Solution Manual for A First Course in Differential Equations with Modeling Applications, 12th Edition Zill (All Chapters 1-9)Latest Edition

A First Course In Differential
Equations With Modeling
Applications, 12th Edition
By Dennis G. Zill

Complete Chapter Solutions Manual
Are Included (Ch 1 To 9)

,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


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)
√
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 Ẋ 2
9. First Order; Nonlinear Because Of Sin (Dy/Dx)
10. First Order; Linear
11. Writing The Differential Equation In The Form X(Dy/Dx) + Y2 = 1, We See That It Is
Nonlinear In Y Because Of Y2. However, Writing It In The Form (Y2 — 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.
1 −X/2
13. From Y = E−X/2 We Obtain Yj = —
2
E . Then 2yj + 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 —
14. From Y = — E 20t We Obtain Dy/Dt = 24e
−20t , So That
5 5
Dy 6 6 −20t
+ 20y = 24e−20t + 20 — e = 24.
dt 5 5

15. From Y = E3x Cos 2x We Obtain Yj = 3e3x Cos 2x—2e3x Sin 2x And Yjj = 5e3x Cos 2x—12e3x Sin
2x, So That Yjj — 6yj + 13y = 0.
J
16. From Y = — Cos X Ln(Sec X + Tan X) We = —1 + Sin X Ln(Sec X + Tan X) And
Obtain Y
Jj jj
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 Yj = 1+2(X+2)−1/2
We Have
J −
(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
Yj 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 /= Π/2 + Nπ}
Or {X X /= Π/10 + Nπ/5}. From Yj = 25 Sec
2 5x We Have

j
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 2 4—X /= 0} Or {X X /= —2 Orj X
/= 2}. From Y = 2x/(4 — X ) We Have
2 2

2
1
Yj = 2x = 2xy2.
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 /= 0 Or Sin X /= 1.
Thus, The Domain Is {X X /= Π/2 + 2nπ}. From jY = —12 (1 — Sin X)
−3/2 (— Cos X) We Have


2yj = (1 — Sin X)−3/2 Cos X = [(1 — Sin X)−1/2]3 Cos X = Y3 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

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

2 Dy Dy 4
—2x — 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 Y2 — 2x2y — 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.




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