Sample for Solution Manual Advanced Engineering Mathematics 7th edition by Dennis Zill

Chapter 1 Introduction to Differential Equations 1.1 Definitions and Terminology 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 ( 6. Second order; nonlinear because of R2 7. Third order; linear 8. Second order; nonlinear because of ˙x2 9. First order; nonlinear because of sin 10. First order; linear in x. 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. 13. From y = e−x/2 we obtain . Then 2y′ + y = −e−x/2 + e−x/2 = 0. 14. From we obtain dy/dt = 24e−20t, so that . 1 15. From y =′′e3x cos2′ x we obtain y′ = 3e3x cos2x−2e3x sin2x and y′′ = 5e3x cos2x−12e3x sin2x, so that y − 6y + 13y = 0. 16. From y = −cosxln(secx + tanx) we obtain y′ = −1 + sinxln(secx + tanx) and y′′ = tanx + cosxln(secx + tanx). Then y′′ + y = tanx. 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 tanx is not defined for x = π/2 + nπ, n an integer, the domain of y = 5tan5x is . From y′ = 25sec2 5x we have ) = 25 + 25tan2 5x = 25 + y2. 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. 19. The domain of the function is . From y′ = 2x/(4 − x2)2 we have . An interval of definition for the solution of the differential equation is (−2,2). Other intervals are (−∞,−2) and (2,∞). 20. The function is , whose domain is obtained from 1 = 0 or sin Thus, the domain is . From ) we have 2y′ = (1 − sinx)−3/2 cosx = [(1 − sinx)−1/2