MAT 274 HW 2 Solutions c Bin Cheng. Due 11:59pm, W 9/07, Points

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1 MAT 274 HW 2 Solutions Due 11:59pm, W 9/07, oints 1. (30 ) The last two problems of Webwork Set 03 Modeling. Show all the steps and, also, indicate the equilibrium solutions for each problem. Instructor s version of roblem 4 of Set 03 Modeling. (Notes for grader: regardless of different numbers, the answer should be similar to (1), (2),(3), (4),(5)) Biologists stocked a lake with 500 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be The number of fish doubled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation ( d = k 1 ), determine the constant k, and then solve the equation to find an expression for the size of the population after t years. Solution. Carrying capacity is given = lug it into the above DE A seperable DE. Rearrange and separate, t Use partial fraction on the LHS d = k d = ( ). k 5300 (5300 ), 5300 d (5300 ) = k d + d (5300 ) = k Integrate both sides or ln ln 5300 = kt + C ln 5300 = kt + C age 1

2 Apply exponential on both sides 5300 = C 1e kt (1) Here, we use another constant C 1 = ±e C so the absolute value function is dropped. This is an implicit form of the general solution. Now, we apply the initial data given in the problem 500 fish at the beginning = (0) = 500 that is, we set t = 0, = 500 in (1) or more generally = C 1e 0 = C 1 = C 1 = fish doubled after a year = (1) = 1000 that is, we set t = 1, = 1000 in (1) initial population carrying capacity initial population = C 1e k Togther with value of C 1 found above, this leads to With the values of k, C 1 found above, we solve for in (1) (2) k = ln C 1 (3) (t) = 5300C 1e kt 1 + C 1 e kt (4) (b) How long will it take for the population to increase to 2650 (half of the carrying capacity)? Solution. Set = 2650 in (1) 1 2 = C 1e kt = t = 1 k ln 1 2C 1 (5) Instructor s version of roblem 5 of Set 03 Modeling. (Notes for grader: regardless of different numbers, the answer should be similar to (6), (7), (8)) age 2

3 Another model for a growth function for a limited pupulation is given by the Gompertz function, which is a solution of the differential equation ( ) d = c ln where c is a constant and is the carrying capacity. (a) Solve this differential equation for c = 0.05, = 3000, and initial population 0 = 600. Solution. Separable equation. Upon rearrangement, it becomes d ln ( ) = c Integrate both sides 1 ln ( ) d = ct + D To integrate the LHS, let u = ln ( ) so that 1 du = d ( / 2 ) = 1 d. lug it back into the previous equation du u = ct + D that is In terms of, we have ln u = ct + D ln ln(/ ) = ct + D Thus, apply exponential on both sides ln(/ ) = D 1 e ct (6) where D 1 = ±e D. Now, apply the initial condition (0) = 600 to (6) ln(3000/600) = D 1 = D 1 = ln 5 Finally, by (6), the specific solution is (t) = 3000 e (ln 5e 0.05t ) (7) age 3

4 (b) Compute the limiting value of the size of the population. Solution. In (7), as t +, 0.05t Therefore e 0.05t e = 0 ln 5e 0.05t 0 e (ln 5e 0.05t) e 0 = lim (t) = = carrying capacity (8) t + 1 (c) At what value of does grow fastest? Solution. The growth rate of is d. On the other hand, by the DE, ( ) d = c ln Therefore, the target function we try to maximize is ( ) f( ) = c ln The problem asks at what value of we have the largest d, i.e. the largest f( ) = c ln ( ). To this end, we take the derivative of f( ) w.r.t. Set f ( ) = 0 above f 1 ( ) = c / [ c 1 + ln 2 + c ln ( )] = 0 = ln ( ) [ = c 1 + ln ( ) = 1 = ( )] = carrying capacity = (9) e e This the value of at which grows the fastest, i.e. f( ) reaches its maximum. (Note: more rigorously, f ( )=0 only guarantees a critical point; in addition, one would need to check the sign of f ( ) to determine whether this is a local max, a local min, or neither.) age 4

