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

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)




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

,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS W ith MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
SO SO SO SO SO SO S
O S
O SO SO SO SO SO




Introduction to Differential Equations S O S O S O




SolutionandAnswerGuide O
S O
S O
S




ZILL,DIFFERENTIALEQUATIONSWITHMODELINGAPPLICATIONS2024, 9780357760192;
O
S O
S O
S O
S O
S O
S S O O
S




CHAPTER #1:INTRODUCTION TO DIFFERENTIALEQUATIONSSO O
S SO SO O
S




TABLEOFCONTENTS S
O S
O




End of Section Solutions........................................................................................................................................................................................... 1
S O S O S O




Exercises 1.1 .............................................................................................................................................................................................................................................. 1
S O




Exercises 1.2 .......................................................................................................................................................................................................................................... 14
S O




Exercises 1.3 ..........................................................................................................................................................................................................................................22
S O




Chapter1 in Review Solutions........................................................................................................................................................30
S
O SO SO SO




ENDOFSECTIONSOLUTIONS S
O S
O S
O




EXERCISES 1.1 S O




1. Second order; linear S O S O




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




3. Fourth order; linear S O S O




4. Second order; nonlinear because of cos(r + u) S O S O S O S O S O S O SO




√ SO




5. Second order; nonlinear because of (dy/dx) 2 or S O S O S O S O S O
SO S O



1 + (dy/dx)2 S O S O




6. Second order; nonlinear because of R2 S O S O SO S O S O




7. Third order; linear S O S O




8. Second order; nonlinear because of ẋ2 S O S O SO S O S O O
S




9. First order; nonlinear because of sin ( dy/dx)
S O S O S O S O S O SO




10. First order; linear S O S O




11. Writing the differential equation in the form x(dy/dx) + y2 = 1, we see that it is nonlinear in y beca
SO S O S O SO SO SO S O S O S O
S O



S O SO SO SO SO SO SO S O O
S SO




use of y2. However, writing it in the form (y2 —1 )(dx/dy) + x = 0, we see that it is linear in x.
SO SO SO SO SO SO SO SO S O
S O



O
S SO SO SO S O SO SO SO SO SO S O S O S O




12. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linearinv.
SO SO SO SO SO SO SO SO SO SO SO S O S O
S O



SO SO SO SO S O O
S O
S O
S




However,writingiti nthef orm(v + uv —ueu)(du/dv) +u = 0,w es eethatitis nonlinear in u. O
S O
S O
S O
S O
S O
S O
S SO SO O
S O
S O
S SO SO O
S O
S O
S O
S O
S S O S O S O




13. Fromy=e − O
S O
S O
S
x/2 we obtain yj = — 1e−
SO SO
S O

SO O
S
O
S
x/2
. Then 2yj + y = —e−
O
S SO
S O

O
S S O SO
x/2
+ e− SO
x/2 = 0. SO




2




1

,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS W ith MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
SO SO SO SO SO SO S
O S
O SO SO SO SO SO




Introduction to Differential Equations S O S O S O




6 6—
14. From y = S O S O — e 20t weobtain dy/dt = 24e−20t,s othat SO

SO SO SO SO
O
S
O
S O
S




5 5
dy +20y = 24e−20t 6 6 −20t SO S O




+ 20 — e = 24.
O
S SO SO




S O
SOS O




dt 5 5
S O




15. Fromy = e3x cos2x weobtainyj = 3e3x cos2x—2e3x sin2x andyjj = 5e3x cos2x— O
S SO SO
SO



O
S O
S O
S O
S
S O

SO
SO



O
S
SO



O
S O
S O
S
S O

SO
SO



O
S




12e3x sin2x, so that yjj — 6yj + 13y = 0. SO



O
S S O S O S O
S O

SO
S O

SO S O S O




j
16. From y = —c osxln(sec x + tan x ) we obtain y = —1 + sin xln(secx + tanx) and
SO SO S O O
S O
S O
S O
S SO O
S O
S SO SO SO SO S O
SO O
S SO O
S O
S O
S O
S SO O
S SO




jj jj
y = tanx+ cos xln(secx+ tanx). Then y + y = tanx.
SO S O SO O
S O
S SO SO O
S O
S O
S O
S O
S SO SO SO S O O
S SO S O O
S




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

S O




we have S O




j 1/2 −
(y —x)y = (y — x)[1+ (2(x+ 2)
SO
S O SO SO SO O
S O
S O
S SO S OS O
SO




]

=y—x+2(y—x)(x+ 2)−1/2
O
S O
S O
S O
S O
S O
S
O
S O
S




= y — x + 2[x + 4(x + 2) 1/2 —x](x + 2)−1/2
SO S O SO SO S O SO S O SO S O
S OS O




S O SO




=y —x + 8(x + 2)1/2(x + 2)−1/2 = y — x + 8.
O
S SO O
S O
S SO O
S O
S SO SO
S O



S O SO SO O
S SO




An interval of definition for the solution of the differential equation is (—
S O S O S O S O S O S O S O S O S O S O S O S O




