# Homework 03

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Math 22B: Homework 3: Part A Due: before August 23 at 11:59 pm Submit on Gradescope (Make sure to select pages!) 1. Solve the following Bernoulli differential equations: (a) y = 5 y + e 2 x y 2 (b) dy dx y x = y 9 (c) dy dx + 2 y x = x 2 y 2 sin( x ) (d) dy dx 2 y x = x 2 y 2 sin( x ) (e) x dy dx (1 + x ) y = xy 2 (f) dy dx = y ( xy 3 1) 2. The following differential equations are homogeneous. Express them in the form dy dx = G ( y x ). Then, make the substitution v = y x and solve them as linear DE. (a) dy dx = x +2 y x y (b) dy dx = x 2 + y 2 2 x 2 (c) dy dx = y ( x y ) x 2 (d) x 2 dy + y ( x + y ) dx = 0 3. Find and classify the equilibrium solutions of the following DE: (a) y = y 2 y 6 (b) y = ( y 2 4)( y + 1) 2 4. For each of the following, sketch the slope field and phase line, then identify the equilibrium solutions and classify them asymptotically. (a) dy dt = 3( y 1) 2 (b) dy dt = 3 y 7 y (c) dy dt = y 2 ( y 2 1) (d) dy dt = y 3 3 y 2 + 2 y 5. Does the following IVP necessarily have to have a unique solution with the given initial conditions? dy dx = x p y 3 (a) y (4) = 3 (b) y ( 2) = 28 6. Consider the IVP y = 10 3 xy 2 / 5 , y (0) = 1 . It ha a unique solution on some open interval containing x 0 = 0. Find the solution and determine the largest open interval on which it is unique. 1
7. Suppose that the populations y of a certain species of fish in a given area of the ocean is described by the logistic equation dy dt = r (1 y K ) y. Although it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level and possibly even driven to extinction. The next few problems explore some of the questions involved in formulating a rational strategy for managing the fishery. (a) At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population y : the more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by Ey , where E is a positive constant, with units of 1/time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is modified and replaced by dy dt = r (1 y K ) y Ey. This equation is known as the Schaefer model after the biologist M. B. Schaefer, who applied it to fish populations. i. Show that if E < r , then there are two equilibrium points, y 1 = 0 and y 2 = K (1 E/r ) > 0. ii. Show that y 1 is unstable and y 2 is asymptotically stable. iii. A sustainable yield Y of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort E and the asymptotically stable population y 2 . Find Y as a function of the effort E ; the graph of this function is known as the yield-effort curve. iv. Determine E so as to maximize Y and thereby find the maxi- mum sustainable yield Y m . (b) In this problem we assume that fish are caught at a constant rate h independent of the size of the fish population. Then y satisfies dy dx = r (1 y/K ) y h. The assumption of a constant catch rate h may be reasonable when y is large but becomes less so when y is small. i. If h < rK/ 4, show we have two equilibrium points y 1 and y 2 with y 1 < y 2 ; determine these points. ii. Show that y 1 is unstable and y 2 is asymptotically stable. iii. From a plot of f ( y ) versus y , show that if the initial population y 0 > y 1 , then y y 2 and t → ∞ , but that if y 0 < y 1 , then y decreases as t increases. increases. Note that y = 0 is not an equilibrium point, so if y 0 < y 1 , then extinction will be reached in a finite time. 2
iv. If h > rK/ 4, show that y decreases to zero as t increases, regardless of the value of y 0 . v. If h = rK/ 4, show that there is a single equilibrium point y = K/ 2 and that this point is semistable (In cases where solutions on one side of an equilibrium solution move towards the equilibrium solution and on the other side of the equi- librium solution move away from it we call the equilibrium solution semi-stable). Thus the maximum sustainable yield is h m = rK/ 4, corresponding to the equilibrium value y = K/ 2. Observe that h m has the same value as Y m in Part (a). The fishery is considered to be overexploited if y is reduced to a level below K/ 2. Note: this last one is from the textbook: section 2.5 problems 19 and 20. 3