analyze a system that can be modeled by a linear, constant coefficient differential equations

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For the project this semester, you will choose a physical, biological, economic, or social system that can be modeled by a linear, constant coefficient 2-by-2 system of differential equations. Your project will have two main pieces:

  • Analysis of a single differential equation - either a (1) single first-order linear differential equation with constant coefficients OR (2) a single second-order linear differential equation with constant coefficients.
  • Analysis of a system of differential equations - linear, constant coefficient 2-by-2 system of differential equations.Most applications are of the first type: when one unknown quantity is considered, the model is a single first-order linear differential equation with constant coefficients; when that quantity interacts with another quantity, the model is a 2-by-2 linear, constant coefficient 2-by-2 system of differential equations. Examples include:
  • • Pond or Lake Pollution
    • Home Heating
    • Drug delivery/diffusion (Compartmental Model)• Pesticide in Trees and Soil
    • Price-Inventory
    • Chemical Reactions
    • Any other that fits the above description exactly.You can use this chapter of a differential equations text by G.B. Gustafson from the University of Utah to get descriptions and details of some of these systems: www.math.utah.edu/~gustafso/2250systems-de. pdfSome applications are of the second type: the model is a single second-order linear differential equation with constant coefficients. It can be analyzed as such with the methods of Lebl Chapter 2. However, the second-order equation can be converted to a 2-by-2 system and analyzed with the methods of Lebl Chapter 3. Examples include:• Spring-mass systems (used in sample project; cannot be chosen for project)• LRC Circuits
    • Any other that fits the above description exactly.1. Outline of Project Write-up(1) Introduction (2) Single Equation(a) Derivation of equation and meaning of parameters (include units) (b) Meaning and Relevance of Homogeneous vs Non-Homogenous(c) General Solution
    (d) An Initial Value Problem and a Particular Solution (Typically Non-Homogeneous)(e) Behavior: Discuss solution in the language of the application (3) System of ODEs(a) Derivation of equation and meaning of parameters (including units) (b) Meaning and Relevance of Homogeneous vs Non-Homogenous(c) Homogenous System
    1. (i) Pick two sets of values for parameters that will give two of the three possibilities foreigenvalues: (1) distinct, real; (2) complex conjugate pair; (3) repeated (or multiple)eigenvalues (I suggest using computational aids for this.)
    2. (ii) For each of the two situations:(A) Discuss how realistic the parameter values are. (At least one set of values should be realistic.)(B) Give general solution

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(C) Draw a phase portrait
(D) Give two (meaningfully different) initial conditions and their particular solutions

• Describe the particular solutions of each IVP in language of the application (d) Non-homogeneous

  1. (i) Meaning of the inhomogeneity
  2. (ii) Find the general solution using Wolframalpha or another symbolic computational aid.
  3. (iii) Plot the general solution on Desmos.com
  4. (iv) Give two situations (by changing initial conditions or the forcing) that lead to meaning-fully different behaviors
    • Draw the trajectory of each of the two particular solutions (using Desmos plots)• Describe the particular solution of each IVP in language of the application

(4) Appendix
(a) Hand-written work for finding general solution of single equation (b) Hand-written work for finding general solutions of 2-by-2 systems

2. Dates

  • Tues. Nov. 27th: Project is due in recitation.3. Comments
  • A sample project, using the spring-mass system, will be posted. It will include examples of the inputs for 3 d (ii) and (iii).
  • All work for finding solutions should be put in a neat, hand-written appendix.
  • Phase plots may be neatly hand-drawn or copied and pasted from a computer plotting tool.
  • Discussions should be typed, and manageable mathematical symbols should be typed. However, longor complicated mathematical expressions can be hand-written.
  • The responses for many parts listed in the outline should be brief. A sentence or two will sufficefor many pieces. Derivations in 2 (a) and 3 (a) should be a short paragraph. Descriptions of the behavior of solutions in the language of your application (2(e), 3 c (ii) (D), 3 d (iv)) should be your longest written sections, but still do not need to be longer than 4-5 sentence paragraphs.
  • Derivations and realistic values for parameters should be cited.
  • For 3 c (i), you may want to use a computational phase portrait creator, like those found at http://mathlets.org/mathlets/linear-phase-portrait... and http://parasolarchives. com/tools/phaseportrait. You can try different parameters and see the phase portraits associated with them to choose your two sets of parameters.
  • I use the phrase ‘meaningfully different’ in a couple of places. What I mean, for example, is not to choose initial conditions (0, 1) and (0, 1.1). The behaviors will be nearly identical. Choose (0, 1) and (−1,0) or (1,10) and (10,1). It will depend on your system, but there should be something interesting to say about the behavior for the two ‘meaningully different’ choices.

