APPM 2360: Section exam 1 7:00pm 8:30pm, February 12, 2014.

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1 APPM 2360: Section exam 1 7:00pm 8:30pm, February 12, ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your student ID number, (3) recitation section (4) your instructor s name, and (5) a grading table. Text books, class notes, and calculators are NOT permitted. A one-page crib sheet is allowed. Start each problem on a new page. Problem 1: (30 points) Consider the following differential equation dy = (1 y2 ) 1/3 1 t (a) Answer the following questions for the DE above: (i) What is the order? (ii) Is it linear or nonlinear? (iii) Are there any equilibrium solutions? Identify them if there are any. (iv) Is it homogeneous? (v) Is it autonomous? (vi) Is it separable? (b) Sketch the direction field in the first quadrant of the t-y plane (t 0, y 0) and determine the stability of all equilibrium solutions. (c) Using Picard s theorem, what can you say about the existence and uniqueness given an initial value of y(0) = 1? Solution 1: (a) (i) It is a first order DE. (ii) It is nonlinear. (iii) There are two equilibrium solutions: y(t) = 1 and y(t) = 1. (iv) It is not homogeneous. (v) It is not autonomous. (vi) It is separable. (b) See figure further down. The solution y(t) = 1 for t < 1 is a stable solution. The solution y(t) = 1 for t > 1 is an unstable solution. (c) Using Picard s theorem, what can you say about the existence and uniqueness given an initial value of y(0) = 1? In a small region around (t, y) = (0, 1) the function f(t, y) = (1 y2 ) 1/3 1 t is well defined and continuous which gives us existence by Picard s theorem. Taking a derivative with respect to y gives: df dy = 1 2y 3 (1 y 2 ) 2/3 (1 t) which is undefined for y = 1. This means that Picard s theorem does not guarantee uniqueness.

Problem 2: (30 points) Consider the following two problems (a) Consider the differential equation y 6t 2 y = e 2t3 cos(πt). (i) Find the homogeneous solution to this differential equation.

2 Problem 2: (30 points) Consider the following two problems (a) Consider the differential equation y 6t 2 y = e 2t3 cos(πt). (i) Find the homogeneous solution to this differential equation. (ii) Use variation of parameters to find a particular solution. (iii) What is the general solution to this differential equation? (iv) Given y(1) = 1, what is the solution to the initial value problem? (b) Consider a Ricatti equation given by Solution 2: y + 2ty = 1 + t 2 + y 2. (i) Show that y 1 (t) = t is a solution to the Ricatti equation. (ii) Show that the substitution y(t) = t + (1/v(t)) transforms the Ricatti equation into a first order linear differential equation in v. (You are NOT required to solve the differential equation.) (a) Solution to part (a) (i) The solution to the homogeneous equation is given by y 6t 2 y = 0 dy y = 6t 2 ln y = 2t 3 + C 1 y h (t) = Ce 2t3 (ii) Let y p (t) = v(t)e 2t3. Differentiating yields y p(t) = v (t)e 2t3 + 6t 2 v(t)e 2t3. Substitute y p and y p into the original differential equation to obtain v (t)e 2t3 Solving for v yields v(t) = (1/π) sin(πt). Finally, y p (t) = (1/π) sin(πt)e 2t3. (iii) The general solution is then y(t) = y h (t) + y p (t) = Ce 2t3 + (1/π) sin(πt)e 2t3 = e 2t3 cos(πt). (iv) The initial condition y(1) = 1 yields C = e 2. Therefore, the solution to the IVP is y(t) = e 2t3 2 + (1/π) sin(πt)e 2t3

