Math 2250 Lab 08 Lab Section: Class ID: Name/uNID: Due Date: 3/23/2017

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1 Math 2250 Lab 08 Lab Section: Class ID: Name/uNID: Due Date: 3/23/2017 TA: Instructions: Unless stated otherwise, please show all your work and explain your reasoning when necessary, as partial credit will be given where appropriate. You are allowed to use any results from lectures as long as they are stated correctly. You are encouraged to work in groups, but your final written solutions must be in your own words. 1

2 1. Characteristic polynomial We ve been studying how roots of the characteristic polynomial for a constant coefficient homogeneous linear differential equation lead to basis functions for the solution space to the DE (section 5.3). The one case we haven t covered in detail yet is when the characteristic polynomial has complex roots r = a ± bi. In this case the two complex function solutions e (a+bi)x, e (a bi)x yield the two real function solutions e ax cos(bx), e ax sin(bx). If the characteristic polynomial has repeated complex root factors (r (a + bi)) 2 (r (a bi)) 2, then we also get solutions etc. xe ax cos(bx), xe ax sin(bx), (a) Find the general solution y(x) to the differential equation y (3) 5y + 3y + 9y = 0 Hint: Find a real root r 1 of the cubic characteristic polynomial then divide the characteristic polynomial by r r 1 to get the quotient quadratic, which will factor easily. 2

3 (b) Find the general solution x(t) to the differential equation x (t) + 4x (t) + 29x(t) = 0. Hint: completing the square to get the roots of the characteristic polynomial works well here - probably better than the quadratic formula. 3

4 2. From basis to differential equation In each of the following problems you are given a set of basis functions for the general solution to a homogeneous, linear DE with constant coefficients. Work backwards to determine the DE (in unknown function y = y(x)) which they solve. 1. The number of basis functions corresponds to the order of the differential equation (and dimension of the solution space). 2. You can determine all characteristic equation roots and their multiplicities from the basis functions. 3. From the characteristic equation, you can write the DE. (a) {e x, xe x, x 2 e x } (b) {e 2x, e x cos(2x), e x sin(2x)} 4

5 (c) {1, x, x 2, x 3 } (d) {e 2x cos(x), e 2x sin(x), xe 2x cos(x), xe 2x sin(x)} 5

6 3. Euler equation A second-order Euler equation is one of the form ax 2 y + bxy + cy = 0, (1) where a, b, and c are constants. The domain for this differential equation is x > 0, so that the coefficient function in front of y stays non-zero. (a) Show that if x > 0, then the substitution u = ln x transforms Eq(1) into the constant-coefficient, linear differential equation a d2 y + (b a)dy + cy = 0, (2) du2 du with independent variable u. We have learned techniques to solve such constantcoefficient, linear, homogeneous differential equations. Hint: Use d du = d dx dx du (Chain Rule) to transform derivatives of y with respect to x to derivatives of y with respect to u. Take care in doing this for the second derivative. (b) Write the characteristic equation corresponding to (2). 6

7 (c) What condition on a, b, and c guarantees that the roots of the characteristic equation (2) are real and distinct? (d) Find a general solution for x > 0 of the Euler equation 4x 2 y + 8xy 15y = 0. 7

8 4. Where the magic algorithms of 5.3 come from Let D represent differentiation with respect to x, that is D = d, dx D2 = d2, etc. Let dx 2 a 1, a 0 be scalars. Use I for the identity operator, I(y) = y. Then we can rewrite our familiar second order linear operator L as a linear combination of the operators I, D, D 2 (see page 341 of our text): Ly = y + a 1 y + a 0 y = D 2 (y) + a 1 D(y) + a 0 I(y) = [ D 2 + a 1 D + a 0 I ] (y). So, we may rewrite the homogeneous DE y + a 1 y + a 0 y = 0 as [ D 2 + a 1 D + a 0 I ] (y) = 0. If the characteristic polynomial p(r) = r 2 + a 1 r + a 0 factors as p(r) = (r a)(r b), then the operator L factors as a composition of linear first order operators (in either order): D 2 + a 1 D + a 0 I = [D ai] [D bi] = [D bi] [D ai]. For distinct roots a b the basis functions are y 1 = e ax and y 2 = e bx because [D ai]e ax = 0 and [D bi]e bx = 0. (a) Verify that [D ai]e ax = 0. 8