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Sample Project: Spring-Mass via Systems of ODEs and Phase Portraits MATH527 Fall 2018: Sample Produced by Instructor, John McClain 1. Introduction I study a damped spring-mass system as a single second-order equation and as a system of first-order equations. I discuss phase portraits and trajectories in the case of underdamped and critically damped unforced systems and in the case of a forced, underdamped system. 2. Single Equation 2.1. Derivation and Meaning of Parameters. Let us assume that there are at most three forces acting on the bob: the force that is due to gravity, the force produced by the spring, and possibly a drag force. The force that is due to gravity is given by mass × gravity (mg), and the force produced by the spring is proportional to the amount (length) that the spring has been stretched or compressed (k∆L) from its natural length (this is called Hooke’s law ). The constant k in the spring force is called the spring constant, and it has units dyne/cm or Newton/m. The drag or damping force is proportional to the velocity of the mass, but opposes the motion. The proportionality constant, c, has units of N · s/m. According to Newton’s second law of motion, the equation of motion of the bob is mass × acceleration of the bob = sum of all forces acting on the bob (1) md00 = k∆L − mg = k(∆l − d) − mg − cd0 + fext (t) = −kd − cd0 + fext (t) This gives the differential equation: md00 + cd0 + kd = fext (t). 2.2. Meaning and Relevance of Homogeneous vs Non-Homogeneous. If there is an external force fext (t), the equation is non-homogeneous. If there is no external force, so that fext (t) = 0, the equation is homogenous. The external force models a time-dependent force on the bob, like a motor attached to it. 2.3. General Solution. I picked the following values of the parameters to study: m = 1 kg, c = 6 N · s/m, and k = 9 N/m. I picked an external of fext = 12 cos 3t N. The general solution is: d(t) = c1 e−3t + c2 te−3t + 2 cos 3t. 2.4. IVP and a Paricular Solution. I picked the initial condition that the spring is pushed at 10 m/s upward from 2, m above equilibrium. So, d(0) = 2 m. and d0 (0) = 10. The particular solution is: y(t) = 10te−3t + 2 cos 3t. 2.5. Behavior of This Solution. The bob in this case moves upward at first to its highest point of about 2.8 m above equilibrium, then back downward to below equilibrium. It moves back up to just slightly above 2 m above equilibrium. Note that 2 m is the amplitude of the forcing. After only one cycle of the bob moving up, down, and then back up, its motion appears to be completely determined by the forcing. The effect of the initial condition has decayed to a low level, leaving a motion that is dominated by oscillations in sync with the sinusoidal external force. 1 2 3. System of ODEs 3.1. Derivation of the System of Equations Model and Meaning of Parameters. To get a system of equations from the single, second-order differential equation, we define y1 (t) to be the displacement from equilibrium. Then, we define y2 (t) to be the velocity, the derivative of displacement. Then, the single, second-order equation, md00 + cd0 + kd = fext (t), becomes the system of equations: y10 = y2 y20 = − k c fext (t) y1 − y2 + m m m or as a matrix vector system: 0 y = 0 1 k −m c −m ! y+ where y = ! 0 , fext (t) m y1 y2 ! = d ! . v The parameters here have the same meanings as for the single equation. They now show up as ratios only. 3.2. Meaning and Relevance of Homogeneous vs Non-Homogeneous. Again, the inhomogeneity models an external force on the bob, like that supplied by a AC motor producing sinusoidally-varying forces on the bob. The inhomogeneity for the system must have the form ! 0 f (t) = fext (t) m because the force only appears in the equation for y20 which comes from Newton’s 2nd Law. If there is no external force, the system is homogeneous. 