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Unformatted text preview: Chapter 3 to find s ( t ). Here input rate = 4 (gal / min) Â· . 5 (lb / gal) = 2 (lb / min) . Since the flow rate in is the same as the flow rate out, the volume of the solution remains constant (200 gal), we have c ( t ) = s ( t ) 200 (lb / gal) and so output rate = 4 (gal / min) Â· s ( t ) 200 (lb / gal) = s ( t ) 50 (lb / min) . Then (3.1) yields ds dt = 2- s 50 â‡’ ds dt + s 50 = 2 . This equation is linear, has integrating factor Î¼ ( t ) = exp R (1 / 50) dt = e t/ 50 . Integrat- ing, we get d ( e t/ 50 s ) dt = 2 e t/ 50 â‡’ s = 100 + Ce- t/ 50 â‡’ c ( t ) = 1 2 + ( C/ 200) e- t/ 50 , where the constant C depends on s . (We do not need an explicit formula.) Taking the limit yields lim t â†’âˆž c ( t ) = lim t â†’âˆž 1 2 + ( C/ 200) e- t/ 50 = 1 2 . 10. In this problem, the dependent variable is x , the independent variable is t , and the function f ( t,x ) = a- bx . Since f ( t,x ) = f ( x ), i.e., does not depend on t , the equation is autonomous. To find equilibrium solutions, we solve f ( x ) = 0 â‡’ a- bx = 0 â‡’ x = a b . Thus, x ( t ) â‰¡ a/b is an equilibrium solution. For x < a/b , x = f ( x ) > 0 meaning that x increases, while x = f ( x ) < 0 when x > a/b and so x decreases. Therefore, the phase line for the given equation is as it is shown in Fig. 3â€“A on page 100. From this picture, we conclude that the equilibrium x = a/b is a sink. Thus, regardless of an initial point x , the solution to the corresponding initial value problem will approach x = a/b , as t â†’ âˆž . 12. Equating expressions (21) evaluated at times t a and t b = 2 t a yields p a p 1 e- Ap 1 t a p 1- p a (1- e- Ap 1 t a ) = p b p 1 e- 2 Ap 1 t a p 1- p b (1- e- 2 Ap 1 t a ) ....
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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.
- Spring '08