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nagle_differential_equations_ISM_Part10

# nagle_differential_equations_ISM_Part10 - Exercises 2.3 To...

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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 dv v - 196 . 2 = - 1 20 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 μ ( x ) = exp - 1 x dx = exp ( - ln x ) = 1 x . Multiplying the given equation by μ ( x ), we get d dx y x = 2 + 1 x y = x 2 + 1 x dx = x (2 x + ln | x | + C ) . 41

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Chapter 2 10. From the standard form of the given equation, dy dx + 2 x y = x - 4 , we find that μ ( x ) = exp (2 /x ) dx = exp (2 ln x ) = x 2 d dx ( x 2 y ) = x - 2 y = x - 2 x - 2 dx = x - 2 ( - x - 1 + C ) = Cx - 1 x 3 .
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nagle_differential_equations_ISM_Part10 - Exercises 2.3 To...

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