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Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Calculus Differential Equations (II) Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Outline 1 Second-Order Equations with Constant Coefficients Mass Spring System and it’s Formulation 2nd Order ODE with Constant Coefficients 2 Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations Method of Undetermined Coefficients Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Outline 1 Second-Order Equations with Constant Coefficients Mass Spring System and it’s Formulation 2nd Order ODE with Constant Coefficients 2 Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations Method of Undetermined Coefficients Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Mass Spring System and it’s Formulation (I) In this section and section 6.5, we explore the mathematics describing mechanical vibrations. In Figure 6.20, we show a mass hanging from a spring that is suspended from the ceiling. Figure: [6.20] Spring-mass system. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Mass Spring System and it’s Formulation (II) We call the natural length of the spring (the length before the spring is hung from the ceiling) l. Observe that hanging the mass from the spring will stretch the spring a distance ∆l from its natural length. We now measure the displacement u(t) of the mass from this equilibrium position. Figure: [6.20] Spring-mass system. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Mass Spring System and it’s Formulation (III) We consider downward (beyond the equilibrium position) to be a positive displacement, while upward (below the equilibrium position) is considered to be the negative direction. The mass in Figure 6.20 has been displaced from its natural length by a total of u(t) + ∆l. Figure: [6.20] Spring-mass system. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Mass Spring System and it’s Formulation (IV) The motion of a spring-mass system is described by Newton’s second law of motion: F = ma. Primary forces acting on the mass: Gravity force: mg (downward). Spring restoring force: −k(u + ∆l) for some positive constant k (called the spring constant). Damping force: −cv, where v = u0 is the velocity of the mass and c is a constant. Figure: [6.20] Spring-mass system. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Mass Spring System and it’s Formulation (V) Combining these forces, Newton’s 2nd law gives the following: mu00 (t) = mg − k[u(t) + ∆l] − cu0 (t) or mu00 (t) + cu0 (t) + ku(t) = mg − k∆l If the mass is not in motion, i.e., u(t) = 0, then we also have u0 (t) = u00 (t) = 0, and the last equation reduces to 0 = mg − k∆l. While this provides a means of solving for the spring constant k in terms of the mass and ∆l, this also simplifies the equation to mu00 (t) + cu0 (t) + ku(t) = 0. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Mass Spring System and it’s Formulation (VI) The equation mu00 (t) + cu0 (t) + ku(t) = 0. is an example of a second-order differential equation, since there is a second derivative in the equation. In fact, it is an especially simple type of differential equation, because the coefficients of u00 , u0 and u are all constants. By solving the equation we learn how to analyze spring motion. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Outline 1 Second-Order Equations with Constant Coefficients Mass Spring System and it’s Formulation 2nd Order ODE with Constant Coefficients 2 Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations Method of Undetermined Coefficients Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients 2nd Order ODE with Constant Coefficients (I) A second order constant coefficient ode takes the following form: ay00 (t) + by0 (t) + cy(t) = 0, where a, b and c are constants. Notice that equation asks us to find a function whose first and second derivatives are similar enough that the combination ay00 (t) + by0 (t) + cy(t) adds up to zero. One candidate for such a function is the exponential function ert . So, we assume that y = ert , for some constant r. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients 2nd Order ODE with Constant Coefficients (II) Observe that if y(t) = ert , then y0 (t) = rert and y00 (t) = r2 ert . Substituting all of this into the differential equation, we get ar2 ert + brert + cert = 0 or (ar2 + br + c)ert = 0. Since ert > 0, this can only happen if ar2 + br + c = 0. The last equation is called the characteristic equation. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients 2nd Order ODE with Constant Coefficients (III) The solutions of the characteristic equation ar2 + br + c = 0. are given by the quadratic formula as: √ √ −b + b2 − 4ac −b − b2 − 4ac r1 = and r2 = . 2a 2a So, there are three possibilities for solutions of the differential equation: 1 r1 and r2 are distinct real solutions (if b2 − 4ac > 0), 2 r1 = r2 are (repeated) real solutions (if b2 − 4ac = 0) or, 3 r1 and r2 are complex solutions (if b2 − 4ac < 0). All three of these cases lead to different solutions of the differential equation, which we must deal with separately. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients 2nd Order ODE with Constant Coefficients (IV) Case 1: If r1 and r2 are distinct real solutions of ar2 + br + c = 0, then y1 = er1 t and y2 = er2 t are two solutions of ay00 (t) + by0 (t) + cy(t) = 0, and y(t) = c1 er1 t + c2 er2 t is the general solution. We illustrate this in example 4.1. