Solutions for A First Course in Differential Equations with Modeling Applications, 12th Edition Zill (All Chapters included)
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Course
Math
Institution
Math
Complete Solutions Manual for A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill ; ISBN13: 9780357760192. (Full Chapters included Chapter 1 to 9)....1. INTRODUCTION TO DIFFERENTIAL EQUATIONS.
2. FIRST-ORDER DIFFERENTIAL EQUATIONS.
3. MODELING WITH ...
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
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