100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
Previously searched by you
Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill - All Chapters ( 1-9) Latest A++$12.99
Add to cart
Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill - All Chapters ( 1-9) Latest A++
9 views 0 purchase
Course
DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS
Institution
DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS
Book
A First Course in Differential Equations with Modeling Applications
Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill - All Chapters ( 1-9) Latest A++
Solution Manual - for A First Course in Differential Equations with Modeling Applications 12th Edition by Dennis Zill, All Chapters |Complete Guide A+
Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition Dennis G. Zill Completed ISBN:9780357760192 Newest Edition 2024 Instant Pdf Download
Solutions for A First Course in Differential Equations with Modeling Applications, 12th Edition Zill, All Chapters 1-9|Complete Guide A+
All for this textbook (6)
Written for
DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS
All documents for this subject (2)
Seller
Follow
TestsBanks
Reviews received
Content preview
Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
Solution and Answer Guide
ZILL, DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS 2024,
9780357760192; CHAPTER #1: INTRODUCTION TO DIFFERENTIAL EQUATIONS
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 MODELING APPLICATIONS 2024, 9780357760192; 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 MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
21. Writing ln(2X − 1) − ln(X − 1) = t and differentiating x
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
, Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
23. Differentiating P = c1 et / 1 + c1 et we obtain
dP 1 + c1 et c1 et − c1 et · c1 et c1 et 1 + c1 et − c1 et
= =
dt (1 + c1 et )2 1 + c1 et 1 + c1 et
c1 et c1 et
= 1− = P (1 − P ).
1 + c1 et 1 + c1 et
2 dy 2
24. Differentiating y = 2x2 − 1 + c1 e−2x we obtain = 4x − 4xc1 e−2x , so that
dx
dy 2 2
+ 4xy = 4x − 4xc1 e−2x + 8x3 − 4x + 4c1 xe−x = 8x3
dx
dy d2 y
25. From y = c1 e2x + c2 xe2x we obtain = (2c1 + c2 )e2x + 2c2 xe2x and = (4c1 + 4c2 )e2x +
dx dx2
4c2 xe2x , so that
d2 y dy
−4 + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
dx2 dx
26. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
dy
= −c1 x−2 + c2 + c3 + c3 ln x + 8x,
dx
d2 y
= 2c1 x−3 + c3 x−1 + 8,
dx2
and
d3 y
= −6c1 x−4 − c3 x−2 ,
dx3
so that
d3 y 2
2 d y dy
x3 + 2x −x + y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
dx3 dx2 dx
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2 = 12x2
In Problems 25–28, we use the Product Rule and the derivative of an integral ((12) of this section):
x
d
ˆ
g(t) dt = g(x).
dx a
x x
e−3t dy e−3t e−3t 3x
ˆ ˆ
27. Differentiating y = e3x dt we obtain = e3x dt + · e or
1 t dx 1 t x
x
dy e−3t 1
ˆ
= e3x dt + , so that
dx 1 t x
ˆ x −3t ˆ x −3t
dy 3x e 1 3x e
x − 3xy = x e dt + − 3x e dt
dx 1 t x 1 t
ˆ x −3t ˆ x −3t
3x e 3x e
= xe dt + 1 − 3xe dt = 1
1 t 1 t
4
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller TestsBanks. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $12.99. You're not tied to anything after your purchase.
