Solution Manual for A First Course in Differential Equations with Modeling Applications 11th Edition Zill / All Chapters / Full Complete 2023

A First Course in Differential Equations with Modeling Applications 11th Edition Zill Solutions Manual Contents Chapter 1 ................................ ................................ ................................ ................................ .......... 1 Chapter 2 ................................ ................................ ................................ ................................ ........ 36 Chapter 3 ................................ ................................ ................................ ................................ ...... 106 Chapter 4 ................................ ................................ ................................ ................................ ...... 157 Chapter 5 ................................ ................................ ................................ ................................ ...... 285 Chapter 6 ................................ ................................ ................................ ................................ ...... 339 Chapter 7 ................................ ................................ ................................ ................................ ...... 430 Chapter 8 ................................ ................................ ................................ ................................ ...... 512 Chapter 9 ................................ ................................ ................................ ................................ ...... 580 Chapter 10 ................................ ................................ ................................ ................................ .... 606 Chapter 11 ................................ ................................ ................................ ................................ .... 639 Chapter 12 ................................ ................................ ................................ ................................ .... 691 Chapter 13 ................................ ................................ ................................ ................................ .... 791 Chapter 14 ................................ ................................ ................................ ................................ .... 849 Chapter 15 ................................ ................................ ................................ ................................ .... 903 2 Chapter 1 Introduction to Differential Equations 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 x˙2 9. 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. 10. 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. 11. From y = e−x/2 we obtain y′ = − 1 e−x/2. Then 2y′ + y = −e−x/2 + e−x/2 = 0. 12. From y = 6 − 6 e−20t we obtain dy/dt = 24e−20t, so that 5 5 dy −20t 6 6 −20t + 20y = 24e dt + 20 5 − 5 e = 24. 13. 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. 1 Definitions and Terminology 2 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS cos x ln(sec x + tan x) we obtain y = (x + 2) (y − 1/2 x)(x + 2) x](x + 2) 2 2 2 3/2 1/2 3 3 14. From 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. 15. The domain of the function, found by solving x+2 ≥ 0, is [−2, ∞). From y′ = 1+2( x+2)−1/2 we have ′ − x)y = (y − x)[1 + (2(x + 2) ] = y − x + 2(y − −1/2 = y − x + 2[x + 4(x + 2)1/2 − −1/2 = y − x + 8(x + 2)1/2 −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. 16. 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 y′ = 25 sec2 5x we have ′ y = 25(1 + tan 5x) = 25 + 25 tan 5x = 25 + y . An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is (π/10, 3π/10), and so on. . 17. The domain of the function is {x . 4 − x2 y′ = 2x/(4 − x2)2 we have . 0} or {x . x −2 and x /= 2}. From 1 2 ′ 2 y = 2x 4 − x2 = 2xy . An interval of definition for the solution of the differential equation is (−2, 2). Other intervals are (−∞ , −2) and (2, ∞). 18. The function is y = 1/√1 − sin x , whose domain is obtained from 1 − sin x 0 or sin x /= 1. . ′ 1 −3/2 Thus, the domain is {x . x /= π/2 + 2nπ}. From y = − 2 (1 − sin x) (− cos x) we have ′ − − 2y = (1 − sin x) cos x = [(1 − sin x) ] cos x = y 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. ′′