  2    B  1, A  0 and y(t)   cos 2t 4 B  2 A  4 B  0  where S  sin 2t and C  cos 2t. 

4. Let yp (t)  A sin 3t  B cos 3t. Then  9 AS  9BC  6 AC  6BS  AS  BC  C 9 A  6 B  A  0  S  sin 3t 3 2 1 , B and yp (t )  (3sin 3t  4 cos 3t) where  A  9 B  6 A  B  1  C  cos 3t 50 25 50

5. Let yp (t)  A sin t  B cos t. Then  AS  BC  AC  BS  AS  BC  2S  C S: C:

 A  B  A  2   B  2, A  1.  B  A  B  1

yp (t )  sin t  2 cos t

6. Let yp (t)  A sin 2t  B cos 2t. Then  4 AS  4BC  0.2 AC  0.2BS  2 AS  2BC  2S S: C:

 4 A  0.2 B  2 A  2    B  0.099, A  0.99.  4 B  0.2 A  2 B  1 

yp (t)  0.99 sin 2t  0.099 cos 2t

7. Let yp (t)  A1 sin t  B1 cos t  A2 sin 2t  B2 cos 2t. Then

 A1S1  B1C1  9 A2S3  9B2C3  2 A1C1  2B1S1  6 A2C3  6B2S3  5A1S1  5B1C1 5 A2S3  5B2C3  S1  2C3 .

S1 :  A1  2 B1  5 A1  1  A1  2 / 10    C1 :  B1  2 A1  5B1  0  B1  1/ 10  1 1   y p (t )  (2 sin t  cos t )  (2 cos 3t  3 sin 3t) S3 :  9 A2  6 B2  5 A2  0  A2  3 / 13  10 13   C3 :  9 B2  6 A2  5B2  2  B2  2 / 13 