3 (b) Solution to part (b). (i) To verify that y 1 (t) = t satisfies the differential equation, note that y 1 (t) = 1. Substituting into the left hand side of the differential equation gives y 1 + 2ty 1 = 1 + 2t 2. Substituting into the right hand side gives 1 + t 2 + y1 2 = 1 + t 2 + t 2. Since they are equal, y 1 (t) = t is a solution. (ii) Let y(t) = t + (1/v(t)). Then, y = 1 1 v 2 v. Substitute this into the original DE yields: 1 1 ( v 2 v + 2t t + 1 ) ( = 1 + t 2 + t + 1 ) 2 v v Simplifying yields the linear differential equation v (t) = 1. Problem 3: (30 points) Answer TRUE OR FALSE. You do NOT need to justify your answer. (a) y = et+y y is a separable ODE (b) y + e ln y = sin t is a linear ODE for y > 0 (c) Solutions to the ODE y = t/y lie on circles in the t-y plane (d) Picard s Theorem for existence and uniqueness holds for the IVP y = y 1/2, y(0) = 1. (e) Let L[y 1 ] = f(t) and L[y 2 ] = f(t)/2 denote two linear differential equations where L[y] denotes a linear operator. Then L[3y 1 + 4y 2 ] = 5f(t). (f) Euler s method for the IVP y = g(y), y(0) = y 0 is y n+1 = y n + hg(y n ), t n+1 = (n + 1)h Solution 3: (a) True (b) True (c) True (d) True (e) True (f) True Problem 4: (30 points) Long ago, Atlantis, a small village of 100 inhabitants was built at the foot of a volcano. 100 years later, the population was 500 inhabitants. Unfortunately, by the time Atlantis grew to inhabitants the volcano erupted, destroying and burying the city. Luckily, everybody could be evacuated. (a) Assuming exponential growth for the population P, what IVP describes the evolution of P before the eruption? (b) What is the growth constant? (c) How many years does it take for the population P to reach inhabitants? Now, the town was rediscovered during archaeological excavations in the year The concentration in C 14 of trees killed during the eruption was measured to be only 1 16th of the amount of a living tree. C 14 -dating assumes that the concentration of C 14 is described by exponential decay. The half-life for C 14 is 5600 years. (d) What is the decay constant for C 14 -dating? (e) How old are the dead trees according to the C 14 dating? (f) In what year did the first 100 inhabitants build Atlantis? Solution 4: (a) dp = kp, with P (t 0) = = 100

4 (b) At t 1 the solution is P 1 = e kt 1 Dividing though by and taking the logarithm P1 ln = kt 1 k = ln (c) At t 1 the solution is P 2 = e kt 2 ( P1 ) 1 = ln(5) t Dividing though by and taking the logarithm P2 P2 1 ln = kt 2 t 2 = ln k = ln(53 ) 100 = 300 years ln 5 (d) The equation for the half-life t 1/2 is 1 2 Q 0 = Q 0 e kt 1/2 Dividing though by Q 0 and taking the logarithm ln(2) = kt 1/2 k = ln(2) t 1/2 = ln(2) 5600 (e) The equation for t 3 is 1 16 Q 0 = Q 0 e kt 3 Dividing through by Q 0 and taking the logarithm ln(16) = kt 3 t 3 = ln(16) k = ln(24 ) 5600 = year ln(2) (f) The year t 0 is given by t 0 = 2000 t 3 t 2 = = Problem 5: (30 points) A large truck carrying Chemical Z crashes off the edge of a highway and begins spilling the Chemical into nearby Happy Duck pond. Thankfully, the driver escapes unharmed. A small dam at one end of the pond allows water to flow out of the pond at a rate of 10 liters/minute (L/min). The concentration of Chemical Z within the holding tank is 2 grams/liter (g/l) and the driver observes the flow rate of Chemical Z into the pond to be 20 L/min. The volume of the pond prior to the spill is 1000 liters. Assuming that Chemical Z is well-mixed in the pond, answer the following questions: (a) Let V be the volume of water within the pond and write down the equation describing the volume of water within the pond as a function of time. (b) Let M be the amount (mass, in units of grams) of Chemical Z in the pond. Write down the initial value problem describing the amount of Chemical Z in the pond. (c) Are there any equilibrium (i.e. do not depend on time) solutions to the initial value problem obtained in part (b)? Show why or why not. (d) Solve the initial value problem obtained in part (b).

5 Solution 5: (a) The differential equation would be dv = FLOW RATE IN FLOW RATE OUT dv = = 10. Integrating gives V = t (b) The rate of change of Chemical Z is given by dm = RATE IN RATE OUT dm M = 2(20) t (10). Then the initial value problem is dm 1 + M = 40, M(0) = 0. (c) Set dm/ = 0 to get M = 40() This solution is not constant in time and therefore is not an equilibrium solution. (d) Solving by integrating factor we have the equation becomes With general solution M(t) = µ = e t = d (µm) = 40µ 1 [20 (200 + t) t + c 1 ] Using the initial value we have c 1 = 0 such that M = 20(200 + t)t

APPM 2360: Midterm exam 1 February 15, 2017

APPM 2360: Midterm exam 1 February 15, 2017 APPM 36: Midterm exam 1 February 15, 17 On the front of your Bluebook write: (1) your name, () your instructor s name, (3) your recitation section number and () a grading table. Text books, class notes,