7 Questions on Introduction to Financial Accounting - Exam 1 | MATH 3113, Exams of Mathematics

Download 7 Questions on Introduction to Financial Accounting - Exam 1 | MATH 3113 and more Exams Mathematics in PDF only on Docsity! Mathematics 3113-005 Exam I Form B February 18, 2011 Name (please print) Instructions: Give concise answers, but clearly indicate your reasoning. Most of the problems have rather short answers, so if you find yourself involved in a lengthy calculation, it might be a good idea to move on and come back to that problem if you have time. I. (16) For each of the following first-order DE’s, carry out a substitution to put the DE into a form that can be solved by either separation of variables or the method for linear equations, and simplify. If it is separable, write it as an equality of a differential of v and a differential of x (that is, up to the step where you are about to integrate both sides), and if it is linear, find the integrating factor and multiply through to make the left-hand side a derivative. In either case, do not continue on from there to find the solution. (a) 4xy′ + 8y = √ y/x The DE is Bernoulli with n = 1/2, so we will substitute v = y1−1/2 = √ y. We have v′ = y′/2 √ y, so y′ = 2 √ yv′ = 2vv′. Substituting, we have 4xy′ + 8y = √ y/x 8xvv′ + 8v2 = v/x v′ + v/x = 1 8x2 which is linear. An integrating factor is ρ(x) = e ∫ 1 x = eln(x) = x, and multiplying through by ρ(x) gives xv′ + v = 1 8x . (b) xy′ = y + √ x2 − y2 It is not Bernoulli and does not involve a linear expression in x and y, so let’s see if it is homogeneous. xy′ = y + √ x2 − y2 y′ = y x + √ x2 − y2 x = y x + √ x2 − y2 x2 = y x + √ 1− y 2 x2 = y x + √ 1− (y x )2 So it is homogeneous. Putting v = y x , we have y = xv, y′ = v + xv′, and we substitute to obtain v + xv′ = v + √ 1− v2 x dv dx = √ 1− v2 1√ 1− v2 dv = 1 x dx . II. (5) (a) Give a definition of an (ordinary) differential equation. An equation involving an unknown function and its derivatives. [Any reasonable definition is accept- able, as long as it mentions that the equation involves derivatives and indicates that the unknowns (that is, the solutions) are functions.] (b) Define the order of a differential equation. It is the highest order of derivative of the unknown function that appears. (c) Give the general form (not a specific example) of a first-order initial value problem (of the kind that appears in the first Existence and Uniqueness Theorem). y′ = F (x, y), y(a) = b Page 2 III. (4) For the first-order linear homogeneous DE y′ +P (x)y = 0, verify that if y1 and y2 are solutions, then so is Ay1 +By2 for any constants A and B. Since y1 and y2 are solutions, y ′ 1 + P (x)y1 = 0 and y ′ 2 + P (x)y2 = 0. Testing Ay1 +By2, we find (Ay1 +By2) ′ + P (x)(Ay1 +By2) = Ay ′ 1 +By ′ 2 + P (x)Ay1 + P (x)By2 = A (y′1 + P (x)y1) +B (y ′ 2 + P (x)y2) = 0 . IV. (10) A small lake contains 100 billion liters of water, contaminated with a pollutant at 1% concentration. Starting at time t = 0, four billion liters per day of water with the pollutant at 5% concentration enters, mixes instantaneously with all the water in the lake, and the same amount of mixed water leaves. We want to know how many days it will take until the level of contamination reachs 3%. Let x(t) denote the volume of pollutant in the lake, in billions of liters. (a) What is x(0)? In terms of x(t), what does the problem ask us to find? x(0) = (0.01) · 100 = 1. The problem asks us to find t0 for which x(t0) = 3. (b) Write an expression for the volume of pollutant that enters between t and t+ ∆t. (0.05) · 4 = 0.2 billion liters of pollutant enter per day. During a time interval ∆t, 0.2 ∆t billion liters enter. (c) Write an expression for the approximate volume of pollutant that leaves between t and t+ ∆t. The concentration at time t is x(t)/100, so during a time interval ∆t, approximately 3 · (x(t)/100) ·∆t leaves. (d) Use (b), (c), and first-semester calculus to set up a first-order linear initial value problem that models the problem, but do not go on to solve it or attempt to find the answer to the problem. From (b) and (c), ∆x ≈ 0.2 ∆t−3 ·(x(t)/100) ∆t, so ∆x ∆t ≈ 0.2−3x(t)/100. Taking the limit as ∆t→ 0, we obtain dx dt = 0.2− 3x/100. The initial value problem is x′ + 3x/100 = 0.2, x(0) = 1 .