A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill

AFirst Course in Differential
br br br br




EquationswithModeling
b r rb rb




b r Applications,12thEditionby rb rb rb




Dennis G.Zill b r br br




CompleteChapterSolutionsManual rb b
r rb




are included (Ch 1 to 9)
b r br br br br br




** Immediate Download
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** Swift Response
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** All Chapters included
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,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
br br br br br br br br br br br br br




Introduction to Differential Equations b r b r b r




SolutionandAnswerGuide b
r b
r b
r




ZILL,DIFFERENTIALEQUATIONSWITHMODELINGAPPLICATIONS2024,
b
r b
r b
r br b
r br




9780357760192; CHAPTER #1: INTRODUCTION TO DIFFERENTIAL EQUATIONS
b r br br br br br br




TABLEOFCONTENTS br rb




End of Section Solutions..............................................................................................................................................1
br br br




Exercises 1.1 ......................................................................................................................................................................... 1
br




Exercises 1.2 ....................................................................................................................................................................... 14
br




Exercises 1.3 ....................................................................................................................................................................... 22
br




Chapter 1 in Review Solutions .......................................................................................................................... 30
br br br br




ENDOFSECTIONSOLUTIONS rb rb rb




EXERCISES 1.1 b r




1. Second order; linear b r b r




2. Third order; nonlinear because of (dy/dx)4 br br br br br




3. Fourth order; linear br b r




4. Second order; nonlinear because of cos(r + u) br br br br br br br



√ br




5. Second order; nonlinear because of (dy/dx)2 or br br br br br
br b r
1 + (dy/dx)2 b r b r




6. Second order; nonlinear because of R br br br br br
2


7. Third order; linear br br




8. Second order; nonlinear because of ẋ 2 br br br br br br




9. First order; nonlinear because of sin (dy/dx)
br br br br br br




10. First order; linear br br




11. Writing the differential equation in the form x(dy/dx) + y2 = 1, we see that it is nonlinear
br br br br br br br br br
b r
b r br br br br br br




in y because of y2. However, writing it in the form (y2 — 1)(dx/dy) + x = 0, we see that it is
b r br br br br br br br br br br br
b r
br br br br b r br br br br br




linear in x.
b r br br




12. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is
br br br br br br br br br br br b r b r
br
br br br br




linear in v. However, writing it in the form (v + uv — ueu)(du/dv) + u = 0, we s ee that it is
b r b
r b
r br b
r b
r b
r b
r b
r b
r br br br br br br br br b
r b
r b
r b
r b
r




nonlinear in u.
b r br b r




13. Fromy = e− b
r br br
x/2
we obtain yj = —12e−
br br
b r
br rb
b
r x/2
. Then 2yj + y = —e− x/2 + e− x/2 = 0.
br br
b r
br br br br br




1

,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
br br br br br br br br br br br br br




Introduction to Differential Equations b r b r b r




66
14. From y = br b r — e—20t we obtain dy/dt = 24e−20t , sothat br
br br br br
br
br b
r




5 5
dy + 20y = 24e−20t 6 6 −20t br b r




+ 20 — e
br br br



= 24. b r br b r
b r


dt 5 5

15. From y = e3x cos2x we obtain yj = 3e3x cos 2x—2e3x sin2x and yjj = 5e3x cos2x—12e3x sin2x,
br br br
br
b
r br br br
b r
br
br
br
br
b
r br br
b r
br
br
b
r
br
b
r




so that yjj — 6yj + 13y = 0.
b r br br
b r
br
b r
br br b r



j
16. From y = — cos x ln(sec x + tan x) we obtain y br br br br br br br br br br br br br br b r = —1 + sinxln(sec x + tanx) and
br br br br rb br br br br br



jj jj
y = tanx+ cos xln(sec x + tanx). Then y + y = tanx.
br b r br b
r br br br rb br br br br br br br b r br br br br




17. The domain of the function, found by solving x+2 ≥ 0, is [—2, ∞). From yj = 1+2(x+2)−1/2
br br b r br br b r b r br b r b r br br br b r b r
br b r
b r




we have br




j −
(y —x)y = (y — x)[1 + (2(x+ 2)
br
b r br br br br br br br br b r
1/2 br

]

=y —x+ 2(y —x)(x + 2)−1/2
br br b
r rb br br br br




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

br br




= y — x + 8(x + 2)1/2(x + 2)−1/2 = y — x + 8.
br br br br br br br br br
b r

br br br br br




An interval of definition for the solution of the differential equation is (—2, ∞) because yj
br br br br br br br br br br br br br br br




is not defined at x = —2.
b r
b r b r b r b r b r b r




18. Since tan x is not defined for x = br br br br br br br b r b r π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
br br br br br br br br br br b r b r br br br