3.3. Homogeneous Systems. I pick two sets of parameters. One {m = 1 kg, c = 1 N ·s/m, k = 17/4 N/m} leads to complex eigenvalues, while the other {m = 4 kg, c = 12 N · s/m, k = 9 N/m} leads to a repeated eigenvalue. (The second set of parameters is the same as I used in my singl equation.) fext (t) = 0 for all t, so the system is homogeneous. 3.3.1. Complex Eigenvalues. With the parameters {m = 1 kg, c = 1 N · s/m, k = 17 4 N/m}, the system can be written: 0 1 −17/4 −1 0 y = ! y. These parameters are realistic, representing a very large strong spring, with a sizeable bob on it. The damping is fairly strong, but the system is still underdamped. The general solution to this system can be written: y = c1 e−t/2 or cos 2t − 21 cos 2t − 2sin2t ! + c2 e−t/2 sin 2t − 21 sin 2t + 2 cos 2t ! 3 Figure 1. Phase Portrait for Damped System, {m = 1 kg, c = 1 N · s/m, k = y = e−t/2 N/m} ! c1 cos 2t + c2 sin 2t (− c21 + 2c2 ) cos 2t − (2c1 + 17 4 c2 2 ) sin 2t 3.3.2. Phase Portrait. Fig. 1 shows a phase portrait for this system. All solutions spiral toward the origin because the two eigenvalues are complex with negative imaginary parts. 3.3.3. Initial Conditions and Particular Solutions with Description of Behavior: I looked at the two initial conditions: 0 y(0) = ! 2 and 1 2 − 14 y(0) = . ! In the first case, the bob is initially at equilibrium and is pushed upward with a velocity of 2 m/s. The particular solution is y=e ! sin 2t −t/2 2 cos 2t − 1 2 . sin 2t The trajectory is shown in Fig. 2 on the left. The bob initially moves upward above equilibrium. It hits its highest point (of about 0.65 m above equilibrium) when its velocity hits 0. It then moves back toward equilibrium, reaching its largest downward velocity (of about -1.05 m/s) before it hits equilibrium. It then moves past equilibrium, hitting its lowest point (of about 0.35 m) when its velocity hits 0. It moves back toward equilibrium, again hitting a maximum of velocity of about 0.45 m/s) before it reaches equilibrium. It crosses equilibrium and moves up to a maximum of displacment that is much smaller than its previous maximum. It continues oscillating up and down through equilibrium, with its maximum and minimum displacements and velocities in each cycle approacing 0. In the second, case, the bob is initially − 14 1 2 m above equilibrium and is pushed with a downward velocity of m/s. The particular solution is 1 2 −t/2 y=e 1 4 cos 2t cos 2t − sin 2t ! . The trajectory is shown in Fig. 2 on the right. This trajectory is very similar to the one described previously. The bob starts above equilibrium moving in a downward direction. Just like for the other initial condition, 4 Figure 2. Trajectories for {m = 1 kg, c = 1 N ·s/m, k = 17 4 N/m} Left: Initial displacement is 0 and initial velocity is 2 m/s. Right: Initial displacement is 0.5 m and initial velocity is -0.25 m/s. Figure 3. Phase Portrait for Damped System, {m = 4 kg, c = 12 N · s/m, k = 9 N/m} the bob oscillates up and down through equilibrium and its amplitude of oscillation decreases over time, until it is nearly at rest at the equilibrium point. 3.4. Repeated Eigenvalues. With the parameters {m = 4 kg, c = 12 N · s/m, k = 9 N/m}, the system can be written: 0 y = 0 1 −9/4 −3 ! y. These parameters are realistic, representing a very large strong spring, with a sizeable bob on it. The damping is very strong. The system is critically damped. The general solution to this system can be written: −3t/2 y(t) = e c1 + c2 (1 + t) − 23 c1 + c2 (− 12 − 23 t) 3.4.1. Phase Portrait. Fig. 3 shows a phase portrait for this system. ! . 5 Figure 4. Trajectories for {m = 4 kg, c = 12 N ·s/m, k = 9 N/m} Left: Initial displacement is 2 m below equilibrium and initial velocity is 7 m/s. Right: Initial displacement is 10 m below equilibrium and initial velocity is 10 m/s. 3.4.2. Initial Conditions and Particular Solutions with Description of Behavior: I looked at the two initial conditions: −2 y(0) = ! and 7 −10 y(0) = 10 ! . In the first case, the bob starts 2 m below equilibrium with an upward velocity of 7 m/s. The particular solution is: −3t/2 y=e 4t − 2 ! 