5 2. (20 ) It is the pollen season. Consider the air in an apartment with volume V. Let the internal concentration of pollen be x(t) and the ambient (external) concentration a(t). Assume, inside the apartment, pollen is evenly distributed so that the total amount of pollen is given by (t) = V x(t). Let air flow into and out of the apartment at a rate of r(t). a) How much air flows in and out during a time period? How much pollen is in and out, respectively? Solution. Air-in and air-out are both r(t). ollen-in is the ambient concentration times air-in : a(t)r(t). ollen-out is the internal concentration times air-out : x(t)r(t) b) Find the corresponding change in the total amount of pollen d (t). Solution. Change in the total amount of pollen is the difference between pollen-in and pollen-out d (t) = a(t)r(t) x(t)r(t) c) Write down a differential equation for the internal concentration x(t). Solution. Rate of change in the total amount of pollen is d (t)/ and by part ii), we have the equation d (t) = a(t)r(t) x(t)r(t). (10) Now, (t) is related to x(t) as (t) = V x(t) which is given in the problem. So, d/ = V dx/ and plug it into equation (10) and arrive at V dx(t) = a(t)r(t) x(t)r(t). (11) This is a closed equation that has only one unknown x(t) (whereas equation (10) has two: x(t) and (t).) d) Suppose a(t), r(t) are known. ropose an applicable method to solve for x(t). DO NOT actually solve it. age 5

6 Solution. Equation (11) is linear in x(t) and x (t), so we can use integrating factor. That is, rewrite the DE into a standard form and introduce x (t) + r(t) a(t)r(t) x(t) = V V µ(t) = e r(t) V and multiply the DE with µ. The LHS should become a perfect derivative and can be integrated easily. 3. (10 ) Suppose a population is a function of time t. The birth rate is 2 times the square of and the death rate is a constant 8. a) What is the differential equation that models the population dynamics? Is the equation linear or nonlinear? How many initial conditions is needed to determine a specific solution? Solution. (t) = Nonlinear. 1 initial condition needed. b) Without help of computers or calculators, sketch the slope field and several typical solutions of the differential obtained from previous problem. In particualr, plot 3 curves; one satisfying (0) = 1, one satisfying (0) = 2 and one satisfying (0) = 4. Is the population increasing or decreasing according to these 3 curves? (The skill of sketching slope fields by hand is required in this class). Solution. The solution with (0) = 1 is decreasing; the one with (0) = 2 remains constant; the one with (0) = 4 is increasing. 4. (20 ) a) Consider y (x) + 2y (x) 5y(x) = 0. Find two linearly independent solutions and compute the associated Wronskian. Solution. The characteristic equation Two roots Two linearly independent solution λ 2 + 2λ 5 = 0 λ 1 = 1 + 6, λ 2 = 1 6 y 1 = e ( 1+ 6)x, y 2 = e ( 1 6)x age 6

7 The Wronskian ( ) ( y 1 y 2 e λ 1x W [y 1, y 2 ] = det y 1 y 2 = det λ 1 e λ 1x = e λ 1x λ 2 e λ 2x e λ 2x λ 1 e λ 1x e λ 2x λ 2 e λ 2x ) thus W [y 1, y 2 ] = (λ 2 λ 1 )e (λ 1+λ 2 )x (12) lug in the values of λ 1,2, we have W [y 1, y 2 ] = 2 6e 2x b) Consider y (x) + 2y (x) + 2y(x) = 0. Find two linearly independent solutions and compute the associated Wronskian. There may be complex numbers in your solution, for example in the exponents. Solution. The characteristic equation λ 2 + 2λ + 2 = 0 Two roots λ 1 = 1 + i, λ 2 = 1 i Two linearly independent solution y 1 = e ( 1+i)x, y 2 = e ( 1 i)x The Wronskian can be computed in the same fashion as above. Indeed, the same formula (12) is applicable W [y 1, y 2 ] = (λ 2 λ 1 )e (λ 1+λ 2 )x lug in the values of λ 1,2 for this case, we have W [y 1, y 2 ] = 2ie 2x age 7