115.3 Assignment #9 Solutions

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Dominick McDaniel
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1 115. Assignment #9 Solutions Assignment #9 Solutions Solve the differential equation d dx = 2(1 ), where 0 = 2 for x 0 = 0. d 1 = 2dx d 1 = 2dx ln 1 =2x + C Find C b inserting the Initial Conditions: ln 1 2 =2(0) + C or C = 0, and thus ln 1 = 2x Exponentiate: e ln 1 = e 2x or 1 =e 2x or 1 =±e 2x or = 1 ± e 2x. The Initial Conditions are onl satisfied if = 1 + e 2x.

2 115. Assignment #9 Solutions Solve the differential equation dn dx = 5 1 N, where N(2) =. 2 dn Simplif first: dx = 10 N 2 2dN 10 N = dx 2dN 10 N = dx 2ln 10 N =x + C Exponentiate: e 2ln 10 N = e x e C or 10 N 2 = e x e C. Find C b inserting the Initial Conditions: 10 2 = e 2 e C or C = (7e) 2, and thus 10 N 2 = (7e) 2 e x or 10 N =(7e)e x 2 = 7e 1 x 2,or 10 N =±7e 1 x 2 or N = 10 ± 7e 1 x 2. The Initial Conditions are onl satisfied if N = 10 7e 1 x 2.

3 115. Assignment #9 Solutions Denote b L(t) the length of a fish at time t and assume that the fish grows according to the von Bertalanff equation dl = k(4 L(t)) with L(0) = 2. dt (a) Solve the differential equation. First we note that 4 L(t) 0, because live fish don t decrease in length! dl 4 L(t) = kdt dl 4 L(t) = kdt ln 4 L(t) =kt + C or ln 4 L(t) = kt C Insert the Initial Condition to get C: ln 4 2 = k(0) C, soc = ln 2 = 5ln2 and we have ln 4 L(t) = kt + 5ln2. Exponentiate: e ln 4 L(t) = e kt+5ln2 or 4 L(t) = 2e kt or L(t) = 4 2e kt.

4 115. Assignment #9 Solutions-4 (b) Use our solution in (a) to determine k under the assumption that L(4) = 10. Sketch the graph of L(t) for this value of k. We have L(4) = 10 = 4 2e k(4),so2e 4k = 24, or e 4k = 4 = Taking natural logarithms, we have 4k = ln 0.75, or k = 0.25 ln and ( ) t L(t) = 4 2e ( 0.25 ln 0.75)t 4 = (c) Find the length of the fish when t = 10. ( ) 10 ( ) L(10) = 4 2 = (d) Find the asmptotic length of the fish, that is, find lim L(t) t x

5 115. Assignment #9 Solutions Use the partial fraction method to solve d dx = (1 ) where 0 = 2 for x 0 = 0. d (1 ) = dx, ( d 1 (1 ) = dx or + 1 ) d = dx or ln ln 1 =x + C or 1 ln 1 = x + C. Insert the Initial Conditions: ln = 0 + C to get C = ln 2, and ln = x + ln 2. 1 Exponentiate: e ln 1 = e x+ln 2 or 1 = 2ex or 1 =±2ex, so 1 = 2ex (where again we have used the Initial Conditions to find the appropriate sign.) Solve for : = 2e x (1 ) = 2e x + 2e x,2e x = (2e x 1), so = 2ex 2e x 1

6 115. Assignment #9 Solutions Solve the differential equation d dx = d = ln =ln x + 1 +C dx x + 1 Insert the Initial Condition to get C: ln 2 =ln 1 +C, soc = ln 2 and we have ln =ln x + 1 +ln 2. x + 1 with 0 = 2ifx 0 = 0. d = dx x + 1 Exponentiate: e ln = e ln x+1 +ln 2 or =2 x + 1 or =±2(x + 1). Since the Initial Conditions are not satisfied if = 2(x + 1), we must have = 2(x + 1).

7 115. Assignment #9 Solutions Suppose that d = (1 )( 2) dx (a) Find the equilibria of this differential equation. ŷ = 0, 1, 2 (b) Graph d dx as a function of, and use our graph to discuss the stabilit of the equilibria. From the graph we see that g() > 0on(, 0) (1, 2) and g() < 0on(0, 1) (2, ), so = 0 and = 2 are stable equilibria and = 1 is unstable. 2 g() (c) Compute the eigenvalues associated with each equilibrium and discuss the stabilit of the equilibria. With g() = (1 )( 2) = + 2 2, we have g () = , so g (0) = 2, g (1) = 1, and g (2) = 2, so = 0 and = 2 are stable equilibria and = 1 is unstable.

8 115. Assignment #9 Solutions Assume the single compartment model defined in 8.2.2; that is, denote b C(t) the concentration of the solute at time t and assume that dc = (20 C(t)) for t 0. dt (a) Solve when C(0) = 5 dc 20 C(t) = dt dc 20 C(t) = dt ln 20 C(t) =t + C or ln 20 C(t) = t + C Insert the Initial Condition to get C: ln 20 5 = (0) + C, soc = ln 15 and we have ln 20 C(t) = t + ln 15. Exponentiate: e ln 20 C(t) = e t+ln 15 or 20 C(t) =15e t or 20 C(t) =±15e t or C(t) = 20 ± 15e t. Since the Initial Conditions are not satisfied if C(t) = e t, we must have C(t) = 20 15e t. (b) Find lim t C(t) 20 (c) Use (a) to find the t for which C(t) = 10 We solve the equation 10 = 20 15e t for t: ( ) ( ) 10 = 15e t or 2 ( ) 2 = e t or ln = t or t = ln 2 = ln 2 = ln

Math 31S. Rumbos Fall Solutions to Exam 1

Math 31S. Rumbos Fall Solutions to Exam 1 Math 31S. Rumbos Fall 2011 1 Solutions to Exam 1 1. When people smoke, carbon monoxide is released into the air. Suppose that in a room of volume 60 m 3, air containing 5% carbon monoxide is introduced