100% tevredenheidsgarantie Direct beschikbaar na je betaling Lees online óf als PDF Geen vaste maandelijkse kosten
logo-home
Solutions for Differential Equations with Boundary-Value Problems, 10th Edition Zill (All Chapters included) €28,93
In winkelwagen

Tentamen (uitwerkingen)

Solutions for Differential Equations with Boundary-Value Problems, 10th Edition Zill (All Chapters included)

 0 keer verkocht
  • Vak
  • Math
  • Instelling
  • Math

Complete Solutions Manual for Differential Equations with Boundary-Value Problems, 10th Edition by Dennis G. Zill ; ISBN13: 9780357760451. (Full Chapters included Chapter 1 to 15)....1. INTRODUCTION TO DIFFERENTIAL EQUATIONS. 2. FIRST-ORDER DIFFERENTIAL EQUATIONS. 3. MODELING WITH FIRST-ORDER DIF...

[Meer zien]

Voorbeeld 4 van de 1030  pagina's

  • 13 mei 2024
  • 1030
  • 2023/2024
  • Tentamen (uitwerkingen)
  • Vragen en antwoorden
  • Math
  • Math
avatar-seller
Differential Equations with
Boundary-Value Problems
10th Edition by Dennis G. Zill




Complete Chapter Solutions Manual
are included (Ch 1 to 15)




** Immediate Download
** Swift Response
** All Chapters included

,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With BOUNDARY VALUE PROBLEMS 2024, 9780357760451; Chapter #1:
Introduction to Differential Equations




Solution and Answer Guide
ZILL, DIFFERENTIAL EQUATIONS WITH BOUNDARY VALUE PROBLEMS 2024,
9780357760451; CHAPTER #1: INTRODUCTION TO DIFFERENTIAL EQUATIONS


TABLE OF CONTENTS
End of Section Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Exercises 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Exercises 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
Exercises 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Chapter 1 in Review Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30



END OF SECTION SOLUTIONS
EXERCISES 1.1
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
p
5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2
6. Second order; nonlinear because of R2
7. Third order; linear
8. Second order; nonlinear because of ẋ2
9. First order; nonlinear because of sin (dy/dx)
10. First order; linear
11. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear
in y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is
linear in x.
12. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is
linear in v . However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is
nonlinear in u.
13. From y = e−x/2 we obtain y ′ = − 12 e−x/2 . Then 2y ′ + y = −e−x/2 + e−x/2 = 0.




1

,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With BOUNDARY VALUE PROBLEMS 2024, 9780357760451; Chapter #1:
Introduction to Differential Equations


6 6 −20t
14. From y = − e we obtain dy/dt = 24e−20t , so that
5 5
 
dy −20t 6 6 −20t
+ 20y = 24e + 20 − e = 24.
dt 5 5

15. From y = e3x cos 2x we obtain y ′ = 3e3x cos 2x−2e3x sin 2x and y ′′ = 5e3x cos 2x−12e3x sin 2x,
so that y ′′ − 6y ′ + 13y = 0.
16. From y = − cos x ln(sec x + tan x) we obtain y ′ = −1 + sin x ln(sec x + tan x) and
y ′′ = tan x + cos x ln(sec x + tan x). Then y ′′ + y = tan x.
17. The domain of the function, found by solving x+2 ≥ 0, is [−2, ∞). From y ′ = 1+2(x+2)−1/2
we have

(y − x)y ′ = (y − x)[1 + (2(x + 2)−1/2 ]

= y − x + 2(y − x)(x + 2)−1/2

= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2

= y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8.

An interval of definition for the solution of the differential equation is (−2, ∞) because y ′ is
not defined at x = −2.
18. Since tan x is not defined for x = π/2 + nπ , n an integer, the domain of y = 5 tan 5x is
{x 5x 6= π/2 + nπ}
or {x x 6= π/10 + nπ/5}. From y ′ = 25 sec2 5x we have

y ′ = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 .

