4.2 Homogeneous Linear Equations
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1 4.2 Homogeneous Linear Equations Homogeneous Linear Equations with Constant Coefficients Consider the first-order linear differential equation with constant coefficients a 0 and b. If f(t) = 0 then this is called the homogeneous form of the equation. (Note: this is not related to the homogeneous functions we looked at in chapter 2.) Solving for y The only nontrivial elementary function whose derivative is a constant mulitple of itself is e rt. Since e rt 0 for all real values of t, then our last equation is satifised when r is root a of Second-Order Equation We can use this procedure to produce solutions to higher-order homogeneous linear DEs. where a, b, and c are constants. Let y = e rt, find y and y
2 Substituting into the original equation we get Three types of solutions come from solving the auxillary equation. Two distinct real roots r 1 and r 2 when b 2 4ac > 0 Two real and equal roots r 1 = r 2 when b 2 4ac = 0 Two roots that are complex conjugates for r 1 = α + βi and r 2 = α βi when b 2 4ac < 0. Example Find two solutions of the differential equation. 1. 4y + y = 0
3 Existence and Uniqueness: Homogeneous Case Theorem 1. For any real numbers a 0, b, c, t 0, Y 0, and Y 1, there exists a unique solution to the initial value problem This solution is valid for all t in (, ). Linear Independence of Two Functions A pair of functions y 1 (t) and y 2 (t) is said to be linearly independent on the interval I if and only if neither of them is a constant multiple of the other on all of I. We say y 1 (t) and y 2 (t) are linearly dependent on I if one of them is a constant multiple of the other on all of I. Representation of Solutions to Initial Value Problem Theorem 2. If y 1 (t) and y 2 (t) are any two solutions to the differential equation That are linearly independent on (, ), the unique constants c 1 and c 2 can always be found so that satisfies the initial value problem on (, ). Thus in our last example we also have the solution: 2. Determine whether the functions are linearly dependent on the interval (0,1). y 1 (t) = e 3t, y 2 (t) = e 4t
4 A Condition for Linear Dependence of Solutions Lemma 1. For any real numbers a 0, b and c, if y 1 (t) and y 2 (t) are any two solutions to And Holds at any point τ, then y 1 and y 2 are linearly dependent on (, ). Note: This is called the Wronskian of y 1 and y 2 and can be expressed as the determinant of a 2 x 2 matrix. Thus if y 1 (t) and y 2 (t) are linearly independent, then we say the general solution is Distinct real roots: If the auxiliary equation has two distinct roots r 1 and r 2, then the general solution is Repeated roots: If the auxiliary equation has one repeated root r, then the general solution is Find the general solution. 3. y + 5y + 6y = 0
5 4. y + 10y + 25y = 0 Solve the initial value problem. 5. y 4y + 3y = 0 ; y(0) = 1, y (0) = 1 3
6 4.3 Auxiliary Equations with Complex Roots In this section we will look at differential equations with constant coefficients Where the auxiliary equation has two complex roots where α and β > 0 are real and i 2 = 1. These roots would yield us the solutions Using Euler s Formula, e iθ = cos θ + i sin θ, we can manipulate the above solution to get Examples 1. y 6y + 10y = 0
7 2. y + 4y + 6y = 0 Higher-Order Equations In general an n th -order differential equation with constant coefficients a n y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = 0 has solutions y i = e r it for i = 1, 2, 3,, n and the resulting auxiliary equation is a n r n + a n 1 r n a 1 r + a 0 = 0 If all the roots are real and distinct, then the general solution is y = c 1 e r 1t + c 2 e r 2t + + c n e r nt When r 1 is a root of multiplicity k of an n th -degree auxiliary equation, then the general solution is y = c 1 e r 1t + c 2 te r 1t + c 3 t 2 e r 1t + + c n t n 1 e r 1t
8 Examples Find the general solution. 6. y + 5y + 3y + 9y = 0
9 7. 26 d4 y dx d2 y dx 2 + 9y = 0
10 4.5 The Superposition Principle Superposition Principle Theorem 3. Let y 1 be a solution to the differential equation And y 2 be a solution to Then for any constants k 1 and k 2, the function k 1 y 1 + k 2 y 2 is a solution to the differential equation Example 1. Given that y p1 = 3e 2x and y p1 = x 2 + 3x are particular solutions to y 6y + 5y = 9e 2x and y 6y + 5y = 5x 2 + 3x 16 respectively, find a particular solution to a. y 6y + 5y = 5x 2 + 3x 16 9e 2x b. y 6y + 5y = 10x 2 6x e 2x
