This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Exercises 2.3 To find M , we can now use the first two equations in the above system. M + C = 100 M + (3 / 4) C = 80 M = 20 . 38. With m = 10, g = 9 . 81, and k = 5, the equation becomes 100 dv dt = 100(9 . 81) 5 v 20 dv dt = 196 . 2 v. Separating variables and integrating yields Z dv v 196 . 2 = 1 20 Z dt ln  v 196 . 2  = t 20 + C 1 v = 196 . 2 + Ce t/ 20 , where C is an arbitrary nonzero constant. With C = 0, this formula also gives the (lost) constant solution v = 196 . 2. From the initial condition, v (0) = 10, we find C . 196 . 2 + C = 10 C = 186 . 2 v ( t ) = 196 . 2 186 . 2 e t/ 20 . The terminal velocity of the object can be found by letting t . v = lim t ( 196 . 2 186 . 2 e t/ 20 ) = 196 . 2 (m / sec) . EXERCISES 2.3: Linear Equations 2. Neither. 4. Linear. 6. Linear. 8. Writing the equation in standard form, dy dx y x = 2 x + 1 , we see that P ( x ) = 1 x ( x ) = exp Z 1 x dx = exp ( ln x ) = 1 x . Multiplying the given equation by ( x ), we get d dx y x = 2 + 1 x y = x Z 2 + 1 x dx = x (2 x + ln  x  + C ) ....
View
Full
Document
This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

Click to edit the document details