{x br b r 5x /
= π/2+ nπ} br br b
r br




or {x b
r br b r x /= π/10+ nπ/5}. From y j= 25sec 25x w e have
br br br br br br b r br b
r br br br




j
y = 25(1 + tan2 5x) = 25 + 25tan2 5x = 25 + y 2 .
b r
br br

br br br br br br br b
r br br br br




An interval of definition for the solution of the differential equation is (—π/10,π/10). An- other
b
r br br br br br br br br br br br rb br b r




interval is (π/10, 3π/10), and so on.
br br br br br br br




19. The domain of the function is {x br br br br br b r br br b r 4— br /
= 0} or {x b r br br x / = 2}. From y j =
= —2 orx /b r b r br b
r b r b r br b
r
b r




br x2
2x/(4 — x2)2 we have b r


1 2
= 2xy2.
br br br


b r




yj = 2x 4—x
2 br rb
br br
br




An interval of definition for the solution of the differential equation is (—2, 2). Other
br br b r br br br b r br br br br br br br




inter- vals are (—∞, —2) and (2, ∞).
√
br b r br br b r br br b r br




20. The function is y = br br br b r 1 — sinx , whose domain is obtained from 1 — sin x /= 0 or sinx /= 1.
br br br br br br br br br br br br br b r b r br br br b r b r




b r 1/
= π/2 + 2nπ}. From y j= — (11 —
Thus, the domain is {x x / br

2
sinx) −3/2 (—cos x) we have br br br br b r br br br br br br b r br b r br br br b r
b r
rb br br br




2yj = (1 —sinx)−3/2 cosx = [(1 —sinx)−1/2]3 cosx = y3 cosx.
br
br br rb b
r
br
br br br br br br
br
b
r br br
br
b
r




An interval of definition for the solution of the differential equation is (π/2, 5π/2).
br b r br br br br br br br b r br b r br




Another one is (5π/2, 9π/2), and so on.
br b r b r b r br b r b r b r




2

, Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
br br br br br br br br br br br br br




Introduction to Differential Equations b r b r b r




21. Writing ln(2X — 1) — ln(X — 1) = t and differentiating
br b r br b r b r b r b r br b r br b r br br x

implicitly we obtain br br 4


— = 1 b r 2
2X — 1 dt br br b r X —1 dt br b
r b r




t
2 1 dX
— = 1 –4 –2 2 4
br b r
br



2X —1 X —1 dt
b r

br b
r br br




–2


–4 br




dX
= —(2X — 1)(X — 1) = (X — 1)(1 — 2X). br br br br


dt
br br br br br br br


b r




Exponentiating both sides of the implicit solution we br br br br br br br




obtain br




2X—1 t rb


=e
b
r br




X —1
br


br br
b r




2X — 1 = Xet — et br br br br
b r
br




(et — 1) = (et — 2)Xbr
br br br
br
br




et 1
X= .
et — 2
br b r
b r
br br




Solving et — 2 = 0 we get t = ln 2. Thus, the solution is defined on (—∞, ln 2) or on (ln 2, ∞).
br
b r
br br br br br br br br br br br br br br br br br br br br br br br




bThe graph of the solution defined on (—∞, ln 2) is dashed, and the graph of the
r br br br br br br b r br br br b r br br br b r br




solution defined on (ln 2, ∞) is solid.
br b r br br br br br br




22. Implicitly differentiating the solution, we obtain b r b r b r b r br y

2 br b r
dy dy 4

—2x — 4xy + 2y =0 br


dx dx
br b r br br br br




2
br br




—x2 dy — 2xydx + ydy = 0 br
br br br br br rb br br




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

–2
Using the quadratic formula to solve y2 — 2x2y — 1 = 0
br br br br br br
br b r
b r b r b r br br br b r



√ br √ br



fory, we get y = b
r b
r b
r b
r br 2x2 br br

4x24 + 4 /2 = br
br br b r b r
± x4 +1. br
br b
r

–4
br


± x b r
br




√ br



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

b r

+
√ brbr




y2 = x2 — x4 + 1 . Both solutions are defined on (—∞, ∞).
br b r b r
br b r
b r
br br b r br br br br br br




The graph of y1(x) is solid and the graph of y2 is dashed.
br br br br br br br br br br br b r br




3