MAT 311 Midterm #1 Show your work! 1. The existence and uniqueness theorem says that, given a point (x 0, y 0 ) the ODE. y = (1 x 2 y 2 ) 1/3

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1 MAT 3 Midterm # Show your work!. The existence and uniqueness theorem says that, given a point (x 0, y 0 ) the ODE y = ( x 2 y 2 ) /3 has a unique (local) solution with initial condition y(x 0 ) = y 0 as long as certain conditions are satisfied. Find all points (x 0, y 0 ) where the conditions are satisfied. (Show your work! You don t need to solve the differential equation.) The ODE is y = f(x, y) where f(x, y) = ( x 2 y 2 ) /3. f is continuous everywhere and f = y 2y( x2 y 2 ) 2/3 is continuous when x 2 +y 2. Therefore the ODE has a unique solution with initial condition y(x 0 ) = y 0 as long as x 2 0+y Let x be a solution to the ODE ẋ = t x 2 satisfying the initial condition x( ) = 0.5. (a) The figure shows a direction field (slope field) for the ODE. Graph the solution x on the figure, with as large a domain as possible. You don t need to copy the figure, just draw your graph on top of it.

(b) Estimate x(0) and x(0.5). x(0) 0, x(.5) 0.8 3. Consider the ODE where k is a constant. dt = x(k x2 ) (a) Sketch the phase line for the ODE for the case where k =.

2 (b) Estimate x(0) and x(0.5). x(0) 0, x(.5) Consider the ODE where k is a constant. dt = x(k x2 ) (a) Sketch the phase line for the ODE for the case where k =. Be sure to show the stationary points and indicate whether they are stable, semi-stable, or unstable. (b) Again assuming that k =, sketch some solution curves for the ODE in the t,x plane. (You don t need to solve the equation to do this.) You should sketch enough curve to show the general behavior of the solutions for any initial condition. Page 2

3 (c) Now letting k vary, say over the range k, sketch the bifurcation diagram for the ODE. Be sure to indicate clearly the value(s) of k where the bifurcation(s) occur. (d) Find the solution to the ODE satisfying the initial condition x(.3) = 0. (Hint: this does NOT require a lot of computation!) For maximum credit you should be able to do this without picking a particular value of k but if you do need a particular value of k then say so and tell me what value of k you used. The constant function x(t) = 0 for all t is the unique solution satisfying this initial condition. 4. Given the ODE (a) Find the general solution. + y2 e 3x = 0 This is a separable equation = y2 e 3x = e 3x y 2 Integrating we obtain y = 3 e3x + C Page 3

4 so y = 3 e3x + C. (b) Find the particular solution satisfying the initial condition y(0) = 6. 6 = + C. 3 Solving for C we obtain C = /6 so y = 3 e3x Given the ODE (a) Find the general solution. + 2xy = + x (Corrected solution: my former integrating factor was wrong, I left out the e). This is a linear equation. The integrating factor is e R 2x = e x2 (we don t need the C) so, multiplying through by e x2 we obtain After integrating d(e x2 y) e x2 y = = e x2 + x ex2 e x2 + x ex2 Rats! It s impossible to integrate the right hand side using elementary functions so all we can say is y = e x2 e x2 + x ex2 (b) Find the particular solution satisfying the initial condition y() = 2. Page 4

5 (Corrected.) Since we can t integrate the right hand side in part (a) about all we can say is that y = 2 + e x2 x eˆx2 + ˆx eˆx2 dˆx. 6. A 500 gal. mixing tank initially contains a solution that is 0% pure alcohol and 90% pure water. Then two valves are opened. One valve allows a solution that is 40% alcohol and 60% water to pour into the tank at a rate of 0 gal. per minute. The other valve allows the mixture in the tank to drain out at a rate of 0 gal. per minute. Assume that the solution in the tank is well-mixed, so that at any particular time the concentration of alcohol is the same throughout the tank. Set up a differential equation with initial conditions that enables you to predict the volume (in gallons) of alcohol in the tank at any time t 0 minutes after the valves are opened. You do not need to solve the differential equation. Let x(t) be the number of gallons of alcohol in the tank t minutes after the valves are open. At time t = 0, 0% of the liquid in the tank is alcohol so the initial condition is x(0) = (0.0)(500) = 50. Now (rate of change of x) =(rate at which alcohol enters tank) (rate at which alcohol leaves tank). The rate at which alcohol enters the tank is (0.40)(0) gallons of alcohol per minute. The concentration of alcohol in the tank is (x gallons alcohol)/(500 gallons solution) so the rate at which alcohol leaves the tank is ( x 500) (0) gallons per minute. Therefore the rate of change of x is ( x ) dt =(0.40)(0) (0) 500 =4 x 50. Page 5

(a) x cos 3x dx We apply integration by parts. Take u = x, so that dv = cos 3x dx, v = 1 sin 3x, du = dx. Thus

(a) x cos 3x dx We apply integration by parts. Take u = x, so that dv = cos 3x dx, v = 1 sin 3x, du = dx. Thus Math 128 Midterm Examination 2 October 21, 28 Name 6 problems, 112 (oops) points. Instructions: Show all work partial credit will be given, and Answers without work are worth credit without points. You