APPM 60 Exam1 Spring 01 Problem 1: (0 points) Consider the following initial value problem where y is the radius of a raindrop and y = y / ; y 0 (1) y(0) = 1, () models the growth of the radius (as water condenses out of the air, the droplet gets bigger). (a) [5] Does this initial value problem (1)-() have a unique solution? Why or why not? (b) [5] Consider the differential equation in (1) with a different initial condition y() = 0. Can we guarantee that this initial value problem has a unique solution? Why or why not? (c) [10] Find all the solutions to the differential equation (1) with the boundary condition from (b) y() = 0. (Hint: Don t forget that only solutions with y 0 make sense for this differential equation.) (a) Yes. The function f(t, y) = y / and its partial derivative f y (t, y) = y 1/ are both continuous at t = 0 and y = 1 and so satisfy the conditions for Picard s theorem. (b) No information. At y = 0, Picard s theorem indicates the existence of a solution, but does not tell us if the solution is unique or not. (c) Since we can t guarantee only one solution, we ll search for a family of solutions, all of which satisfy y() = 0. First we separate variables y / = dt and apply the initial condition y 1/ = t + C ( 1 t + C y(t 0 ) = 0 = ( ) t0 + C where t 0 will have some relationship (,, etc.) to. This leads to C = t 0 and a solution of ( ) t t0. However, for any t < t 0, the solution is zero and thus the full solution is ) for any t 0. {( t t0 ) t t 0 0 t < t 0
Problem : (0 points) (a) [10] Solve the following initial value problem using the integrating factor method. (1 + 4t ) dt + 4ty = t 1 + 4t ; y(0) = 1 (b) [10] Find the general solution of the following differential equation using the method of Euler-Lagrange (i.e. variation of parameters). (a) y + y t = 1 t ; t > 0 (1 + 4t ) dt + 4ty = t 1 + 4t y + 4t 1 + 4t y = t/ 1 + 4t Using the integrating factor method, p(t) = 4t. Solving for the integrating factor, 1+4t we will use the following substitution, u = 1 + 4t. 4t 1 + 4t dt = 1 1 u du = 1 ln u = 1 ln 1 + 4t = 1 ln(1 + 4t ) = ln( 1 + 4t ) Thus, the integrating factor is exp(ln( 1 + 4t )) = 1 + 4t. 1 + 4t (y + 4t 1 + 4t y) = (t/ 1 + 4t ) 1 + 4t d dt (y 1 + 4t ) = t d dt (y 1 + 4t ) = t Solving for the initial condition: Thus, t +1 1+4t y 1 + 4t = t + C (t + C)/ 1 + 4t y(0) = 1 = C C = 1
(b) y + y t = 1 t () (i) Variation of parameters method y + 1 t y = 1 t, t > 0 We first solve the homogeneous problem, then allow the constant of integration to vary with time to find a particular solution. The homogeneous problem is: y h + 1y t h = 0. 1 y = 1 t dt ln y = ln t + c 1 ( ) 1 y h = C t Now, we either derive the formula for v(t) or use the formula directly (without derivation) to find y p (t). y p = v(t) 1 ( t y p(t) = v(t) 1t ) ( ) 1 + v (t) t ( Then, v(t) 1t ) ( ) 1 + v + 1 t t v(t)1 t = 1 t, by plugging y p(t) into the ODE Then, v (t) = 1 v(t) = t ( ) 1 y p (t) = t = 1 t y h + y p so, C t + 1 Problem : (0 points) Suppose a colony of cells grows exponentially, doubling its weight every day. (a) [5] Let y 0 e kt be the weight of the cells (grams) at time t (days). Find the constant k. (The answer here may be left in terms of elementary functions, such as 80 sin() or e.) (b) [10] Suppose now that, in addition to cells reproducing, 1 gram of cells is continuously drained out every hour. Write down a differential equation for y(t), where t is in units of days (remember to express all quantities in units of days). (c) [5] Solve the equation created in (b) and find y(t), if the weight of the colony at time t = 0 is 40 grams.
(a) The doubling time for the colony is one day. Therefore, we know that (b) The differential equation is y 0 e k 1 = y 0 (4) e k = (5) k = (6) = y 4 (7) dt (c) The quickest way to solve this equation is via separation of variables = dt (8) y 4 y 4 = dt (9) ln y 4 = t + C 1 (10) y 4 = C e t (11) C e t + 4 (1) C t + 4 (1) for unknown constant C. And, from the initial condition we obtain y(0) = 40 = C 0 + 4 C = 40 4 and thus the full solution is (14) (15) ( 40 4 ) t + 4. (16) Problem 4: (0 points) True/False. If the statement is always true, mark true. Otherwise, mark false. You do not need to show your work. (Any work will not be graded.) (a) [4] Suppose y 1 and y are both solutions to a linear, nonhomogeneous differential equation. Then y = y 1 + y is also a solution to the same differential equation. (b) [4] Let y(t) be a solution of the differential equation, y = r ( 1 y L ) y; r, L > 0. If y(0) > 0, then lim t L. (c) [4] For some first-order, linear differential equations, the solutions obtained by the Euler-Lagrange method and the integrating factor method are different. (d) [4] The following differential equation is separable. y ty + t = y 1
(e) [4] Let L be a linear operator and suppose y 1 and y are both solutions to the differential equation L(y) = f(t). Then (y 1 y ) is a solution of L(y) = 0. (a) False. The equation is nonhomogeneous, therefore the superposition property for linear homogeneous equations (page 59) does not apply. (b) True. Solutions of the logistic equation approach the carrying capacity for all initial conditions y 0 > 0. (c) False. The Euler-Lagrange method and the Integrating Factor method will result in the same solutions. (d) True. This equation is separable. To see, consider the following steps y 1 y = ty t + y 1 (17) y = t(y 1) + y 1 (18) y = (t + 1)(y 1) (19) = (t + 1)dt (0) (e) True. Since y 1 and y are both solutions, this means that L(y 1 ) = f(t) and L(y ) = f(t). Since L is linear, L(y 1 y ) = L(y 1 ) L(y ) = f(t) f(t) = 0. Problem 5: (0 points) Consider the following differential equation, which contains the free parameter a. For parts (a) and (b), set a =. dt = (y 1)(y a) (a) [6] Determine all the equilibrium points of the system. (b) [9] Draw a direction field in the ty-plane for this differential equation and use your picture to determine the stability of any equilibria found in part (a). (c) [5] Consider the parameter a. A bifurcation point is the specific value of a parameter (say a = ã) at which the solutions to a differential equation undergo a dramatic qualitative change. That is, solutions for a < ã are very different than solutions for a > ã. What is the bifurcation point for a? (a) Equilibria occur when y = 0. In this case, at y = 1 and y =. (b) The equilibrium solution y = is unstable because nearby solutions tend away from y =, and the solution y = 1 is stable because nearby solutions tend towards y = 1 as t increases.
4 1 0 1 1 0 1 (c) The differential equation undergoes a qualitative change when a = 1. In particular, the system always has equilibria at y = 1 and y = a. When a < 1, y = a is a stable equilibrium and y = 1 is unstable. when a > 1, y = a is unstable and y = 1 is stable. When a = 1, the equilibria coincide and so the system has a single semi-stable equilibrium.