Math 0290 Midterm Exam

Transcription

1 ath 0290 idterm Exam JAKE IRRA University of Pittsburgh July 11, 2016 Directions 1. The purpose of this exam is to test you on your ability to analyze and solve differential equations corresponding to real-world applications, as well as your ability to interpret those results. The credit given for each problem will be proportional to the understanding and ability demonstrated by your solutions. 2. You are permitted to use atlab or athematica. If desired, you may use a word processor such as icrosoft Word or LaTeX to write solutions. 3. You are not permitted to use notes or the internet. 4. It is not required, nor desirable, for your work to include explanations of how you compute integrals or derivatives, how you simplify algebraic expressions, or how you solve algebraic equations or systems thereof. You may include the results of such computations without further explanation. 5. Since this is not a programming course, I am willing to help with rudimentary trouble-shooting of code, so feel free to raise your hand and ask questions.

2 Problem 1 [25 points] At time t = 0 (in seconds) a tank with volume V = 5 m 3 is full of salt-water with concentration c 0 = 15 kg/m 3 (so that the original amount of salt in the tank is A(0) = A 0 = 75 kg. Starting at this time, water flows into the tank at a rate of R = m 3 /s with concentration c in (t) = c 0 e kt where k = s 1. The well-mixed solution flows out of the tank at an equal rate R. (part a [10 bonus points]) Write the differential equation for the amount A(t) of salt in the tank. If you wish not to do this yourself, you may ask me and I will provide it for you. (part b [20 points]) Solve the differential equation and initial value problem. (part c [5 points]) When will the salt concentration reach 0.2 kg/m 3. Recall concentration equals amount of salt (in kg) divided by the tank volume (in m 3 ). (part d [5 bonus points]) Write, but don t solve, the differential equation for this problem if the outflow rate were R 2 = m 3 /s (so the tank were emptying). Solution (a) Begin with A (t) = rate in rate out = c in R c tank R = c 0 Re kt A R. Rewrite this: V (b) dt + R V A = c 0Re kt, A(0) = A 0 = 75 This is a first-order linear equation, with integrating factor e R V t. Integrate both sides. e R t V dt + R V er V t A = c 0 Re ( d dt [er V t A] = c 0 Re ( R V k)t R V k)t e R V t A = c R 0RV R kv e( V k)t + C Initial conditions give C = A 0 c 0RV R kv. Divide by er V t to obtain (c) Solving A(t) = = 1 kg, we find A = c 0RV R kv e kt + Ce R V t t = s = 5 hours 23 minutes 44 seconds

3 (d) Now V is not constant, but is given by V(t) = V 0 (R 2 R 1 )t Rearrange: dt = rate in rate out = c 0 Re kt A V R 2 dt = c 0R 1 e kt A V 0 (R 2 R 1 )t R 2 dt + R 2 V 0 (R 2 R 1 )t A = c 0R 1 e kt

4 Problem 2 [25 points] The vibrations of a guitar string subject to the vibrations of a tuning fork is modeled by the differential equation mx + kx = sin(ωt), x(0) = 0, x (0) = 0 where m = , k = 85 N/m, = m, and ω = k/m = (part a [15 points]) Solve the differential equation and initial value problem. (part b [10 points]) What will be the vibration amplitude at time t = 2 s? Solution (a) The homogeneous solution is x c (t) = C 1 cos(ω 0 t) + C 2 sin(ω 0 t) where ω 0 = k/m. For the particular solution we guess x p (t) = At cos(ω 0 t) + Bt sin(ω 0 t) Plug into the differential equation, noting that k mω 2 = 0 to obtain Combine x = x c + x p 2Bmω 0 cos(ω 0 t) 2Amω 0 sin(ω 0 t) = sin(ωt) A = = 2mω 0 2 mk, B = 0 x(t) = C 1 cos(ω 0 t) + C 2 sin(ω 0 t) 2 mk t cos(ω 0t) Initial conditions x(0) = 0 and x (0) = 0 yield C 1 = 0, C 2 = solution, then, is (b) The amplitude when t = 2 is Also an acceptable answer is x(t) = 2k cos(ω 0t) 2 mk t cos(ω 0t) t 2 mk = m 2mω 0 2 = 2k The final

5 t 2 mk = k

6 Problem 3 [25 points] (part a [15 points]) Find the general solution for x(t) for the differential equation x + 3x + 9x + 27x = 5e 3t (part b [10 points]) In an engineering application, you come across a differential equation you ve never seen before, that you wish to solve for x(t). t 2 x tx + 2x = 0 After perusing a Differential Equations reference, you find that this is called an Euler- Cauchy equation. The reference states that the solutions will be of the form x(t) = t r for some appropriate choices of r. Using this, find all solutions to the differential equation. (part c [5 bonus points]) Find a solution to part (b) satisfying the initial conditions x(1) = 1 and x (1) = 0. (part d [5 bonus points]) What technique would you use to solve the differential equation Solution (a) t 2 x + tx x = e 3t Complimentary homogeneous equation first: r 3 + 3r 2 + 9r + 27 = 0 gives roots r = 3 and r = ±3i. These yield the complimentary solution x c (t) = C 1 e 3t + C 2 cos(3t) + C 3 sin(3t) For the particular solution, we would guess x p (t) = Ae 3t, but we tack on a t: x p (t) = Ate 3t Plug in to find A: A = 5/18. The general solution, then, is (b) x(t) = 5 8 te 3t + C 1 e 3t + C 2 cos(3t) + C 3 sin(3t) We are guessing the solution x(t) = t r. We plug this into the differential equation to obtain t r + rt r + ( 1 + r)rt r Divide by t r and expand: ( 1 + r + (r 1)r)t r = 0 r 2 1 = 0 r = ±1 Thus the solutions are x(t) = t and x(t) = 1 t.

7 (c) Any linear combination of these two solutions will be a solution. x(t) = C 1 t + C 2 1 t Solve the initial conditions for C 1 and C 2 : C 1 = 1 2, C 2 = 1 2. (d) x(t) = 1 2 (t + 1 t ) Variation of parameters can be used to find a particular solution to this inhomogeneous differential equation.