An interval of definition for the solution of the differential equation is (−π/10, π/10). An-
other interval is (π/10, 3π/10), and so on.
19. The domain of the function is {x 4 − x2 6= 0} or {x x 6= −2 or x 6= 2}. From y ′ =
2x/(4 − x2 )2 we have
 2
1

y = 2x = 2xy 2 .
4 − x2
An interval of definition for the solution of the differential equation is (−2, 2). Other inter-
vals are (−∞, −2) and (2, ∞).

20. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x 6= 0 or sin x 6= 1.
Thus, the domain is {x x =6 π/2 + 2nπ}. From y ′ = − 12 (1 − sin x)−3/2 (− cos x) we have

2y ′ = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x.

An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another
one is (5π/2, 9π/2), and so on.


2

, Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With BOUNDARY VALUE PROBLEMS 2024, 9780357760451; Chapter #1:
Introduction to Differential Equations




21. Writing ln(2X − 1) − ln(X − 1) = t and differentiating x

implicitly we obtain 4

2 dX 1 dX
− =1 2
2X − 1 dt X − 1 dt
 
2 1 dX t
− =1 –4 –2 2 4
2X − 1 X − 1 dt
–2
2X − 2 − 2X + 1 dX
=1
(2X − 1) (X − 1) dt
–4
dX
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X).
dt

Exponentiating both sides of the implicit solution we obtain

2X − 1
= et
X −1
2X − 1 = Xet − et

(et − 1) = (et − 2)X

et − 1
X= .
et − 2

Solving et − 2 = 0 we get t = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞).
The graph of the solution defined on (−∞, ln 2) is dashed, and the graph of the solution
defined on (ln 2, ∞) is solid.

22. Implicitly differentiating the solution, we obtain y

dy dy 4
−2x2 − 4xy + 2y =0
dx dx
2
−x2 dy − 2xy dx + y dy = 0
x
2xy dx + (x2 − y)dy = 0. –4 –2 2 4

–2
Using the quadratic formula to solve y 2 − 2x2 y − 1 = 0
√  √
for y , we get y = 2x2 ± 4x4 + 4 /2 = x2 ± x4 + 1 . –4

Thus, two explicit solutions are y1 = x2 + x4 + 1 and

y2 = x2 − x4 + 1 . Both solutions are defined on (−∞, ∞).
The graph of y1 (x) is solid and the graph of y2 is dashed.




3

Dit zijn jouw voordelen als je samenvattingen koopt bij Stuvia:

Bewezen kwaliteit door reviews

Bewezen kwaliteit door reviews

Studenten hebben al meer dan 850.000 samenvattingen beoordeeld. Zo weet jij zeker dat je de beste keuze maakt!

In een paar klikken geregeld

In een paar klikken geregeld

Geen gedoe — betaal gewoon eenmalig met iDeal, creditcard of je Stuvia-tegoed en je bent klaar. Geen abonnement nodig.

Direct to-the-point

Direct to-the-point

Studenten maken samenvattingen voor studenten. Dat betekent: actuele inhoud waar jij écht wat aan hebt. Geen overbodige details!

Veelgestelde vragen

Wat krijg ik als ik dit document koop?

Je krijgt een PDF, die direct beschikbaar is na je aankoop. Het gekochte document is altijd, overal en oneindig toegankelijk via je profiel.

Tevredenheidsgarantie: hoe werkt dat?

Onze tevredenheidsgarantie zorgt ervoor dat je altijd een studiedocument vindt dat goed bij je past. Je vult een formulier in en onze klantenservice regelt de rest.

Van wie koop ik deze samenvatting?

Stuvia is een marktplaats, je koop dit document dus niet van ons, maar van verkoper mizhouubcca. Stuvia faciliteert de betaling aan de verkoper.

Zit ik meteen vast aan een abonnement?

Nee, je koopt alleen deze samenvatting voor €28,93. Je zit daarna nergens aan vast.

Is Stuvia te vertrouwen?

4,6 sterren op Google & Trustpilot (+1000 reviews)

Afgelopen 30 dagen zijn er 69052 samenvattingen verkocht

Opgericht in 2010, al 15 jaar dé plek om samenvattingen te kopen

Begin nu gratis
€28,93
  • (0)
In winkelwagen
Toegevoegd