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 1u x 2 1u 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 1u Thus, ln ln x ln C. ln ln Cx 2 1u 1u 1 u Kx2 1 Hence, (Cx)2 Kx2 . u( x) 1u 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 x2 1 1 x 1 1 Cx 2 Thus, ln ln C ln . u( x) Cx u 2 x2 2 2 x2 ln x2
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 vx 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 xK xK 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 u1 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)); 42 x2 1 u( x ) x 2 2 ( 2x2 ) 2x
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 x21 ), u( x )( x I 2 x21 ) 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 x21 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
ex .
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
x2 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
ex .
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 xe b x
In order for u(0) to exist, b = e – 1. > subs(b=exp(1)-1, %); u( x )
e x1 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. 104 i 10i 12 or
which gives 105 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 2 106 , i 1.2(1 e 0.2 ) 0.2175 Amp di 2 10 4 i 200e 2t dt
2. Li Ri v. 103 i 20i 0.2e 2t or
F(t) e
2104 dt
e 210 t . i(t) e 210 t 200e 2t e 210 t dt C 4
4
4
4 4 200 20002t e 210 t e C 0.01e 2t Ce 210 t . i(0) 0 0.01 C 20002
C 0.01.
i(t) 0.01 e 2t e 210 t 4
5 dq dq 1 12 7 q or 2 105 q sin 2t. F(t) e 210 t dt 10 dt 50 5 5 12 q(t ) e 210 t sin 2te 210 t dt C . This integral requires integration by parts 15 10 5 5 10 105 210 t twice: sin 2te dt cos 2t sin 2t e 210 t . 2 2
3. 12 sin 2t 50
5 5 12 105 1010 q(t ) sin 2t cos 2t Ce 20 t 12 1012 (cos 2t 105 sin 2t) Ce 210 t . 50 2 2
i(t) 24 012 (sin 2t 105 cos 2t) 2 105 Ce 210 t 5
dq 6 5000q . F(t) e 5000t dt 100 q(t ) e 5000t 0.06e 5000t dt C 12 106 Ce 5000t . q(0) 0. C 12 10 6
4. 12 200q q / 106 or
q(t) 12 106 (1 e 5000t ). as t , q 12 06. Let q 6 10 6 1 Then 1 e 5000t . t 1.386 104 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 1.111 104 ( 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 1.11110 t a. At t 0, C(0) 0.004. K 0.03056 4
C(t) 0.03456 0.03056e 1.11110 t 7. Exiting velocity v 2 gh . Flow out q Av 9 106 2 gh . V D 2h and V (t)
3
h3 . V qt 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 106 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 h3 / 2 dh 6.29 0 4 dt dt dt dt
2h1/ 2 6.29 104 t K. h(0) 2. K 1.414. h(t) (3.143 10 4 t .707)2
For h 1, 1 (3.143 104 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 4 105 dt or 2 5 dt V 2.453 10 1 1 dV V 495.3 4 105 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) . x0
2. A discontinuity at 0, not a jump because lim(ln x) . x0
3. A jump discontinuity. 1 4. Not a jump discontinuity because lim 2 . x0 x
5. No discontinuity at x = 0.
sin x sin x 6. Not a jump discontinuity because lim 1 lim . x0 x x0 x x 0 x 0
7. Not a discontinuity because lim x sin / x 0 . x0
8. Not sectionally continuous because lim ln x . x0
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 x0, , x0, 0, 0x, 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(22 2 ) x c2 e(22 2 ) x
8. m2 4m 4 0. m 2 2 2.
u( x) c1 e (22 2 ) x c2 e (22 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(22i ) x c2 e(22i ) x u( x) c1e (13i ) x c2 e (13i ) x