Week #4 : Applications of Linear Higher-Order DEs Goals: Solving Homogeneous DEs with Constant Coefficients - Second and Higher Order Applications - Damped Spring/Mass system Applications - Pendulum 1
Homogeneous Equations - Review and Generalizing to nth Order - 1 Homogeneous Equations - Review We saw last week how to solve second-order homogeneous linear equations. Note 1: we can make linear combinations of individual solutions to make more general solutions. Lemma. If both y 1 and y 2 are solutions to y + p(x)y + q(x)y = 0, then any linear combination y = C 1 y 1 + C 2 y 2 is also a solution. In other words, the solutions form a vector space.
Homogeneous Equations - Review and Generalizing to nth Order - 2 Note 2: If the equation has constant coefficients, an informed guess that solutions will be of the form y = e rt will always get us started. If we start with the DE ay + by + cy = 0 where a, b, c R and a 0, then assuming that y = e rt, any valid r must statisfy the DE s characteristic equation: ar 2 + br + c = 0.
Homogeneous Equations - Review and Generalizing to nth Order - 3 Homogeneous DEs - Arbitrary order A linear equation of order n can be expressed in the form A n (x)y (n) + A n 1 (x)y (n 1) + + A 0 (x)y = F (x). Assuming that A n (x) 0, we can rewrite the equation in the standard form: y (n) + p 1 (x)y (n 1) + + p n (x)y = G(x)
Homogeneous Equations - Review and Generalizing to nth Order - 4 y (n) + p 1 (x)y (n 1) + + p n (x)y = G(x) In this n th order DE, the general solution will be the span of n linearly independent solutions, y 1,..., y n. It therefore behooves us to develop a test for linear independence for sets of n functions.
The Wronskian - n functions - 1 The Wronskian - n functions For n differentiable function y 1, y 2,..., y n, the Wronskian is y 1 (x)... y n (x) y 1 W [y 1,..., y n ] := det (x)... y n(x)..... y (n 1) 1 (x)... y n (n 1) (x) Lemma. If the Wronskian is nonzero at some point, then y 1,... y n are linearly independent.
The Wronskian - n functions - 2 Problem. Use the Wronskian to show that the functions x, x 2, and x 1 are linearly independent.
The Wronskian - n functions - 3 Problem. Given that x, x 2, and x 1 are solutions to x 3 y +x 2 y 2xy + 2y = 0 where x > 0, find the general solution.
The Wronskian - n functions - 4 Concept check: why weren t the solutions x, x 2, and x 1 to x 3 y + x 2 y 2xy + 2y = 0 in the form of e rx?
Homogenous DEs with Const Coeffs - Cases - 1 Homogenous DEs with Constant Coefficients - Cases Proposition. If the characteristic equation has distinct roots r 1,..., r n, then the general solution is C 1 e r 1x + C 2 e r 2x + + C n e r nx. Problem. Solve y (4) + y 7y y + 6y = 0.
Homogenous DEs with Const Coeffs - Cases - 2 Proposition. If the characteristic equation has a root r of multiplicity m, then part of the general solution is C 1 e rx + C 2 xe rx + + C m x m 1 e rx. Problem. Solve y (4) y (3) 3y + 5y 2y = 0.
Homogenous DEs with Const Coeffs - Cases - 3 Proposition. If the characteristic equation has a pair of conjugate roots a ± b 1, then the part of the general solution is C 1 e ax cos(bx) + C 2 e ax sin(bx). Problem. Solve y (4) y = 0.
Homogenous DEs with Const Coeffs - Cases - 4 Summary Form of r. Form of solution term(s) Unique real root r 1 y = Duplicated real roots. E.g. r = r 1, r 1 y 1 = and y 2 = Unique pair of complex conjugate roots y 1 = and r 1,2 = a ± b 1 y 2 = Duplicated complex conjugate pair. E.g. r 1,2,3,4 = a ± b 1, a ± b 1 (4 roots)
Homogenous DEs with Const Coeffs - Example - 1 Homogenous DEs with Constant Coefficients - Example Problem. Solve y (4) 8y (3) + 26y 40y + 25y = 0. Hint. Expand (r 2 4r + 5) 2.
