nagle_differential_equations_ISM_Part10

nagle_differential_equations_ISM_Part10 - Exercises 2.3 To...

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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 ) ....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part10 - Exercises 2.3 To...

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