9 (b) Now suppose instead that the characteristic polynomial has a double root a = b, i.e. p(r) = (r a) 2. Then L = D 2 + a 1 D + a 0 I = [D ai] [D ai]. Show that [D ai]xe ax = e ax. (c) Use part (a) and composition of operators to deduce that [D ai] [D ai]xe ax = 0 (This explains why for double roots our homogeneous DE solution space basis is {e ax, xe ax }.) (d) Show that for any differentiable function f(x), [D ai]f(x)e ax = f (x)e ax. (e) Explain using (d) why if the operator L of degree at least three has a factor [D ai] [D ai] [D ai] = [D ai] 3 Then the three functions e ax, xe ax, x 2 e ax all solve L(y) = 0. 9

10 5. Buoy Consider a floating, cylindrical body with radius r, height h, and uniform density ρ <= 0.5 (recall that the density of water is 1 g/cm 3 ). The buoy is initially suspended at rest with its bottom at the top surface of the water and is released at time t = 0. Thereafter, it is acted on by two forces: a downward gravitational force equal to its weight mg = (ρπr 2 h)g and (by Archimedes principle of buoyancy) an upward force equal to the weight, (πr 2 x)g of water displaced, where x = x(t) is the depth of the bottom of the buoy beneath the surface at time t as indicated in the figure above (positive axis is downward). (a) Derive the differential equation satisfied by the depth function x = x(t). Write your DE in a form so that the only constants that appear are ρ, h, and g. Do not leave any m, π or r appearing in your DE. (b) Write the two initial conditions. 10

11 (c) Let s take a trial particular solution of the form x p (t) = C eq, where C eq is some constant. Find C eq. This constant solution gives the equilibrium position of the buoy. Refer to Theorem 5 on pg 335 for additional details on solving non-homogeneous DEs. (d) Now solve the associated homogeneous DE corresponding to the DE in part (a). You are seeking a general solution. (e) Summing the general solution from part (d), x h (t), with the particular solution x p (t) yields the general solution to the nonhomogeneous DE in part (a). Use the initial conditions to solve the initial value problem. What type of motion does the buoy undergo about its equilibrium position? Find lim t x(t) or show that it does not exist. 11

12 (f) Show that the period is p = 2π ρh/g. (g) Compute the amplitude of the motion if ρ = 0.5 g/cm 3 and h = 100 cm. (h) Now suppose you alter the buoy system so that your DE from part (a) has the additional term + c d x on the left-hand side (where you have written the DE in part (a) as instructed and written it so that the left-hand side features only terms involving x and the right-hand side features only a constant), where c d > 0 is a damping constant with units 1/s. Suppose c d = 9 s 1, ρ = 0.5 g/cm 3, h = 100 cm, and g = 980 cm/s 2. The particular solution found in (c) remains unchanged. Now re-solve the initial value problem (notice your associated homogeneous DE changes). Find lim t x(t) or show that it does not exist. How is this altered buoy system different from the original buoy in terms of the way it behaves for large time values? 12

13 6. Energy in a mass-spring damper Let x(t) be the position of a mass m attached to a spring with Hooke s constant k and damping piston with constant c, yielding the differential equation mx + cx + kx = f(t), where f(t) is an external forcing on the mass. We wish to account for the total energy of the mass-spring configuration, neglecting the heat energy loss due to damping. We define the total energy E(t) to be the sum of kinetic and potential energy. Potential energy P E(t) is stored by the compressed or stretched spring, and is the work done to stretch/compress it as the mass moves from from equilibrium x = 0, to position x: P E(t) = As usual, the kinetic energy of the mass is The sum is the total energy x 0 kudu = k 2 x2 KE(t) = m 2 (x ) 2. E(t) = P E(t) + KE(t) = 1 2 (kx 2 + m(x ) 2 ) (a) Take the derivative of E(t) with respect to time. Use the chain rule on the right side of the equation. Then simplify your result, so that you get a formula for E (t) that only depends on the forcing f(t), the velocity x (t), and the damping coefficient c. 13

14 (b) Assume there is no external forcing, i.e. f = 0. In this case, what condition(s) on m,c,k guarantee that the energy in the system is constant (i.e., de/dt = 0)? (c) Set f(t) = 0, m = 1, c = 2, and k = 5. Set initial conditions to be x(0) = 2 and x (0) = 0. How long will it take for the system to loose 80% of its initial energy? To solve, find the solution x(t) to the DE and use this solution to compute the energy function E explicitly. (d) Plot the energy curve E(t) and describe its behavior. Explain in words the mechanism that drives the energy picture that you observe, and why the energy is not decreasing initially, but then decreases more rapidly later on. 14