Mechanical Vibrations - Spring-mass system - 1 Mechanical Vibrations - Spring-mass system We now consider the spring/mass system seen earlier, but with more detail. Consider a mass m hanging on the end of a vertical spring of unstretched length l. Using Newton s second law, build a differential equation that governs the system. k c m
Mechanical Vibrations - Spring-mass system - 2 k c m
Mechanical Vibrations - Spring-mass system - 3 Problem. Consider a mass of 0.5 kg with spring constant k = 2 N m 1 in an undamped unforced system. Assume the mass is displaced 0.4 m from equilibrium and released. Describe the longterm behaviour of the system.
Oscillating Solutions: Sine/Cos and Shifted-Cos Forms - 1 Oscillating Solutions: Sine/Cos and Shifted-Cos Forms Remark. We can always write the general solution of an undamped unforced spring system in either of two equivalent forms: To convert between them, A = C 2 1 + C2 2 and α (in radians) satisfies cos(α) = C 1 A and sin(α) = C 2 A The parameter α is called the phase angle.
Oscillating Solutions: Sine/Cos and Shifted-Cos Forms - 2 Problem. Explain how this relationship helps us understand the behaviour of solutions when r comes in complex conjugate pairs.
Oscillating Solutions: Sine/Cos and Shifted-Cos Forms - 3 Problem. Consider a spring/mass system with - mass of 0.5 kg, - spring constant k = 2 N m 1 and - damping constant c = 0.5 N s m 1 and no external force applied. Assume the mass is displaced 0.3 m from equilibrium and released. Describe the long-term behaviour of the system.
Oscillating Solutions: Sine/Cos and Shifted-Cos Forms - 4 - mass of 0.5 kg, - spring constant k = 2 N m 1 and - damping constant c = 0.5 N s m 1
Unforced Spring/Mass System - Patterns of Behaviour - 1 Unforced Spring/Mass System - Patterns of Behaviour k m c my = ky cy or my + cy + ky = 0 r = c ± c 2 4km 2m Damping c 2 4km r Description None c = 0 Light c 2 < 4km Heavy c 2 > 4km
Unforced Spring/Mass System - Patterns of Behaviour - 2 Problem. Sketch possible solutions for all four spring/mass cases. y y t t y Undamped y Damped t t Critically Damped Overdamped
Homogeneous DEs - Arbitrary Order - 2 d n y dx n + p 1(x) dn 1 y dx n 1 + + p n(x)y = f(x). In this n th order DE, the general solution will be the span of n linearly independent solutions y 1,..., y n. It therefore behooves us to develop a test for linear independence for sets of more than 2 functions.
Demonstration - Spring/Mass - 1 Demonstration - Spring/Mass We will demonstrate how the solutions to the damped spring/mass DE change as the damping is gradually increased. my + cy + ky = 0 In this demonstration, we will use m = 1 kg, and k = 25 N/m.
Demonstration - Spring/Mass - 2 y + cy + 25y = 0 Problem. What damping level will produce critical damping? What will the form of the solutions be when damping is below critical? What will the form of the solutions be when damping is above critical?
Demonstration - Spring/Mass - 3 During demonstration, try to ask yourself the following questions: As damping increases in general, does the graph of the solution change gradually or dramatically? What about the mathematical form of the solution? Near critical damping specifically, does the graph of the solution change gradually or dramatically? What about the mathematical form of the solution?
Applications - Pendulum - 1 Applications - Pendulum Problem. Consider the simple pendulum (mass at the end of a rod) shown below. If we start from the rotational (torque) version of Newton s Second Law, (moment of inertia) (angular accel) = torques we obtain (ml 2 ) (θ ) = mgl sin(θ) or, in (almost) standard form:
Applications - Pendulum - 2 θ + g L sin(θ) = 0 Use a well-known approximation from calculus to simplify this to a linear DE: What limitations does this put on our interpretation of the solution?
Applications - Pendulum - 3 Find the general solution to the linearized differential equation θ + g L θ = 0 Use your solution to predict the period of the oscillations of a pendulum, given g and the length of the rod, L.
Applications - Pendulum - 4 Comment: what does this mean about the swinging of a pendulum for larger angles?