11 Existence and Uniqueness: Nonhomogeneous Case Theorem 4. For any real numbers a 0, b, c, t 0, Y 0, and Y 1, suppose y p (t) is a particular solution to the nonhomogeneous differential equation In an interval I containing t 0 and that y 1 (t) and y 2 (t) are linearly independent solutions to the associated homogeneous equation In I. Then there exists a unique solution in I to the initial value problem And it is given by For the appropriated choice of the constants c 1 and c 2. Example 1. Find the general solution y y 2y = 1 2t, y p (t) = t 1
12 6.3 Undetermined Coefficients and the Annihilator Method Notation An n th -order differential equation can be written as It can also be written even more simply as where L denotes the linear n th -order differential operator or characteristic polynomial In this section, we will look for an appropriate linear differential operator that annihilates f(x). Factoring Operators Example 1. Rewrite the differential equation using operator notation and factor.
13 Annihilator Operators If L is a linear differential operator with constant coefficients and f is a sufficiently differentiable function such that then L is said to be an annihilator of the function. Examples Find the differential operator that annihilates each function. 2. y = k 3. y = x 4. y = x 4 The function x n is annihilated by the differential operator Note: This annihilator will also annihilate any linear combination of these functions: The function x n e αx is annihilated by the differential operator If L 1 annihilates y 1 (x) and L 2 annihilates y 2 (x), then the product of the differential operators L 1 L 2 will annihilate the linear combination Note: The differential operator that annihilates a function is not unique.
14 Example 5. Verify that the given differential operator annihilates the indicated function 2D 1; y = 4e x 2 6. x 3 5x e 2x 8. 2e 3x 3xe 3x
15 The functions x n e αx cos βx and x n e αx sin βx are annihilated by the differential operator Special Case: when α = 0 and n = 0, then Example sin x 10. 8x sin x + 10 cos 5x Method of Undetermined Coefficients: Annihilator 1. Solve the associated homogeneous equation to find y c. 2. Find the annihilator for g(x) and apply it to both sides of the differential equation. 3. Solve the now homogeneous DE to find the general solution. 4. Identify the particular solution, y p, and find its derivatives. 5. Plug y p and its derivatives into the original equation to find the unknown constant.
16 Examples Solve 11. y 3y 4y = 3e 2x
17 12. y + 4y = 4 cos x + 3 sin x 8
18 4.6 Variation of Parameters Cramer s Rule The solution to a system of equations Is given by Solving Second-Order DE s For a second-order DE of the form Where the solution to the related nonhomogeneous equation is Can we vary the paramters of c 1 and c 2 to be function u 1 (x) and u 2 (x)? Let a particular solution be Where y 1 and y 2 form a fundamental set of solutions on I of the associated homogeneous equation. Then And Substituting into our DE y + P(x)y + Q(x)y = f(x) And after a lot of simplification, we get
19 d dx [u 1 y 1 + u 2 y 2 ] + P[u 1 y 1 + u 2 y 2 ] + u 1 y 1 + u 2 y 2 = f(x) In order to come up with a system of equations to solve for u 1 and u 2 we will make the assumption that u 1 y 1 + u 2 y 2 = 0 Which gives us a second equation u 1 y 1 + u 2 y 2 = f(x) This system of equations can be solved using Cramer s Rule and gives us Where Method of Variation of Parameters 1. For a given equation Find the complementary function 2. Compute the Wronskian W(y 1 (x), y 2 (x)) 3. Put the DE in standard form by dividing by a 2 4. Find u 1 and u 2 from 5. Then and the general solution is
20 Examples Solve. 1. y + y = sec x
21 2. y 9y = 9x e 3x
22 Third- Order Equations When n = 3, Where y 1, y 2, and y 3 are a linearly independent set of solutions of the associated homogeneous DE and u 1, u 2, and u 3 are determined by Higher-Order Equations For a n th -order DE in standard form The complementary function is The particular solution is And Cramer s Rule gives us
23 4.7 Variable Coefficient Equations Existence and Uniqueness of Solutions Theorem 5. Suppose p(t), q(t), and g(t) are continuous on an interval (a, b) that contains the point t 0. Then, for any choice of the initial values Y 0 and Y 1, there exists a unique solution y(t) on the same interval to the initial value problem Example 2. x dy = y, y(0) = 1 dx The Cauchy-Euler Equations A linear second-order differential equation that can be expressed in the form Where a, b, and c are constants, is called a Cauchy-Euler, or equidimensional, equation. Notice the exponent of the variable coefficient matches the degree of the derivative for each term.