2, ∞) because yj is not defined at x = —2.
SO S O S O
S O

S O S O S O S O S O S O




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




{x 5x /= π/2+nπ} SO S O O
S O
S O
S SO




or{ x x /= π/10+nπ/5}. Fromy j= 25s ec 25xw ehave
O
S SO S O SO SO O
S O
S O
S SO S O SO O
S SO O
S SO




j
y = 25(1 + tan2 5x)= 25 + 25 tan2 5x = 25 + y 2.
S O
S O S O



SO SO SO O
S SO O
S SO O
S SO SO O
S SO




An interval ofdefinition forthe solutionofthedifferential equationis (—π/10,π/10). An-
O
S O
S O
S O
S O
S O
S O
S O
S O
S O
S O
S O
S O
S O
S




Sother interval is (π/10, 3π/10), and so on.
O SO SO SO O
S S O SO SO




19. The domain of the function is {x 4 — x2 SO S O S O S O SO S O SO SO S O S O SO /= 0} or {x
S O SO SO x /= —2 orx /= 2} .Fromy j =
S O S O O
S O
S S O S O O
S O
S
S O




2x/(4 — x2)2 we have S O SO
S O



S O




1 2
= 2xy2.
yj = 2x SOS O S O




4—x2
S O




OS OS




An interval of definition for the solution of the differential equation is (—2, 2) . Other inter-
S O S O S O S O S O S O S O S O S O S O S O S O SO S O S O




vals are (—∞, —2) and (2, ∞) .
√
S O S O S O S O S O S O S O SO




20. The function is y = 1/ 1 — sin x , whose domain is obtained from 1 — sin x /= 0 or sin x /= 1.
SO S O S O S O S O SO SO O
S SO S O S O S O S O SO S O S O SO O
S S O S O SO S O O
S S O S O




Thus, the domain is {x x/= π/2+ 2nπ}. From y j= — (11 —sinx) −3/2
2
(— cosx) we have
SO SO O
S O
S SO S O O
S S O O
S SO SO SO S O SO S O SO O
S O
S S O
S O



O
S O
S SO SO




2yj = (1 —sinx) −3/2 cosx= [(1 —sinx) −1/2] 3 cosx = y3 cosx.
S O

O
S O
S O
S O
S
SO



O
S O
S SO O
S O
S O
S
SO



O
S SO O
S
SO



O
S




An interval of definition for the solution of the differential equation is (π/2, 5π/2) . Another one is
S O S O S O S O S O S O S O S O S O S O S O S O SO S O S O S O S O




(5π/2, 9π/2), and so on. SO S O S O S O




2

, Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS W ith MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
SO SO SO SO SO SO S
O S
O SO SO SO SO SO




Introduction to Differential Equations S O S O S O




21. Writing ln(2X — 1) — ln(X — 1) = t and differentiating
SO S O S O S O S O S O S O SOS O SOS O SO SO
x

implicitly we obtain S O S O 4


— =1 S O 2
2X — 1 dt SO O
S S O
X —1 dt O
S O
S S O




t
2 1 dX –4 –2 2 4
— = 1
SO S O




X —1 dt
O
S




2X —1
S O




O
S O
S SO SO




–2


–4 O
S




dX
= —(2X —1)(X — 1) = (X — 1)(1 — 2X).
dt
SO SO O
S SO O
S SO SO SO O
S SO O
S




S O




Exponentiating both sides of the implicit solution we obtain S O S O S O S O S O S O S O S O




2X— O
S




1 t X —1
=e
O
S O
S SO O
S
S O



O
S




2X — 1 = Xet — et SO SO SO SO
S O



SO




(et —1) =(et —2)X SO



O
S O
S O
S
SO



O
S




et 1
X= .
et — 2
S O S O


S O

SO SO




Solving et — 2 = 0 we get t = ln 2. Thus, the solution is defined on (—
SO
S O



SO SO S O SO SO SO S O S O O
S S O SO SO SO SO S O SO




∞, ln 2) or on (ln 2,∞). The graph of the solution defined on (—
SO SO SO SO SO SO O
S S O S O S O S O S O S O S O S O




∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid.
SO SO S O S O S O S O S O S O S O S O S O S O S O S O S O S O S O




22. Implicitly differentiating the solution, we obtain S O S O S O S O S O
y

2 SOS O
dy dy 4

—2x — 4xy + 2y = 0
dx dx
SO S O SO O
S SO S O SO




S O S O


2
—x2 dy — 2xydx+ ydy = 0 S O



O
S O
S O
S O
S SO O
S SO SO




x
2xydx + (x2 —y)dy = 0. O
S O
S SO
S O



O
S SO SO
–4 O
S –2 2 4

–2
Using the quadratic formula to solve y2 — 2x2 y — 1 = 0
SO SO S O SO SO SO
SO S O



S O S O S O SOSO SOS O




√ √
fory,w egety = 2x2 ± 4x 4 + 4 /2 = x2 ± x4 +1.
SO SO



SO SO SO
S O SO




–4
O
S O
S O
S O
S S O SO SOS O S O S O O
S O
S




√
O
S




Thus, two explicit solutions are y1 = x2 +
SO




x4 + 1 and
S O


S O
SO SO SO SO SO S O
SOS O




√
SO S O




OO
S




y2 = x2 — x4 + 1 . Both solutions are defined on (—∞, ∞).
SOS O
S O
SOS O

S O


S O O
S S O S O S O S O S O S O SO




The graph of y1(x) is solid and the graph of y2 is dashed.
S O S O S O S O S O S O S O S O S O S O
SOS O
S O




3