Math 392 Exam 1 Solutions Fall (10 pts) Find the general solution to the differential equation dy dt = 1

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1 Math 392 Exam 1 Solutions Fall (10 pts) Find the general solution to the differential equation = 1 y 2 t + 4ty = 1 t(y 2 + 4y). Hence (y 2 + 4y) = t y y2 = ln t + c. 2. (8 pts) Perform Euler s method with the given step size t on the given initial-value problem over the time interval given. Your answer should be given in the form of a table. Recall the formula for Euler s method is: y k+1 = y k + f(t k, y k ) t. = 2t + y, y(0) = 1, 0 t 1, and t = 0.5. k t k y k f(t k, y k ) ( 1) = (0.5) + ( 1.5) = (1) + ( 1.75) = 0.25 In general y k+1 = y k + f(t k, y k ) t. So y 0 = y(0) = 1, y 1 = y 0 + f(t 0, y 0 ) t = 1 1(0.5) = 1.5, and y 2 = y 1 + f(t 1, y 1 ) t = 3 0.5(0.5) = = (10 pts) Find the bifurcation values for = y2 ay and also draw the bifurcation diagram. Show your work. We begin by finding the equilibrium points. Then = 0 y2 ay = 0 y(y a) = 0. So the equilibrium points are y = 0 and y = a. Rightaway we notice that if a = 0 we only have one equilibrium point and if a 0 we have 2 equilibrium points. Therefore a = 0 is a bifurcation value. Next if a > 0 we have the following sign chart The phase line is then which remains similar for all a > 0. 1

2 Now if a < 0 we have the following sign chart The phase line is then which remains similar for all a < 0. bifurcation diagram is Hence the only bifurcation value is a = 0 and the 4. (10 pts) Find a particular solution to the differential equation + y = t + e t. We first find the homogeneous solution for = y. Then y h = ke = ke t. Since the nonhomogeneous part includes part of the homogeneous solution then a particular solution is of the form y p = at + b + cte t. So p = a + ce t cte t. Hence a + ce t cte t + at + b + cte t = t + e t a + b + at + ce t = t + e t. Thus a = 1, c = 1 and a+b = 0 b = a = 1. Therefore a particular solution is y p = t 1+e t. 2

3 5. ( 10 pts) Find the solution to the initial value problem = 2y + e5t, y(0) = 1. This is a first order linear non-homogeneous equation. We will use the integrating factor method. We first rewrite the differential equation: 2y = e5t. Then µ(t) = e 2 = e 2t. Hence the general solution is y = 1 e 2t e 2t e 5t = e 2t e 3t = e 2t ( e3t 3 + c) = e5t 3 + ce2t. Using the initial condition y(0) = 1 we have 1 = c c = 2. Therefore the solution to the 3 initial value problem is y = e5t e2t. 6. (10 pts) A 200-gallon tank initially contains a mixture of 150 gallons of sugar water containing 1 pound of sugar per gallon. Sugar is added at a rate of 3 pounds per minute. Suppose that the mixture is kept well mixed and that sugar water is draining out at rate of 2 gallons per minute. Write an initial value problem that models the amount of sugar in the tank. Simplify your answer, but do not solve the problem. Note that there is no liquid added in. Solution: Let Q(t) be the amount of sugar in lbs at time t. Then Q(0) = 1 1lbs 150gal = 150 gal gallons. The rate the sugar is coming in is: Rate In= 3 lbs min. The rate the sugar is coming out is given by: Rate Out=2 gal min Qlbs ygal = 2Q y, where y is the number of gallons of sugar water in the tank at time t. To compute y notice that y is changing at a constant rate and thus the equation of y is linear with respect to time. We have that y(0) = 150 and y(1) = = 148. Notice that no liquid is coming in. So the slope of the line is m = 2 and since y(0) = 150 then y = 2t Therefore the initial value problem that models the amount of sugar in the tank is given by dq = 3 2Q dq, Q(0) = 150 or 2t = 3 Q, Q(0) = 150. t

4 7. (15 pts) Consider the differential equation = f(y), where the graph of f(y) is given below. (a) (8 pts) Find the equilibrium points, sketch the phase line for this equation and classify the equilibrium points as sources, sinks or nodes. The equilibrium points satisfy = 0 f(y) = 0. Hence y = 2, 1, 3 are equilibrium points. The sign chart is and thus the phase line is 4

5 (b) (7 pts) Draw a sketch of the solution to this differential equation with initial value y(1) = 0 and discuss the long term behavior of this solution. Let y(t) be the solution of the initial value problem. Then y(t) 2 as t. 8. (12 pts) Consider the differential equation = y2. (a) (6 pts) Show that y 1 (t) = 1 1 t and y 2(t) = 1 are both solutions to = y2. 2 t Notice that y 1 = 1 ( ) 2 1 (1 t) = = y and (1 t) y 2 = 1 ( ) 2 1 (2 t) = = y 2 2 (2 t) 2. Therefore both y 1 and y 2 satisfy the differential equation and thus are solutions. 5

6 (b) (6 pts) What can you say about solutions of = y2 for which the initial condition y(0) satisfies 1 < y(0) < 1? 2 Notice that the function f(y) = y 2 is continuous everywhere and that f = 2y is also continuous everywhere. Therefore the hypothesis of the Existence and Uniqueness Theorem y are satisfied. Now, y 1 (0) = 1 1 = 1 and y 2(0) = 1 2. Thus y 2(0) < y(0) < y 1 (0) and by the Uniqueness Theorem we have that y 2 (t) < y(t) < y 1 (t) for all t. 9. (15 pts) Six differential equations and three phase lines are given below. Determine the equation that corresponds to each phase line and state briefly how you know your choice is correct. (i) = y2 y 1, (ii) (vi) = y y2. = y 1 y, (iii) = y2 y, (iv) = y2 2y, (v) = y3 y, (a) (b) (c) (a): We have 2 equilibrium points y = 1 and y = 0. Hence the options are (i), (ii), (iii), and (vi). Examining the sign charts for each one of them we see that (i) is always positive, so it can t be that one. Also (vi) has opposite signs than what we were given. Equation (ii) has a positive sign between 0 and 1 instead of a negative as given. Morevoer, (iii) has the right sign chart and thus equation (iii) has phase line (a). (b): The only equation that has 3 equilibrium points is (v) since y 3 y = y(y 2 1) = y(y 1)(y + 1). Moreover, the sign chart for (v) agrees with the phase line (b) and hence (v) has phase line (b). (c): We have 2 equilibrium points y = 1 and y = 0. From the analysis we did in (a) we know (ii) is the equation that has positive sign between y = 0 and y = 1 and it is the only one like that. Hence equation (ii) has phase line (c). 6