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Finding General Solutions Example (4.1) Find the general solution of 1 y00 − y0 − 6y = 0 and 2 y00 + 4y0 − 2y = 0. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients 2nd Order ODE with Constant Coefficients (V) Case 2: If r1 = r2 are repeated roots of the equation ar2 + br + c = 0, then we have found only one solution of: y1 = er1 t . The second solution is y2 = ter1 t . The general solution of ay00 (t) + by0 (t) + cy(t) = 0, is then y(t) = c1 er1 t + c2 ter1 t . We illustrate this case in example 4.2. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Finding General Solutions (Repeated Roots) Example (4.2) Find the general solution of y00 − 6y0 + 9y = 0. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients 2nd Order ODE with Constant Coefficients (VI) Case 3: If r1 and r2 are complex roots of the equation ar2 + br + c = 0, we can √ write these as r1 = u + vi and r2 = u − vi, where i = −1. To interpret a complex exponential like e(u+vi)t we rely on Euler’s formula: eiθ = cos θ + i sin θ. The solution corresponding to r = u + vi is then e(u+vi)t = eut+vti = eut evti = eut (cos vt + i sin vt). It can be shown that y1 = eut cos vt and y2 = eut sin vt are solutions of the differential equation. So, the general solution of the differential equation is y1 (t) = c1 eut cos vt + c2 eut sin vt. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Finding General Solutions (Complex Roots) Example (4.3) Find the general solution of the equations 1 y00 + 2y0 + 5y = 0 and 2 y00 + 4y = 0. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Initial Value Problem We can now find the general solution of any equation of the form ay00 (t) + by0 (t) + cy(t) = 0. Notice that the general solution of a second-order differential equation always involves two arbitrary constants. To determine the value of these constants, we specify two initial conditions, most often y(0) and y0 (0) A second-order differential equation, plus two initial conditions is called an initial value problem. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Solving an Initial Value Problem (I) Example (4.4) Find the solution of the initial value problem y00 + 4y0 + 3y = 0, y(0) = 2, y0 (0) = 0. Figure: [6.21] y = −e−3t + 3e−t . Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Solving an Initial Value Problem (II) Example (4.5) Find the solution of the initial value problem y00 + 9y = 0, y(0) = 4, y0 (0) = −6. Figure: [6.22] y = 4 cos 3t − 2 sin 3t. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Spring-Mass System with No Damping (I) Example (4.6) A spring is stretched 6 inches when an 8-pound weight is attached. The mass is then pulled down an additional 4 inches and released. Neglect damping. Find an equation for the position of the mass at any time t and graph the position function. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Spring-Mass System with No Damping (II) Figure: [6.23] u(t) = 1 3 cos 8t. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Spring-Mass System with Damping (I) For a real spring-mass system, there is always some damping, so that the idealized perpetual motion of example 4.6 must be modified somewhat. In example 4.7, again take note of the steps required to obtain the equation of motion. Example (4.7) A spring is stretched 5 cm when a 2-kg mass is attached. The mass is set in motion from its equilibrium position with an upward velocity of 2 m/s. The damping constant equals c = 4. Find an equation for the position of the mass at any time t and graph the position function. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Spring-Mass System with Damping (II) √ 2 Figure: [6.24] u(t) = − √195 e−t sin 195t. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients RLC Electric Circuit Model (I) The charge in simple electrical circuits can be modeled with the same equation we just used for the motion of a spring-mass system. An RLC-circuit illustrated in the figure, consists of a resistor R, a capacitor C, an inductor L and a voltage source V. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients RLC Electric Circuit Model (II) Assume that there is no impressed voltage. Let Q(t) (coulombs) be the total charge on the capacitor at time t and I(t) be the current. Then I = Q0 (t). It is known that The voltage drop across the resistor is IR. Q The voltage drop across the capacitor is . C The voltage drop across the inductor is LI 0 (t). Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients RLC Electric Circuit Model (III) These voltage drops must sum to the impressed voltage. If there is none, then LI 0 (t) + RI(t) + 1 Q(t) = 0 C or LQ00 (t) + RQ0 (t) + 1 Q(t) = 0. C Observe that this is the same as the equation for describing the motion of a spring-mass system, except for the names of the constants. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Finding the Charge in an Electrical Circuit Example (4.8) A series circuit has an inductor of 0.2 henry, a resistor of 300 ohms and a capacitor of 10−5 farad. The initial charge on the capacitor is 10−6 coulomb and there is no initial current. Find the charge on the capacitor and the current at any time t. Figure: [6.25] Q(t) = 10−6 2e−500t − e−1000t . Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Outline 1 Second-Order Equations with Constant Coefficients Mass Spring System and it’s Formulation 2nd Order ODE with Constant Coefficients 2 Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations Method of Undetermined Coefficients Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations (I) Imagine yourself trying to tape an important event on a video camera. You might be more concerned with keeping a steady hand than with understanding the mathematics of motion control, but mathematics plays a vital role in helping you produce a professional-looking tape. In section 6.4, we modelled mechanical vibrations when the motion is started by an initial displacement or velocity. In this section, we extend that model to cases where an external force such as a shaky hand continues to affect the system. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations (II) As usual, the starting place for our model is Newton’s second law of motion: F = ma. We now add an external force to the spring force and damping force considered in section 6.4. If the external force is F(t) and u(t) gives the displacement from equilibrium, as defined before, we have mu00 (t) + cu0 (t) + ku(t) = F(t). The only change from the spring model in section 6.4 is that the right-hand side is no longer zero. If F(t) = 0 then the equation is homogeneous. For F(t) 6= 0, we call the equation nonhomogeneous. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations (III) Consider the nonhomogeneous equation of the form mu00 (t) + cu0 (t) + ku(t) = F(t). Our goal is to find the general solution of such an equation Let up (t) denote one particular solution of the nonhomogeneous equation. Notice that if u(t) is any other solution, then we have that m(u − up )00 + c(u − up )0 + k(u − up ) = (mu00 + cu0 + ku) − (mu00p − cu0p + kup ) = F − F = 0. That is, the function u − up is a solution of the homogeneous equation mu00 + cu0 + ku = 0. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations (IV) So, if the general solution of the homogeneous equation mu00 + cu0 + ku = 0. is c1 u1 + c2 u2 , then u − up = c1 u1 + c2 u2 for some constants c1 and c2 and u = c1 u1 + c2 u2 + up . is the solution of the nonhomogeneous equation. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations (V) Theorem (5.1) Let u = c1 u1 + c2 u2 be the general solution of mu00 + cu0 + ku = 0 and let up be any solution of mu00 + cu0 + ku = F. Then the general solution of mu00 + cu0 + ku = F is given by u = c1 u1 + c2 u2 + up . Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Solving a Nonhomogeneous Equation Example (5.1) Find the general solution of u00 + 4u0 + 3u = 30e2t given that up = 2e2t is a solution. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Outline 1 Second-Order Equations with Constant Coefficients Mass Spring System and it’s Formulation 2nd Order ODE with Constant Coefficients 2 Nonhomogeneous Equations: Undetermined Coefficients Nonhomogeneous Equations Method of Undetermined Coefficients Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Method of Undetermined Coefficients (I) While example 5.1 showed us how to piece together the solution of a nonhomogeneous equation from a particular solution and the general solution of the corresponding homogeneous equation, we still do not know how to find a particular solution. To find a particular solution we present a method here, called the method of undetermined coefficients. It works for equations with constant coefficients and where the nonhomogeneous term is not too complicated. The method essentially relies on our ability to make an educated guess about the form of a particular solution. We illustrate this technique for example 5.1. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Method of Undetermined Coefficients (II) Consider the equation u00 + 4u0 + 3u = 30e2t , The most likely candidate for the form of u(t) is a constant multiple of e2t . (How else would u00 , 4u0 and 3u all add up to 30e2t ?) Let A be a constant. Our educated guess is up (t) = Ae2t , We next substitute this into the differential equation and solve for A. (If it turns out to be impossible to solve for A, then we have made a bad guess.) Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Method of Undetermined Coefficients (III) Here, we have u0p = 2Ae2t and u00p = 4Ae2t and so, requiring up to be a solution of the nonhomogeneous equation gives us 30e2t = u00p + 4u0p + 3up = Ae2t + 4(2Ae2t ) + 3(Ae2t )30e2t = 15Ae2t . So, 15A = 30 or A = 2. A particular solution is then up (t) = 2e2t , as desired. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Solving a Nonhomogeneous Equation Example (5.2) Find the general solution of u00 + 2u0 − 3u = −30 sin 3t. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Rules for Making Good Guess (I) In general, we start with the function F(t) and then add terms for each derivative. In example 5.2, we started with A sin 3t and then added its derivative B cos 3t. We do not need to add other terms, because all other derivatives involve either sin 3t or cos 3t. However, suppose that F(t) = 7t5 . The initial guess would include At5 and the derivatives Bt4 , Ct3 and so on. The initial guess is then A5 t5 + A4 t4 + A3 t3 + A2 t2 + A1 t + A0 . Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Rules for Making Good Guess (II) There is one exception to the preceding rule. If any term in the initial guess is also a solution of the homogeneous equation, we must multiply the initial guess by a sufficiently high power of t so that nothing in the modified guess is a solution of the homogeneous equation. (In the case of second-order equations, this means multiplying the initial guess by either t or t2 .) To see why, consider u00 + 2u0 − 3u = 4e−3t . The initial guess Ae−3t won’t work: for u = Ae−3t we have u00 + 2u0 − 3u = 0, so there’s no choice of A to make u00 + 2u0 − 3u = 4e−3t . However, Ate−3t does work, as seen in example 5.3. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Modifying an Initial Guess Example (5.3) Show that u(t) = Ae−3t is not a solution of u00 + 2u0 − 3u = 4e−3t , but that there is a solution of the form u(t) = Ate−3t . Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients The Form of up for au00 + bu00 + cu = F(t) A summary of rules for making good guesses is given in the table below. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Finding the Form of Particular Solutions Example (5.4) Determine the form for a particular solution of the following equations: 1 y00 + 4y0 = t4 + 3t2 + 2e−4t sin t + 3e−4t and 2 y00 + 4y = 3t2 sin 2t + 3te2t . Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients The Motion of a Spring Subject to an External Force (I) Example (5.5) A mass of 0.2 kg stretches a spring by 10 cm. The damping constant is c = 0.4. External vibrations create a force of F(t) = 0.2 sin 4t newtons, setting the spring in motion from its equilibrium position. Find an equation for the position of spring at any time t. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients The Motion of a Spring Subject to an External Force (II) Notice in Figure 6.26 that after a very brief time the motion appears to be simple harmonic motion. We can verify this by a quick analysis Figure: [6.26] Spring motion with an external force. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients The Motion of a Spring Subject to an External Force (III) Recall that the solution comes in two pieces, the particular solution up (t) = 41 1 sin 4t − cos 4t 6756 1689 and the solution of the homogeneous equation √ √ uh (t) = c1 e−t cos 97t + c2 e−t sin 97t. As t increases, e−t → 0 and uh (t) → 0 regardless of the value of the constants c1 and c2 . Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients The Motion of a Spring Subject to an External Force (IV) This says that as t → ∞ u = up (t) + uh (t) → up which is a simple oscillation. For this reason, the solution of the homogeneous equation is called the transient solution and the particular solution is called the steady-state solution. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Finding a Steady-State Solution Example (5.6) For u00 + 3u0 + 2u = 20 cos 2t, find the steady-state solution. Figure: [6.27] Steady-state solution. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Resonance and Beats (I) There are several possibilities for the steady-state motion of a mechanical system. Two interesting cases, called resonance and beats, are introduced here. In their pure forms, both occur only when there is no damping and the external force is a sine or cosine. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Resonance and Beats (II) In these cases, then, the equation of motion is mu00 + ku = F(t). The characteristic equation for the homogeneous equation is mr2 + k = 0, which has solutions r r=± k i m and the solution of the homogeneous equation is c1 cos ωt + c2 sin ωt, r where ω = k is called the natural frequency of the system. m Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Resonance and Beats (III) Resonance occurs in a mechanical system when the external force is a sine or cosine whose frequency exactly matches the natural frequency of the system. For example, suppose that F(t) = sin ωt. Then our initial guess up (t) = A sin ωt + B cos ωt matches the homogeneous solution and must be modified to the guess up (t) = t(A sin ωt + B cos ωt). Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Resonance and Beats (IV) The graph of up (t) = t(A sin ωt + B cos ωt). would oscillate, but the presence of the t factor would cause the oscillations to grow larger and larger without bound. The graph of y = 2t sin 4t in Figure 6.28 illustrates this behavior. Figure: [6.28] y = 2t sin 4t. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Resonance and Beats (V) Physically, resonance can cause impressive disasters: A singer hitting a note (thus producing an external force) at exactly the natural frequency of a wineglass can shatter it. Soldiers marching in step across a bridge at exactly the natural frequency of the bridge can create large oscillations in the bridge that can cause it to collapse. These phenomena are well known and it is for this reason that soldiers are ordered to break stride while marching across a bridge. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Resonance and Beats (VI) The phenomenon of beats occurs when the forcing frequency is close to the natural frequency. For example, consider u00 + 4u = 2 sin 2.1t with u(0) = u0 (0) = 0. The homogeneous solution is c1 sin 2t + c2 cos 2t and so, the forcing frequency of 2.1 is close to the natural frequency of 2. The solution is u(t) = 5.1219 sin 2t − 4.878 sin 2.1t. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Resonance and Beats (VII) The graph in Figure 6.29 illustrates the beats phenomenon of periodically increasing and decreasing amplitudes. This can be heard when tuning a piano. If a note is slightly off, its frequency is close to the frequency of the external tuning fork and you will hear the amplitude variation shown in the figure. Figure: [6.29] Beats. Second-Order Equations with Constant Coefficients Nonhomogeneous Equations: Undetermined Coefficients Resonance and Beats (VIII) Example (5.7) For the system u00 + 5u = 3 sin ωt, find the natural frequency, the value of ω that produces resonance and a value of ω that produces beats.