24 Solving a Cauchy-Euler Equation Given an homogeneous second-order DE In Section 4.3 we let y = e rt, for a Cauchy-Euler Equation let y = t r Thus y = t r is a solution of the differential equation whenever r is a solution to the auxiliary equation As in Section 4.2, there are three types of solutions come from solving the auxiliary equation. Two distinct real roots r 1 and r 2 when b 2 4ac > 0 o General Solution: Two real and equal roots r 1 = r 2 when b 2 4ac = 0 o General Solution: Two roots that are complex conjugates for r 1 = α + βi and r 2 = α βi when b 2 4ac < 0. o General Solution:
25 Example Solve. 1. 2x 2 y + xy y = x 2 y + y = 0
26 3. x 2 y + xy + 4y = 0 Solve the nonhomogeneous Cauchy-Euler equation using variation of parameters. 4. x 2 y 2xy + 2y = x 4 e x
27
28 4.9 Free Mechanical Vibrations Spring-Mass Oscillator When the spring is not stretched and the mass m is at rest, the system is at equilibrium. Forces Acting in the System When the mass m is displaced from equilibrium the spring exerts a force, F spring, against the displacement where y is the displacement of the mass, and k is the spring constant or stiffness. The system also experiences friction, given by Where b 0 is the damping coefficient. Any other external forces such as gravitational, electrical, or magnetic forces will be lumped together as the known function F ext (t). Using Newton s Second Law, we get the second order differential equation
29 Undamped, Free Motion If we consider the simplest case, where there is no friction (b = 0) and no external forces (F ext (t) = 0) acting on the system then we get the equation To put in standard form, we divide by m to get where ω 2 = k. This equation describes undamped free motion or simple harmonic motion m The resulting auxiliary equation is Which has complex roots Thus the general solution is Notes: Type equation here. When initial conditions are used to find c 1 and c 2, the resulting particular solution is called the equation of motion. The period, measured in seconds, is then The frequency, or cycles per second, is (aka natural frequency) The circular frequency, is
30 Examples 1. A 20 kg mass is attached to a spring. a. If the frequency of simple harmonic motion is 2 π cycles/s, what is the spring constant k? b. What is the frequency of simple harmonic motion if the original mass is replaced with an 80 kg mass? Alternate Form of y(t) A simpler form of y(t) that makes it easier to determine amplitude is Where A = c c 2 2 and φ is the phase angle defined by tan φ = c 1 c 2 Note: Be careful when solving for the phase angle. Remember that the domain of tan 1 x is restricted to the first and fourth quadrants.
31 2. A 3-kg mass is attached to a spring with stiffness k = 48 N m. The mass is displaced 1 2 m to the left of the equilibrium point and given a velocity of 2 m/sec to the right. The damping force is negligible. Find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position?
32 3. Express the equation in the alternate form. y(t) = 2 3 cos 2t 1 sin 2t 6 Free damped motion A mass will only experience free undamped motion in a perfect vacuum and is therefore not very realistic. Thus on a damped system with no external forces we have Type equation here. The auxiliary equation associated with this equation is With roots Type equation here. Underdamped or Oscillatory Motion When the discriminant is negative we get two complex roots, α ± βi, where
33 And the general solution is Or more simply Type equation here. Where A = c c 2 2 and φ is the phase angle defined by tan φ = c 1 c 2 This system is called underdamped because there is not enough damping to prevent oscillation. The coefficient Ae λt is called the damped amplitude of vibration or damping factor. Because this y(t) is not periodic, the quasi period, or time between maxima, is given by And the quasi frequency is Overdamped Motion When the discriminant is positive we get two distinct real roots, r 1 and r 2, and a general solution Type equation here. This system is called overdamped because the damping force is great enough to prevent oscillation.