7 − 6t . The trajectory is shown in Fig. 4 on the left. It shows that the bob starts below the equilibrium with upward velocity. It moves upward through the equilibrium to its maximum displacement of about 0.46 m when its velocity is 0. Then it moves back down toward equilibrium, with negative velocity, but its velocity approaches 0 and it nevers reaches equilibrium. In the second case, the bob starts 10 m below equilibrium with an upward velocity of 10 m/s. The particular solution is: −5t − 10 y = e−3t/2 10 + ! . 15 2 t The trajectory is shown in Fig. 4 on the right. The bob starts far (10 m) below equilibrium with an upward velocity of 10 m/s. It approaches equilibrium from below as its velocity drops to 0, never reaching equilibrium. 3.5. Non-homogeneous System. 3.5.1. Meaning of Inhomogeneity: I used the same parameters as in my first homogeneous system: {m = 1 kg, c = 1 N · s/m, k = 17/4 N/m}. I chose a sinusoidal external forcing for the inhomogeneity, fext (t) = 16 cos 4t, modeling a motor attached to the bob pushing it up and down smoothly over time. ! ! 0 0 f (t) = = fext (t) 16 cos 4t m The system in matrix-vector form is then: 0 y = 0 1 − 17 4 −1 ! y+ 0 16 cos 2t ! . 6 Figure 5. Trajectories for {m = 1 kg, c = 1 N · s/m, k = 17 4 N/m, fext (t) = 16 cos 4t N } Left: Initial displacement is 2 m below equilibrium and initial velocity is -7 m/s. Right: Initial displacement is 0 and initial velocity is 0. From wolframalpha.com, a particular solution is: yp = 64(16 sin 4t−47 cos 4t) 2465 4t+16 cos 4t − 256(47 sin2465 ! . Adding this to my general solution to the homogeneous system, the general solution to the non-homogeneous system is: −t/2 y(t) = e 64(16 sin 4t−47 cos 4t) 2465 256(47 sin 4t+16 cos 4t c2 ) sin 2t + 2 2465 c1 cos 2t + c2 sin 2t + (− c21 + 2c2 ) cos 2t − (2c1 + ! Fig. 5 shows, on the left, a trajectory satisfying the initial condition ! −2 . y(0) = −7 In this case, the bob starts below equilibrium with a downward velocity. It moves down almost 1 m to its lowest point (when the velocity is 0 as was always the case without forcing). It starts moving back toward equilibrium with positive velocity but before it reaches it, its velocity becomes negative and it starts moving down away from equilibrium again. But then the velocity increases and it begins moving toward equilibrium again. From then on, it moves up and down past equilibrium approaching a perfectly sinusoidal motion (like an undamped system) with angular frequency matching that of the motor, 4 rad/s. The effect of the initial conditions decay, and the system follows the motor. Fig. 5 shows, on the right, a trajectory satisfying the initial condition ! 0 y(0) = . 0 In this case, the bob is released from rest at the equilibrium position. So, the motor has no initial conditions to ‘overcome.’ Thus, the motion just follows the motor from the start; the bob moves is perfect sinusoidal motion with the same frequency as the motor. 4. Appendix Handwritten calculations. Full detail is not required, but it needs to be enough for us to see that you have solved the systems. The calculations can be in done with parameters, rather than specific values.
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Report: project system of secondary order differential equations (2)

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RLC CIRCUITS VIA SYSTEMS OF ODEs AND PHASE PORTRAITS
MATH527 FALL 2018: PRODUCED BY (NAME)

1. INTRODUCTION
The project studies the RLC electric circuit. This circuit contains an inductor (L), a resistor (R), and a
capacitor (C). Since it is a closed circuit, the application of Kirc...


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