34 Critically Damped Motion When the discriminant is zero we get one real root, r 1, and a general solution This system is said to be critically damped because any slight decrease in the damping force would result in oscillatory motion. Critically damped motion is very similar to overdamped motion. Example 4. A 1 kg mass is attached to a spring whose constant is 16 N/m, and the entire system is submerged in a liquid that imparts a damping force numerically equal to 10 times the instantaneous velocity. Determine the equation of motion if a. The mass is initially released from rest from a point 1 meter to the right of the equilibrium position. b. The mass is initially released from a point 1 meter to the right of the equilibrium position with a velocity of 12 m/s to left.
35
36 4.10 Forced Mechanical Vibrations We now consider a spring-mass system with an external force, F ext (t), applied to it. This gives us a differential equation for forced mechanical motion In particular, we will look at the response of an underdamped system (i.e. complex roots) to a sinusoidal force. Where F 0 and γ are nonnegative constants. If we were to solve this nonhomogeneous differential equation, the general solution, y(t) = y h + y p, would be (b 2m The first term of the solution is called the transient term because Ae )t 0 as t +. Over time, the equation of motion essentially becomes y p. This is called the steady-state solution or the steady-state response to F 0 cos γt.
37 Notice that the amplitude of Are off by a factor of Called the frequency gain or gain factor. The graph of M(γ) is called the frequency response curve, or resonance curve. In an underdamped system, the resonance curve, M(γ), has a maximum at And a resonance frequency of When a system is stimulated by an external force at this frequency, it is said to be at resonance. If the damping constant b is very small and the forcing function has a frequency near the resonance frequency then the system will be subject to increasingly large oscillations that can lead to the breakdown of the system. i.e. when a bridge collapses or a glass shatters. Examples 1. A 20-kg mass is attached to a spring with stiffness k = 49 N m. At t = 0, an external force f(t) = 10 cos 2t N is applied to the system. The damping constant for the system is 3 N sec m. Determine the steady-state solution for the system.
38 2. Determine the equation of motion for an undamped system at resonance governed by d 2 y dt 2 + y = 5 cos t, y(0) = 0, y (0) = 1
39 3. A 2-kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 20 cm upon coming to rest at equilibrium. At time t = 0, the mass is displaced 5 cm below the equilibrium position and released. At this same instant, an external force F(t) = 0.3 cos t N is applied to the system. If the damping constant for the system is 5 N-sec/m, determine the equation of motion for the mass. What is the resonance frequency for the system?
40
41 5.7 Electrical Systems Series Circuit Analogue For an LRC series electrical circuit, using Kirchoff s second law, the sum of the voltages of the inductor, resistor, and capacitor equals the voltage E(t) impressed on the circuit. Which becomes the second-order DE when I(t) = dq dt is substituted in. Notice, this equation is very similar to the equation used to describe forced motion spring/mass systems in sections 4.9 and Moreover, the terms transient, steady-state, overdamped, underdamped, critically damped, and resonance frequency are analogous as well.
42 So the steady-state charge could be found using And a resonance frequency of, γ r 2π could be found using If we take the original equation and differentiate it with respect to t, it becomes This equation allows us to solve for current directly.
43 Example 1. Find the charge on the capacitor in an LRC series circuit when L = 1 4 h, R = 20 Ω, C = f, E(t) = 0 V, q(0) = 4 C, and i(0) = 0 A. Is the charge on the capacitor ever equal to zero?
44 2. An RLC series circuit has a voltage source given by E(t) = 10 cos 50t V, a resistor of 120 Ω, an inductor of 4 H, and a capacitor of (2200) 1 F. Find the steady-state current (solution) for this circuit. What is the resonance frequency of the circuit?
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